7.       ARITHMETIC  &  NUMBER‑THEORETIC RECREATIONS

 

          7.A.    FIBONACCI NUMBERS

 

          We use the standard form:  F₀ = 0,  F₁ = 1,  F_n+1 = F_n + F_n-1,  with the auxilliary Lucas numbers being given by:  L₀ = 2,  L₁ = 1,  L_n+1 = L_n + L_n-1.

 

Parmanand Singh.  The so-called Fibonacci numbers in ancient and medieval India.  HM 12 (1985) 229-244.  In early Indian poetry, letters had weights of  1  or  2  and meters were classified both by the number of letters and by the weight.  Classifying by weight gives the number of sequences of  1s  and  2s  which add to the weight  n  and this is  F_n+1. 

                    Pingala (c-450) studied prosody and gives cryptic rules which have been interpreted as methods for generating the next set of sequences, either classified by number of letters or by weight and several later writers have given similar rules.  The generation implies  F_n+1 = F_n + F_n-1.  Virah_nka (c7C) is slightly more explicit.  Gop_la (c1134) gives a commentary on Virah_nka which explicitly gives the numbers as  3, 5, 8, 13, 21.  Hemacandra (c1150) states "Sum of the last and last but one numbers ... is ... next."  This is repeated by later authors. 

                    The Pr_krta Paingala (c1315) gives rules for finding the  k-th sequence of weight  n  and for finding the position of a particular sequence in the list of sequences of weight  n  and the positions of those sequences having a given number of  2s  (and hence a given number of letters).  It also gives the relation   F_n+1 = Σ_i BC(n-i,i).

                    N_r_yana Pandita's Ganita Kaumudi (1356) studies additive sequences in chap. 13, where each term is the sum of the last  q  terms.  He gives rules which are equivalent to finding the coefficients of  (1 + x + ... + x^q-1)^p  and relates to ordered partitions using  1, 2, ..., q.

 

                    WESTERN HISTORIES

 

H. S. M. Coxeter.  The golden section, phyllotaxis, and Wythoff's game.  SM 19 (1953) 135‑143.  Sketches history and interconnections. 

H. S. M. Coxeter.  Introduction to Geometry.  Wiley, 1961.  Chap. 11: The golden section and phyllotaxis, pp. 160-172.  Extends his 1953 material.

Maxey Brooke.  Fibonacci numbers:  Their history through 1900.  Fibonacci Quarterly 2:2 (1964) 149‑153.  Brief sketch, with lots of typographical errors.  Doesn't know of Bernoulli's work.

Leonard Curchin & Roger Herz-Fischler.  De quand date le premier rapprochement entre la suite de Fibonacci et la division en extrême et moyenne raison?  Centaurus 28 (1985) 129-138.  Discusses the history of the result that the ratio  F_n+1/F_n  approaches  _.  Pacioli and Kepler, described below, seem to be the first to find this.

Roger Herz‑Fischler.  Letter to the Editor.  Fibonacci Quarterly 24:4 (1986) 382.

Roger Herz-Fischler.  A Mathematical History of Division in Extreme and Mean Ratio.  Wilfrid Laurier University Press, Waterloo, Ontario, 1987.  Retitled: A Mathematical History of the Golden Number, with new preface and corrections and additions, Dover, 1998.  Pp. 157-162 discuss early work relating the Fibonacci sequence to division in extreme and mean ratio.  15 pages of references.

Georg Markovsky.  Misconceptions about the Golden Ratio.  CMJ 23 (1992) 2-19.  This surveys many of the common misconceptions _ e.g. that  _  appears in the Great Pyramid, the Parthenon, Renaissance paintings and/or the human body and that the Golden Rectangle is the most pleasing _ with 59 references.  He also discusses the origin of the term 'golden section', sketching the results given in Herz-Fischler's book.

 

Fibonacci.  1202.  Pp. 283‑284: Quot paria coniculorum in uno anno ex uno pario germinentur.  Rabbit problem _ the pair propagate in the first month so there are  F_n+2  pairs at the end of the  n-th month.  (English translation in:  Struik, Source Book, pp. 2‑3.)  I have a colour slides of this from L.IV.20 & 21 and Conventi Soppresi, C. I. 2616.  This is on ff. 130r-130v of L.IV.20,  f. 225v of L.IV.21,  f. 124r of CS.C.I.2616.

Unknown early 16C annotator.  Marginal note to II.11 in Luca Pacioli's copy of his 1509 edition of Euclid.  Reproduced and discussed in Curchin & Herz-Fischler and discussed in Herz-Fischler's book, pp. 157-158.  II.11 involves division in mean and extreme ratio.  Uses  89, 144, 233  and that  144² = 89 * 233 + 1.  Also refers to  5, 8, 13.

Gori.  Libro di arimetricha.  1571.  F. 73r (p.81).  Rabbit problem as in Fibonacci.

J. Kepler.  Letter of Oct 1597 to Mästlin.  ??NYS _ described in Herz-Fischler's book, p. 158.  This gives a construction for division in extreme and mean ration.  On the original, Mästlin has added his numerical calculations, getting  1/_ = .6180340,  which Herz-Fischler believes to be the first time anyone actually calculated this number.

J. Kepler.  Letter of 12 May 1608 to Joachim Tanckius.  ??NYS _ described in Herz-Fischler (1986), Curchin & Herz-Fischler and Herz-Fishler's book, pp. 160-161.  Shows that he knows that the ratio  F_n+1/F_n  approaches  _  and that  F_n² + (-1)ⁿ  =  F_n-1F_n+1.

J. Kepler.  The Six‑Cornered Snowflake.  Op. cit. in 6.AT.3.  1611.  P. 12 (20‑21).  Mentions golden section in polyhedra and that the ratio  F_n+1/F_n  approaches  _.  See Herz-Fischler's book, p. 161.

Albert Girard, ed.  Les Œuvres Mathematiques de Simon Stevin de Bruges.  Elsevier, Leiden, 1634.  Pp. 169‑170, at the end of Stevin's edition of Diophantos (but I have seen other page references).  Notes the recurrence property of the Fibonacci numbers, starting with  0,  and asserts that the ratio  F_n+1/F_n  approaches the ratio of segments of a line cut in mean and extreme ratio, i.e.  _,  though he doesn't even give its value _ but he says  13, 13, 21  'rather precisely constitutes an isosceles triangle having the angle of a pentagon'.  Herz-Fischler's book, p. 162, notes that Girard describes it as a new result and includes  0  as the starting point of the sequence.

Daniel Bernoulli.  Observationes de seriebus quae formantur ex additione vel substractione quacunque terminorum se mutuo consequentium, ubi praesertim earundem insignis usus pro inveniendis radicum omnium aequationum algebraicarum ostenditur.  Comm. Acad. Sci. Petropolitanae 3 (1728(1732)) 85‑100, ??NYS.  = Die Werke von Daniel Bernoulli, ed. by L. P. Bouckaert & B. L. van der Waerden, Birkhäuser, 1982, vol. 2, pp. 49+??.  Section 6, p. 52, gives the general solution of a linear recurrence when the roots of the auxiliary equation are distinct.  Section 7, pp. 52‑53, gives the 'Binet' formula for  F_n.  [Binet's presentation is so much less clear that I suggest the formula should be called the Bernoulli formula.]

R. Simson.  An explication of an obscure passage in Albert Girard's commentary upon Simon Stevin's works.  Phil. Trans. Roy. Soc. 48 (1753) 368‑377.  Proves that  F_n² + (‑1)ⁿ  =  F_n-1F_n+1.  This says that the triple  F_n-1, F_n, F_n+1  "nearly express the segments of a line cut in extreme and mean proportion, and the whole line;" from which he concludes that the ratio  F_n+1/F_n  does converge to  _.  Herz-Fischler's book, p. 162, notes that his proof is essentially an induction.  (He also spells the author Simpson, but it is definitely Simson on the paper.)

Ch. Bonnet.  Recherches sur l'usage des feuilles dans les plantes.  1754, pp. 164‑188.  Supposed to be about phyllotaxis but only shows some spirals without any numbers.  Refers to Calandrini.  Nice plates.

Eadon.  Repository.  1794.  P. 389, no. 47.  Cow calves at age two and every year thereafter.  How many offspring in 40 years?  Answer is:  0 + 1 + 1 + 2 + 3 + ... + F₃₉.  We would give this as:  F₄₁ - 1,  but he gets it as:  2F₃₉ + F₃₈ - 1.

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions, no. 32, pp. 22 & 82.  Similar to Fibonacci, with a cow and going for 20 generations.

Martin Ohm.  Die reine Elementar-Mathematik.  2nd ed., Jonas Verlags-Buchhandlung, Berlin, 1835.  P. 194, footnote to Prop. 5.  ??NYS _ extensively discussed in Herz-Fischler's book, p. 168.  This is the oldest known usage of 'goldene Schnitt'.  It does not appear in the 1st ed. of 1826 and here occurs as:  "... nennt man wohl auch den goldenen Schnitt" (... one also appropriately calls [this] the golden section).  The word 'wohl' has many, rather vague, meanings, giving different senses to Ohm's phrase.  Herz-Fischler interprets it as 'habitually', which would tend to imply that Ohm and/or his colleagues had been using the term for some time.  I don't really see this meaning and interpreting 'wohl' as 'appropriately' would give no necessity for anyone else to know of the phrase before Ohm.  However the term is used in several other German books by 1847.   [Incidentally, this is not the Ohm of Ohm's Law, but his brother.] 

A. F. W. Schimper & A. Braun.  Flora.  1835.  Pp. 145 & 737.  ??NYS

J. Binet.  Mémoire sur l'integration des équations linéaires aux différences finies, d'un ordre quelconque, à coefficients variables.  (Extrait par l'auteur).  CR Acad. Sci. Paris 17 (1843) 559‑567.  States the Binet formula as an example of a general technique for solving recurrences of the form:  v(n+2) = v(n+1) + r(n)v(n),  but the general technique is not clearly described, nor is the linear case.

B. Peirce.  Mathematical investigation of the fractions which occur in phyllotaxis.  Proc. Amer. Assoc. Adv. Sci. 2 (1849) 444‑447.  Not very interesting.

G. T. Fechner.  Vorschule der Ästhetik.  Breitkopf & Härtel, Leipzig, 1876.  ??NYS.  Origin of the aesthetic experiments on golden rectangles.

Édouard Lucas.  Théorie des fonctions numériques simplement périodiques.  AJM 1 (1878) 184-240 (Sections 1-23)  &  289-321 (Sections 24-30).  [There is a translation by Sidney Kravitz of the first part as:  The Theory of Simply Periodic Numerical Functions, edited by Douglas Lind, The Fibonacci Association, 1969.  Dickson I 400, says this consists of 7 previous papers in Nouv. Corresp. Math. in 1877-1878 with some corrections and additions.  Robert D. Carmichael; Annals of Math. (2) 15 (1913) 30-70, ??NX, gives corrections.]  The classic work which begins the modern study of recurrences.

Pearson.  1907.  Part II, no. 63: A prolific cow, pp. 126 & 203.  Same as Fibonacci's rabbits, but wants the total after 16 generations.

There is a magic trick where you ask someone to pick two numbers and extend them to a sequence of ten by adding the last two numbers each time.  You then ask him to add up the ten numbers and you tell him the answer, which is 11 times the seventh number.  In general, if the two starting numbers are  A  and  B,  the  n-th  term is  F_n-2A + F_n-1B  and the sum of the first  2n  terms is  F_2nA + (F_2n+1-1)B  =  L_n (F_nA + F_n+1B),  but only the case  n = 5  is interesting!  I saw Johnny Ball do this in 1989 and I have found it in: Shari Lewis; Abracadabra!  Magic and Other Tricks; (World Almanac Publications, NY, 1984);  Puffin, 1985;  Sum trick!, p. 14, but it seems likely to be much older.

 

          7.B.    JOSEPHUS OR SURVIVOR PROBLEM

 

          See Tropfke 652.

          This is the problem of counting out every  k‑th from a circle of  n.  Early versions counted out half the group;  later authors and the Japanese are interested in the last man _ the survivor.  Euler (1775) seems to be the first to ask for the last man in general which we denote as  L(n, k).  Cardan, 1539, is the first to associate this process with Josephus.  Some later authors derive this from the Roman practice of decimation. 

          For last man versions, see the general entries and:  Michinori?,  Kenk_,  Cardan,  Coburg,  Bachet,  van Etten,  Yoshida,  Muramatsu,  Schnippel,  Ozanam (1696 & 1725),  Les Amusemens,  Fujita,  Euler,  Miyake,  Matuoka,  Boy's Own Book,  Nuts to Crack,  The Sociable,  Indoor & Outdoor,  Leske,  Secret Out,  Le Vallois,  Hanky Panky,  Kemp,  Mittenzwey,  Gaidoz,  Ducret,  Lemoine,  Akar et al,  Lucas,  Schubert,  Busche,  Tait,  Ahrens,  Rudin,  MacFhraing,  Mendelsohn,  Barnard,  Zabell,  Richards,  Dean,  Richards, 

 

 2 to last, counted by 9s:  Boy's Own Book, 

 3 to last, counted by 9s:  Boy's Own Book, 

 4 to last, counted by 9s:  Boy's Own Book, 

 5 to last, counted by 9s:  Boy's Own Book, 

 6 to last, counted by 9s:  Boy's Own Book, 

 7 to last, counted by 9s:  Boy's Own Book, 

 9 to last, counted by 9s:  Boy's Own Book, 

10 to last, counted by 9s:  Boy's Own Book, 

11 to last, counted by 9s:  Boy's Own Book, 

12 to last, counted by 9s:  Boy's Own Book,  Secret Out, 

12 to last, counting number unspecified:  Coburg, 

13 to last, counted by 2s:  Ducret,  Leeming, 

13 to last, counted by 9s:  Boy's Own Book,  Leske,  Secret Out,  Rudin, 

14 to last, counted by 10s:  Mittenzwey, 

17 to last, counted by 3s:  Barnard, 

21 to last, counted by 5s:  Hyde, 

21 to last, counted by 7s:  Nuts to Crack,  21 to last, counted by 10s:  Hyde,  The Sociable,  Indoor & Outdoor,  Hanky Panky, 

21 to last, counted by 8s:  Mittenzwey, 

24 to last, counted by 9s:  Kemp, 

28 to last, counted by 9s:  Kemp, 

30 to last, counted by 9s:  Schnippel, 

30 to last, counted by 10s:  see entries in next table for  15 & 15 counted by 10s

40 to last man, counted by 3s:  van Etten (erroneous), 

41 to last man, counted by 3s:  van Etten,  Ozanam (1725),  Ducret,  Lucas (1895), 

General case:  Euler,  Lemoine,  Akar et al.,  Schubert,  Busche,  Tait,  Ahrens,  MacFhraing,  Mendelsohn,  Robinson,  Jakóbczyk,  Herstein & Kaplansky,  Zabell,  Richards,  

 

          There are a few examples where one counts down to the last two persons _ see references to Josephus and:  Pacioli,  Muramatsu,  Mittenzwey,  Ducret,  Les Bourgeois Punis.

 

          Almost all the authors cited consider  15 & 15 counted by 9s,  so I will only index other versions.

 

 2 &  2 counted by  3s:  Ball (1911), 

 2 &  2 counted by  4s:  Ball (1911), 

 3 &  3 counted by  7s:  Ball (1911), 

 3 &  3 counted by  8s:  Ball (1911), 

 4 &  4 counted by  2s:  Leeming, 

 4 &  4 counted by  5s:  Ball (1911),  

 4 &  4 counted by  9s:  Ball (1911), 

 5 &  5 counted by  ??:  Dudeney (1905),  Pearson,  Ball (1911),  Ball (1920),  Shaw, 

 6 &  6 counted by  ??:  Dudeney (1900), 

 8 &  2 counted by  ??:  Les Bourgeois Punis, 

 8 &  8 counted by  8s:  Kanchusen, 

 8 &  8 counted by  ??:  Dudeney (1899), 

12 & 12 counted by  6s:  Harrison, 

15 & 15 counted by  3s:  Tartaglia,  Alberti, 

15 & 15 counted by  4s:  Tartaglia, 

15 & 15 counted by  5s:  Tartaglia, 

15 & 15 counted by  6s:  AR,  Codex lat. Monacensis 14908,  Tartaglia, 

15 & 15 counted by  7s:  Tartaglia,  Schnippel, 

15 & 15 counted by  8s:  Codex lat. Monacensis 14836,  AR,  Codex lat. Monacensis 14908,  Tartaglia,  Alberti, 

15 & 15 counted by 10s:  Michinori?,  Reimar von Zweiter,  AR,  Codex lat. Monacensis 14908,  Chuquet,  Tartaglia,  Buteo,  Yoshida,  Muramatsu,  Wingate/Kersey,  Schnippel,  Alberti,  Shinpen Kinko-ki,  Fujita,  Miyake,  Matuoka,  Sanpo Chie Bukuro,  Hoffmann,  Brandreth,  Benson,  Williams,  Collins,  Dean.  (Almost all of these actually continue to the last person.) 

15 & 15 counted by 11s:  Tartaglia,  Schnippel, 

15 & 15 counted by 12s:  AR,  Codex lat. Monacensis 14908,  Tartaglia, 

15 & 15 counted by other values, not specified _ ??check:  Codex lat. Monacensis 14836,  Meermanische Codex,  at‑Tilimsâni,  Bartoli,  Murray 643,  Chuquet,  Keasby, 

17 & 15 counted by 10s:  Schnippel, 

17 & 15 counted by 12s:  Mittenzwey, 

18 &  2 counted by 12s:  Pacioli,  Rudin, 

18 &  6 counted by  8s:  Manuel des Sorciers, 

18 & 18 counted by  9s:  Chuquet, 

24 & 24 counted by  9s:  Chuquet, 

30 &  2 counted by  7s:  Pacioli, 

30 &  2 counted by  9s:  Pacioli, 

30 &  6 counted by 10s:  Ducret,  30 & 10 counted by 12s:  Endless Amusement II,  Magician's Own Book,  The Sociable,  Boy's Own Conjuring Book,  Lucas (1895), 

36 &  4 counted by 10s:  Jackson, 

n²-n+1  &  n-1  counted by  n:  Lucas (1894),  Cesarò,  Franel,  Akar, 

 

          Many authors provide a mnemonic for the case of  15 and 15 counted by 9s.  In this case, the longest group of the same type is five, so a common device is to encode the numbers  1, 2, 3, 4, 5  by the vowels  a, e, i, o, u  and then produce a phrase with the vowels in the correct order.  I will call this a vowel mnemonic.  The most popular form is:  Populeam virgam Mater Regina ferebat,  giving the numerical sequence:  4, 5, 2, 1, 3, 1, 1, 2, 2, 3, 1, 2, 2, 1.  The first group of 4 are good guys, followed by 5 bad guys, etc.  Below I list the mnemonics and where they occur, but I did not always record them in my notes below, so I must check a number of the sources again ?? _ the classification was inspired by seeing that Franci (op. cit. in 3.A) describes a vowel mnemonic in P. M. Calandri which I had overlooked.  Ahrens gives many more verse and vowel mnemonics _ to be added below.  Hyde gives an Arabic mnemonic due to al-Safadi using the first letters of the Arabic alphabet: a, b, gj, d, h.

 

Dahbagja Ababgja Baba  [=  Da h b a gj a  A ba b gj a  Ba b a]:  Hyde from al-Safadi.

Unspecified(?) verse mnemonic:  ibn Ezra

Populeam jirgam mater Regina ferebat:  Badcock

Populeam virgam Mater Regina tenebat:  Hyde from Wit's Interpreter;  Murphy;  Schnippel/Bolte

Populeam virgam mater Regina ferebat:  van Etten;  Schnippel;  Ozanam 1725;  Les Amusemens;  Hooper;  Jackson;  Manuel des Sorciers;  Boy's Own Book;  The Sociable;  Le Vallois;  Gaidoz;  Lucas

Populeam virgam mater regina ferebat:  Manuel des Sorciers; 

Populea virga pacem regina ferebat:  Minguét

Mort tu ne failliras pas en me liurant le trespas:  van Etten

Mort, tu ne falliras pas   En me livrant au trépas:  Manuel des Sorciers; 

Mort, tu ne falliras pas.  En me livrant le trépas:  Schnippel/Bolte;  Ozanam 1725;  Les Amusemens;  The Sociable;  Le Vallois (without the first comma);  Ducret;  Lucas

On tu ne dai la pace ei la rendea:  Schnippel/Bolte

Gott schuf den Mann in Amalek, der (or den) Israel bezwang:  Schnippel

Gott schlug den Mann in Amalek, den Israel bezwang:  Schnippel/Bolte

So du etwan bist gfalln hart, Stehe widr, Gnade erwart:  Schnippel/Bolte

Non dum pena minas a te declina degeas:  Schnippel/Bolte

Nove la pinta dà e certi mantena:  P. M. Calandri

From member's aid and art, Never will fame depart:  Schnippel/Bolte

From numbers, aid and art / Never will fame depart:  Wingate/Kersey

From numbers, aid, and art, Never will fame depart:  Ingleby;  Jackson

From number's aid and art, Never will fame depart:  Gaidoz

From numbers aid and art / Never will fame depart:  The Sociable

 

          See 5.AD for the general problem of stacking a deck to produce a desired effect.

 

Josephus.  De Bello Judaico.  c80.  Book III, chap. 8, sect. 7.  (Translated by Whiston or by Thackeray (Loeb Classical Library, Heinemann, London, 1927, vol. 2, pp. 685‑687.))  (Many later authors cite Hegesippus which is a later version of Josephus.)  This says that Josephus happened to survive  "by chance or God's providence".

H. St. J. Thackeray.  Josephus, the Man and the Historian.  Jewish Institute Press, NY, 1929, p. 14.  Comments on the Slavonic text, which says that Josephus  "counted the numbers with cunning and therby misled them all"  but gives no indication how.

 

Ahrens.  MUS II.  1918.  Chap. XV: Das Josephsspiel, pp. 118-169.  This is the most extended and thorough discussion of this problem and its history.  I have used it as the basis of this section.  He gives a rather complex method, based on work of Busche, Schubert and Tait, for determining the last man, or any other man in the sequence of counting out, which I never worked through, but which is clearly explained under Richards (1999/9).

 

Gerard Murphy.  The puzzle of the thirty counters.  Béaloideas _ The Journal of the Folklore of Ireland Society XII (1942) 3‑28.  In this work and the material cited (mostly ??NYS), the problem of  15 and 15 counted by 9s  is shown to have the medieval name  Ludus Sancti Petri  = St. Peters‑Spiel  = St. Peter's Lake (lake being an Old English word for game [or Anglo-Saxon for 'to play'])  = Sankt Peter Lek or Sankt Päders Lek (in Swedish).  Murphy cites Schnippel & Bolte to assert that it was known to the Arabs in the 14C _ cf below.  He says the usual European version has Christians and Jews on a ship, with St. Peter present and suggesting the counting out process.  [I had forgotten that such versions occur in Ahrens, MUS II 130.]  However, Murphy was unable to consult MUS, so his background is not as complete as it might be.

                    Murphy demonstrates that the problem was recently well‑known in both Scots and Irish Gaelic in a form where a woman has to choose between two groups of warriors seated in a circle, with emotional reasons for her preference.  The solution is given in a vernacular mnemonic, using actual numbers as in early Latin forms, while later Latin and vernacular forms used vowel mnemonics.  He gives an Irish reconstruction, with English translation, based on several 18C MSS whose texts he estimates as 13C to 17C, probably 16C.  This is titled:  Goid Fhinn Agus Dubháin Anso (Here is the Thieving of Fionn and Dubhán).  One MS has the Latin subtitle:  Populeam virgam Mater Regina tenebat,  which is a common Latin vowel mnemonic.  One of Murphy's sources says this refers to the Queenly Mary appearing to the ship's captain and holding a poplar rod.

                    Murphy also gives an extended Irish story (3pp) built around the problem:  Ceann Dubhrann na Ndumhchann Bán (Ceann Dubhrann of the White Sandhills).  The Gaelic names  Fionn  and  Dubhán  are derived from  'fionn'  and  'dub'  meaning  'white'  and  'black'.  Murphy gives a contemporary Irish version on board a ship with a white captain and a black wife and a crew of 15 and 15, with half having to go overboard due to lack of food.  He sketches numerous other Irish and Scots version with varying combinations of details, but using essentially the same verse mnemonic.

                    Murphy cites a study by Manitius of a 9C MS (Bib. Nat. Paris, No. 13029) where the problem begins  "Quadam nocte niger dub nomine, candiduus alter" (One night a black man named Dub and another [named] White).  They have to choose between the blacks and the whites to keep watch.  Cf. Codex Einsidelensis No. 326 below.  The MS ends with the prose line  "These two Irish soldiers, one named 'Find' the other 'Dub', were engaged in hunting.  'Find' means "white", 'dub' "black"."  The 12C Rouen MS No. 1409 attributes the problem to a Clemens Scottus, which Murphy interprets as Clement the Irishman.  The 12C MS Bib. Nat. Paris No. 8091 attributes it to a Thomas Scottus.

                    Murphy concludes that the problem has an Irish origin, c800.  He gives what he believes to be the earliest Latin form, basically Bib. Nat. Paris No. 13029, and opines there must have been an Irish predecessor.

Codex Einsidelensis No. 326.  10C.  F. 88'.  Latin verse.  Published by Th. Mommsen, Handschriftliches.  Zur lateinischen Anthologie.  Rheinischen Museum für Philologie (NS) 9 (1854) 296‑301, with material of interest on pp. 298-299.  Latin given in:  M. Curtze, Bibliotheca Math. (2) 9 (1895) 34‑35.  Latin & German in MUS II 123‑125.  Begins:  "Quadam nocte niger dux nomine, candidus alter".  15 white  &  15 black soldiers, half to keep watch, counted off by 9.  The colours refer to clothing, not skin!

Codex lat. Monacensis 14836.  11C.  F. 80' gives rules for  15 and 15 counted by 9 (though this value is not specified)  and mentions counting by 8 and other values.  No mention of what is being counted.  Quoted and discussed by:  M. Curtze; Zur Geschichte der Josephspiels; Bibliotheca Math. (2) 8 (1894) 116  and in:  Die Handschrift No. 14836 der Königl. Hof‑ und Staats‑bibliothek zu München; AGM 7 (1895) 105 & 111‑112 (Supplement to  Zeitsch. für Math. und Physik 40 (1895)).

Codex Bernensis 704.  12C.  Published by:  Hermann Hage; Carmina medii aevi maximam partem inedita; Ex Bibliothecis Helveticus collecta; Bern, 1877; no. 85, pp. 145‑146.  ??NYS.   Latin in:  Curtze, op. cit. at Codex Einsidelensis, pp. 35‑36;  and in:  MUS II 127.  Jews  &  Christians.

Ahrens, MUS II 118‑147, gives many further references from 10‑13C.  Originals ??NYS.

Meermanische Codex, 10C.  Mentions counting by other values.

Leiden Miscellancodex, 12C

Basel Miscellancodex, 13C

Michinori Fujiwara (1106-1159).  This work is lost, but has been conjectured to contain a form of the problem _ see under Kenk_, c1331, and Yoshida, 1634.

Rabbi Abraham ben Ezra.  Ta'hbula (or Tachbûla), c1150.  ??NYS _ described in:  Moritz Steinschneider; Abraham ibn Ezra (Abraham Judaeus, Avenare); Zur Geschichte der mathematischen Wissenschaft im XII Jahrhundert;  Zeitschr. für Math. und Physik 25 (1880): Supp: AGM 3, Part II (1880).  The material is Art. 20, pp. 123‑124.  15 students and 15 good-for-nothings on a ship, counted by 9s.  This seems to be the first extant example on board a ship.  Verse mnemonic, which Steinschneider says is not original.  Steinschneider cites further sources.  Smith & Mikami, p. 84, say ben Ezra died in 1067 _ ??

Reinmar von Zweter.  Meisterlied:  "Ander driu, wie man juden und cristen ûz zelt".  13C.  (In MUS II 128.)  Jews and Christians on a ship, counts by 10.

Kenk_, also known as  Kenk_ Yoshida  or  Urabe no Kaneyoshi  or  Yoshida no Kaneyoshi  (1283-1350).  Tsurezuregusa.  c1331.  Translated by Donald Keene as: Essays in Idleness  The Tsurezuregusa of Kenk_; Columbia Univ. Press, NY, 1967.  (Kenk_ was the author's monastic name.  His lay name was  Urabe no Kaneyoshi.  He lived for a long time at Yoshida in Kyoto.) 

                    This book is one of the classics of Japanese literature, consisting of 243 essays, ranging from single sentences to several pages.  The most common themes of these relate to the impermanence of life and the vanity of man. 

                    In Japanese, the Josephus problem is called  Mamakodate  or  Mamako-date San  (or  Mama-ko tate no koto - cf Mat_-oka, 1808)  or  Mamagodate  (Scheme to benefit the step-children  or  Stepchild disposition).  It is said to have been in the lost work of Michinori Fujiwara (1106‑1159), qv.  The word  Mamagodate  first occurs in essay 137 of Kenk_, pp. 115-121 in Keene's version (including a double-page illustration which doesn't depict the problem), whose beginning is characteristic of Kenk_'s style:  "Are we to look at cherry blossoms only in full bloom, the moon only when it is cloudless?  To long for the moon while looking on the rain, to lower the blinds and be unaware of the passing of the spring _ these are even more deeply moving."  The passage of interest is toward the end, on p. 120 of Keene: "When you make a mamagodate¹ with backgammon counters, at first you cannot tell which of the stones arranged before you will be taken away.  Your count then falls on a certain stone and you remove it.  The others seem to have escaped, but as you renew the count you will thin out the pieces one by one, until none is left.  Death is like that."  The footnote refers to counting 15 and 15 by 10s, so that 14 white stones are eliminated, then the counting is reversed and all the black stones are eliminated.  "The Japanese name mamagodate (stepchild disposition) derives from the story of a man with fifteen children by one wife and fifteen by another; his estate was disposed of by means of the game, one stepchild in the end inheriting all."  Kenk_'s text clearly shows he was familiar with the process of counting to the last man and the use of the name indicates that he was familiar with the version mentioned in the footnote, though its earliest explicit appearance in Japan is in Yoshida, 1634, qv.  My thanks to Takao Hayashi for the reference to Keene.

Thomas Hyde.  Mandragorias seu Historia Shahiludii, ....  (= Vol. 1 of De Ludis Orientalibus, see  4.B.5 for vol. 2.)  From the Sheldonian Theatre (i.e. OUP), Oxford, 1694.  Prolegomena curiosa.  The initial material and the Prolegomena are unpaged but the folios of the Prolegomena are marked (a), (a 1), ....  The material is on  (e 1).v - (e 2).v, which are pages 34-36 if one starts counting from the beginning of the Prolegomena.  Cited by Bland (loc. cit. in 5.F.1 under Persian MS 211, p. 31);  Ahrens (MUS II 136)  &  Murray 280.  Several citations are to  ii.23,  which may be to the 1767 reprint of Hyde's works.

                    Hyde asserts that the problem of the ship  with 15 Moslems and 15 Christians on a ship, counted by 9s, was given by al-Safadi (Salâhaddîn as‑Safadî  = al-Sâphadi  = AlSáphadi) (d. 1363) in his  Lâmiyato ’l Agjam  (variously printed in the text).  This must be his  Sharh Lâmîyat al‑‘Ajam  of c1350.  Hyde gives an Arabic mnemonic using the first five letters of the Arabic alphabet, which he transliterates as:  Dahbagja Ababgja Baba  [=  Da h b a gj a  A ba b gj a  Ba b a].  He says the problem occurs in an English book called Wit's Interpreter (??NYS) (8^oS.87.Art) where the mnemonic  Populeam virgam mater Regina tenebat  is given.  He then says that the problem is also described in 'Megjdium & Abulphedam' _ p. 43 of his main text identifies Abulpheda as a prince born in 672 AH _ ??

Shihâbaddîn Abû’l‑‘Abbâs Ahmad ibn Yahya ibn Abî Hajala at‑Tilimsâni alH‑anbalî (??sp).  Kitâb ’anmûdhaj al‑qitâl fi la‘b ash‑shatranj [Book of the examples of warfare in the game of chess].  c1370.  Copied by  Muhammed ibn ‘Ali ibn Muhammed al‑Arzagî  in 1446.

                    This is the second of Dr. Lee's MSS, described in 5.F.1, denoted Man. by Murray.  Murray 280 says "Man. 36‑45 relate to  as‑Safadî's  problem of the ship (see Hyde, ii.23)", described by Murray as 15 Christians and 15 Muslims counted by  n.  Bland has "the well-known problem of the Ship, first as described by Safadi, and then in other varieties.  (Hyde, p. 23.)"

Murray 620 says the problem is of Muslim origin and says it appears in the c1530 Italian version of the Bonus Socius collection which Murray denotes It.  (See 5.F.1 under Bonus Socius.)  Murray refers to  15 & 15 counted by 9s,  but it is not clear if this refers to this particular MS. 

Murray 622 cites  MS Sloan 3281 in the BM, 14C, as giving the Latin mnemonic solution.

Bartoli.  Memoriale.  c1420.  F. 100r (= Sesiano p. 135).  Il giuocho de' Cristiani contra Saracin.  15 Christians and 15 Saracens  _ the text ends in the middle of the statement of the problem.

AR.  c1450.  Prob. 80, pp. 52, 181‑182 & 229.  15 Christians and 15 Jews.  Gives only mnemonics for counting by  10, 9, 8, 6 or 12.

Codex lat. Monacensis 14908.  c1460.  F. 76 gives mnemonics for 15 Jews and 15 Christians counted by  6, 8, 9, 10, 12.  Quoted and discussed by Curtze, opp. cit. under Cod. lat. Mon. 14836, above.  [In the first paper, the codex number is misprinted as 14809.]

Murray 643 says the MS Lasa version, c1475, of the Civis Bononiae collection (described in 5.F.1 under Civis Bononiae) has  "16 diagrams of the 'ship' puzzle under different conditions".

P. M. Calandri.  c1480.  Pp. 142‑143.  15 Christians and 15 Jews on a boat counted by 9s.  Vowel mnemonic:  Nove la pinta dà e certi mantena.  Diagrammatic picture on p. 143.

Chuquet.  1484.  Prob. 146.  English in FHM 228-230, with reproduction of the original on p. 229.  15 Jews and 15 Christians on a ship, counting by 9s.  Says one can have 18 or 24 of each and can count by 10s, etc.  The reproduction on FHM 229 shows a circle marked out, with  Populeam virgam matre regina tenebat  written in the middle.  The commentary says this "problem is comparatively rare in fifteenth century texts", which doesn't seem like a fair assessment to me.

Pacioli.  De Viribus.  c1500.  Probs. 56‑60: De giudei christiani in diversi modi et regole a farne quanti se vole.  Agostini quotes Pacioli who seems to be saying that the situation happened to him!??  Does  2 & 30 by 9s,  2 & 18 by 7s,  2 & 30 by 7s,  15 & 15 by 9s.

Calandri.  Aritmetica.  c1500.  Ff. 102v-103v, pp. 205‑207.  Coloured plate opp. p. 192 of the text volume.  (Tropfke 654 gives this in B&W.)  Franciscans and Camoldensians on a boat:  15 & 15 counted by 9s.

Elias Levita der Deutsche.  Ha‑Harkabah.  Rome, 1518.  ??NYS.   Attributes to ben Ezra, c1150??.  Smith & Mikami, p. 84, say this seems to be the first printed version of the problem.

Cardan.  Practica Arithmetice.  1539.  Chap. 61, section 18, ff. T.iiii.r ‑ T.v.v, but the material of interest is just a few sentences on f. T.iv.v (p. 113).  Very brief description of 15 and 15 as 'ludus Josephus'.  MUS says this is first to relate the problem to Josephus as the last man.  (Does he give numbers??)

Hans Sachs (1494-1576).  Meisterleid:  'Historia  Die XV Christen und XV Türcken, so auff dem meer furen'.  (MUS II 132‑133 gives text.)

Tartaglia.  General Trattato, 1556, art. 203, pp. 264v‑265r.  15 whites and 15 blacks (or Turks and Christians) counted out by  3, 4, ..., 12.  No reference to Josephus.

Buteo.  Logistica.  1559.  Prob. 89, pp. 303-304.  15 Christians and 15 Jews on a ship counted by 10s.  [Mentioned in H&S 52.]

Simon Jacob von Coburg.  Ein new und Wolgegründt Rechenbuch ....  1565 or 1612 (in quarto, not to be confused with octavo versions of 1565 and 1613 which do not contain the problem), f. 250v.  ??NYS _ described in MUS II 133-134.  12 drinkers deciding who shall pay the bill.  Ahrens doesn't specify the counting number.  Ahrens & Bolte (below) say this is the earliest example, after Cardan, of finding the last man.  Ahrens describes numerous later examples of this type from 1693 on.

Prévost.  Clever and Pleasant Inventions.  (1584), 1998.  Pp. 183-185.  This seems like a version of the Josephus problem but isn't.  Place ten counters in a circle and then ten on top of them.  Start anywhere and count off five and remove the top counter.  He says to count five again _ starting on the place where the counter was removed, so we now would say he is counting four _ and remove the top counter.  Continue in this way, counting the places where a top counter has been removed and you manage to remove all the top counters.  In fact this is impossible, but after removing five counters, you subtly start counting from the next position rather than where the top counter was removed!  Hence you remove the top counters in the order  5, 9, 3, 7, 1, 6, 10, 4, 8, 2.  Your audience will not observe this and hence cannot reproduce the effect.

                    The mathematical description is simpler if one counts by fours, removing  4, 8, 2, 6, 10, 5, 9, 3, 7, 1.  The first five values are the values of  4a (mod 10)  for  a = 1, 2, 3, 4, 5.  Because  GCD (4, 10) = 2,  this sequence repeats with period  5.  Your trick shifts from the even values to the odd values and then you can count out the five odd values.

Bachet.  Problemes.  1612.  Préface, 1624: A.5.v - A.7.r;  1884: 8-9  &  prob. XX, 1612: 103‑106;  prob. XXIII,  1624: 174-177;  1884: 118‑121.  Turks & Christians _ discusses Josephus as last man.

van Etten.  1624.  Prob. 7 (7), pp. 7‑9 (16‑19).  15 Turks and 15 Christians counted by 9s.  Mnemonics:  Populeam virgam mater Regina ferebat;  Mort tu ne failliras pas en me liurant le trespas.  Discusses other cases, Roman decimation and Josephus as  40  counted by 3s.  In the 1630 edition,  40  is changed to  41.  Henrion's Notte, pp. 9‑10, refers to Bachet's prob. 23 and mentions the correction of  40  to  41.

Yoshida (Shichibei) K_y_ (= Mitsuyoshi Yoshida) (1598-1672).  Jink_‑ki.  Additional problems in the 2nd ed., 1634.  Op. cit. in 5.D.1.  ??NYS.  Shimodaira (see the entry in 5.D.1) discusses the Josephus problem on pp. 12-14.  He gives some of the information on the Japanese names and on Michinori (1106-1159) and Kenk_, c1331, which is presented under them.  I have a transcription of (some of?) Yoshida into modern Japanese which includes this material as prob. 3 on pp. 66-67.

                    15 children (in black) and 15 stepchildren (in white) counted by 10s.  When  14  stepchildren are eliminated, the last stepchild says the arrangement was unfair and requests the counting to go the other way from him (so that he is number 1 in the counting).  His stepmother agrees and thereby eliminates all her own children.

                    This is discussed in Smith & Mikami, pp. 80-84.  They quote a slightly later version by  Seki K_wa (1642-1708)  where the stepmother simply reverses the order due to overconfidence.  (On p. 121, they identify the source as  Sandatsu Kempu,  a MS of K_wa.)  This is also discussed in MUS II 139‑140, where it says that the change in counting was an error on the stepmother's part.  Needham, p. 62, gives a picture from the 1634 ed. of Yoshida, but this is different than the picture in my modern transcription.  I have a xerox from an 1801 ed.  Dean, 1997, gives the picture, discusses this and provides some additional details, citing the Heibonsha encyclopaedia for the version with the intelligent stepchild.  Dean, 1997, also gives an illustration from a 1767 version called Shinpen Jinko-ki, cf at 1767.

Ahmed el-Qalyubi (d. 1659).  Naouadir (or Nauadir), c1650?, published at Boulaq (a suburb of Cairo), 1892, hist. 176, p. 82.  ??NYS _ described in Basset (1886-1887 below) and MUS II 136.  15 Moslems and 15 infidels on a ship counted by 9s.

Muramatsu Kuday_ Mosei.  Mantoku Jink_‑ri.  1665.  ??NYS _ described in MUS II 139 and Smith & Mikami, pp. 80-84.  Smith & Mikami, p. 81, and Dean, 1997, give Muramatsu's schematic diagrams.  The top diagram is for the classic 15 and 15 counted by 10s.  The second has 32 people counted by 10s to the last two, though the first 15 are coloured black and the second 15 are coloured white, with the last two drawn as squares marked by dice patterns for 5 and 6.

Wingate/Kersey.  1678?.  Prob. 3, pp. 531-532.  15 & 15 counted by 9s or 10s or any other.  Christians and Turks.  From numbers, aid and art / Never will fame depart.  Discusses Josephus.

Thomas Hyde.  Historia Nerdiludii, hoc est dicere, Trunculorum; ....  (= Vol. 2 of De Ludis Orientalibus, see above for vol. 1.)  From the Sheldonian Theatre (i.e. OUP), Oxford, 1694.  De Ludo Charîgj seu Charîtch, pp. 225-226.  Says it is an Arabic game, i.e. Ludus Exeundi & Eliminadi.  The description is very vague, but it seems to involve counting out in a circle.  The diagram shows a circle of 21 and the text mentions counting by ten or by five.  No reference to any other version of the process.  ??need to read the Latin more carefully.

Emil Schnippel (& Johannes Bolte).  Das St. Peters-Spiel (with a Nachtrag by Bolte).  Zeitschrift für Volkskunde 39 (1929) 190-192  (&  192-194).  Schnippel describes the appearance of solutions of the  St. Peters-Spiel  = Sankt Päders Lek  = Saint Peter's Lake  on 17-18C rune calendars from Sweden, which mystified academics until identified by G. Stephens in 1866.  He gives the vowel-mnemonic:  Populeam virgam mater regina ferebat.  The rune marks are  X  for  Χριστιαvoι (Xristianoi)  and  I  for  ’Ioυδα_oι (Ioudaioi).  He gives the German vowel-mnemonic:  Gott schuf den Mann in Amalek, der (or den) Israel bezwang.  He cites other writers (??NYS) who describe a 1497 MS with 15 & 15 Christians and Jews counted by 10s,  and versions counted by 7s and 11s,  and a version with 17 & 15 Christians and Jews counted by 12s.  He cites:  a 1604 reference to Josephus but without specific numbers;  a 1703 version with 15 & 15 French and Germans;  and a 1782 version with 30 deserters, 15 to be pardoned. 

                    [Nigel Pennick; Mazes and Labyrinths; Robert Hale, London, 1990, p. 37, says that in Finland, stone labyrinths are sometimes called  "Pietarinleikki (St Peter's Game).  The latter name refers to a traditional numerical sequence which appears to be related to the lunar cycle.  It is known from rock carvings and ancient Scandinavian calendars and as an anti-semitic folk-tale."  Can anyone provide details of a connection to the lunar cycle or its appearance in rock carvings??]

                    Bolte's Nachtrag cites Gaidoz et al. (below at 1886-1887) and MUS and an article by himself in Euphorion 3 (1896) 351-362, ??NYS _ cited MUS II 132.  He sketches the history as given by Ahrens.  Mentions the Japanese versions and reproduces Matuoka's picture.  He adds three citations including a 1908 Indian version with 15 honest men and 15 thieves counted by 9s to the last man (??).  He gives vowel-mnemonics in Latin, French, German, English and Italian as follows.

                    Non dum pena minas a te declina degeas.

                    Populeam virgam mater regina ferebat.

                    Mort, tu ne falliras pas.  En me livrant le trépas.

                    So du etwan bist gfalln hart, Stehe widr, Gnade erwart.

                    Gott schlug den Mann in Amalek, den Israel bezwang.

                    From member's (sic) aid and art, Never will fame depart.

                    On tu ne dai la pace ei la rendea.

Ozanam.  Murphy, note 4, says the problem is not in the 1694 ed. _ but see below which could explain why Murphy didn't find it here.

Ozanam.  1696.  Preface to vol. 2 _ first and second of unnumbered pages, which are pp. 269‑270.  1708: Author's Preface _ second and third of unnumbered pp.   Discusses Josephus, citing Bachet.

Ozanam.  1725.  Prob. 45, 1725: 246‑250.  Prob. 17, 1778: 168-171;  1803: 168-171;  1814: 148-150.  Prob. 16, 1840: 76-77.  15 Turks and 15 Christians counted by 9s.  Gives two verse mnemonics:  Mort, tu ne failliras pas, En me livrant le trépas;  Populeam virgam mater Regina ferebat.  Discusses decimation.  Quotes Bachet on Josephus and asserts Hegesippus says Josephus used the method and suggests  41  counted by  3s  (however, Hegesippus doesn't say this!).

Kanchusen.  Wakoku Chiekurabe.  1727.  Pp. 8 & 35.  8 and 8 counted by 8s.  This is pointing out the remarkable fact that one can count out either set first by starting at different points, in different directions.  See:  Dudeney, 1899 & 1905;  Shaw, 1944?

Minguét.  Engaños.  1733.  Pp. 152-154 (1755: 110-111; 1822: 169-171).  15 & 15  by 9s,  whites and blacks.  Populea virga pacem regina ferebat.

Alberti.  1747.  'Modo di disporre 30 cose ...', pp. 132‑134 (77‑78).  15 Christians and 15 Turks or Jews, counted by  3, 8, 9, 10.

Les Amusemens.  1749.  Prob. 16, p. 138: Tiré de Josephe l'Historien.  15 and 15 counted by 9s.  French and Latin mnemonics:  Mort tu ne failliras pas  En me livrant le trépas;  Populeam Virgam Mater Regina ferebat.

Shinpen Jinko-ki (New Edition of the Jonko-ki), more correctly entitled Sanpo Shinan Guruma (A Mathematical Compass).  1767.  BL ORB 30/3411.  ??NYS _ illustration reproduced in Dean, 1997.

Fujita Sadasuke.  Sandatsu Kaigi.  1774.  ??NYS _ cited in a draft version of Dean, 1997, as a Japanese commentary on the problem.

Hooper.  Rational Recreations.  Op. cit. in 4.A.1.  1774.  Vol. 1, recreation XIII, pp. 42-43.  30 deserters of whom 15 are to be punished, counted by 9s.  Populeam virgam mater regina ferebat.

L. Euler.  Observationes circa novum et singulare progressionum genus.  (Novi Comment. Acad. Sci. Petropol. 20 (1775 (1776)) 123‑139.)  = Opera Omnia (1) 7, (1923) 246‑261.  Gets the recurrence for the last man:  L(n) º L(n-1) + k (mod n).

Miyake Kenry_.  Shojutsu Sangaku Zuye.  1795.  ??NYS.  (Described in MUS II 142‑143.)  First(?) to modify Yoshida's problem (1634?) so that the last stepchild sees his imminent fate and asks for the count to restart with him.  Smith, History II 543 and Smith & Mikami, p. 82, give a poorish picture from this.  Dean, 1997, is a better picture.

Matuoka (= Mat_-oka ??= Matsuoka N_ichi).  1808.  ??NYS _ translated by Le Vallois, with reproductions of the pictures, cf below.  Ahrens, MUS II 140‑142, discusses this, based on Le Vallois and reproduces the main picture from Le Vallois.  Gives Miyake's version.  Le Vallois gives the title as: Mama-ko tate no koto (Problème des beaux-fils (i.e. step-sons)).  There is a diagram showing the counting-out processes.

Ingleby.  Ingleby's Whole Art of Legerdemain, containing all the Tricks and Deceptions, (Never before published) As performed by the Emperor of Conjurors, at the Minor Theatre, with copious explanations; Also, several new and astonishing Philosophical and Mathematical Experiments, with Preliminary Observations, Including directions for practicing the Slight of Hand.  T. Hughes  &  C. Chaple,  London, nd [1815].  Trick L.  The Turks and Christians, pp. 104-106.  15 & 15 counted by 9s.  "This ingenious trick, which is scarcely known, ...."  "From numbers, aid, and art, /  Never will fame depart."

Sanpo Chie Bukuro (A Bag of Mathematical Wisdom).  1818.  BL ORB 30/3411.  ??NYS - illustration reproduced and discussed in Dean, 1997.  Here a man and a woman are studying a set of 29 black and white go stones and the text describes the problem and how to arrange the children.

Badcock.  Philosophical Recreations, or, Winter Amusements.  [1820].  Pp. 83-85, no. 130: Thirty soldiers having deserted, so to place them in a ring, that you may save any fifteen you please, and it shall seem the effect of chance.  15 & 15 by 9s.  Populeam jirgam mater Regina ferebat.  (jirgam  must be a misprint of  virgam.)  Says Josephus and 'thirty or forty of his soldiers' hid in a cave and Josephus arranged to be one of the last.

Jackson.  Rational Amusement.  1821.  Arithmetical Puzzles.

No. 16, pp. 4-5 & 54-56.  15 Turks and 15 Christians counted by 9s.  Solution gives:  From numbers, aid, and art, Never will fame depart  and  Populeam virgam mater regina ferebat.

No. 39, pp. 9-10 & 61-62.  Decimation of a troop of 40 to be counted by 10s _ where to place the four ringleaders so they will be the four to be shot.

Manuel des Sorciers.  1825.  ??NX  Pp. 79-80, art. 40.  15 & 15 by 9s.  Populeam virgam mater regina ferebat.  Mort, tu ne falliras pas   En me livrant au trépas.  Says one can also do  18 & 6 by 8s, etc.  Cf Gaidoz, below, col. 429.

Endless Amusement II.  1826?  P. 117 (misprinted 711 in 1826?): Predestination illustrated.  30 and 10 counted by 12s.

Boy's Own Book.

The slighted lady.  1828: 411‑412;  1828-2: 417-418;  1829 (US): 210-211;  1855: 565‑566;  1868: 670.  13 counted down to last person by 9s.  Before 1868, it gives the survivor for  2, 3, ..., 13,  counted out by 9s.

The partial reprieve.  1828: 417‑418;  1828-2: 422;  1855: 571;  1868: 672-673;  1881: 214.  30 criminals counted by 9s to eliminate 15.  Populeam virgam mater regina ferebat.

Nuts to Crack XIV (1845), no. 72.  21  counted by  7s  to the last man.

Magician's Own Book.  1857.

The fortunate ninth, pp. 221-222.  15 oranges and 15 apples, counted by 9s.  English mnemonics based on vowel coding.

Another decimation of fruit, p. 224-225.  30 apples and 10 oranges, counted by 12s in order to get the oranges first.

The Sociable.  1858.

Prob. 31: The puzzle of the Christians and the Turks, pp. 296 & 312-314.  From numbers aid and art  /  Never will fame depart.  Mort, tu ne faillras pas  /  en me livrant le trepas.  Populeam Virgam Mater regina ferebat.  Then considers counting out 10 from 40, counting by 12s.  Discusses Josephus, citing Hegesippus, and suggests counting by 3s.  = Book of 500 Puzzles, 1859, prob. 31, pp. 14 & 30-32.

Prob. 39: The landlord tricked, pp. 298 & 316.  21 counted by 7s to the last man.  = Book of 500 Puzzles, 1859, prob. 39, pp. 16 & 34.  = Wehman; New Book of 200 Puzzles; 1908, p. 51.

Indoor & Outdoor.  c1859.  Part II, prob. 19: The landlord tricked, p. 136.  Identical to The Sociable.

Boy's Own Conjuring Book.  1860.

The fortunate ninth, pp. 190‑191.  Identical to Magician's Own Book.

Another decimation of fruit, p. 194.  Identical to Magician's Own Book.

Leske.  Illustriertes Spielbuch für Mädchen.  1864?  Prob. 564-30, pp. 254 & 396: Aus 12 Dreizehn machen.  Count 13 by 9s to the last person.

The Secret Out.  Op. cit. in 4.A.1.  1871?  A delicate distribution, p. 12.  Count 13 by 9s to the last person (different context than Leske).  Mentions counting 12 by 9s.

M. le Capitaine Le Vallois.  Les Sciences exactes chez les Japonais.  With comments by Louis de Zélinski & M. Sédillot.  Congrès International des Orientalists (= International Congress of Orientalists).  Compte-Rendu de la première session, Paris, 1873.  Maisonneuve et Cie., Paris, 1874.  T. 1, pp. 289‑299, with comments on pp. 299-303.  The material of interest is on pp. 294-298.  Gives a translation of Mat_-oka, 1808, and reproduces the pictures, cf above.  Discusses Bachet, Ozanam,  Mort tu ne failliras pas  En me livrant le trépas,  Populeam virgam mater Regina tenebat,  Josephus (saying Josephus arranged to be last).

Hanky Panky.  1872.  The landlord tricked, pp. 129-130.  Identical to The Sociable, prob. 39.

Kamp.  Op. cit. in 5.B.  1877.  No. 7, pp. 323‑324.  28 counted by 9s until one is left.  Footnote seems to refer to 24 counted by 9s.

Mittenzwey.  1879?  Prob. 311‑314, pp. 49‑50 & 97‑98.

311:  15 Negroes and 17 Europeans on a ship, counted by 12s.

312:  7 students and a crafty Jew who wishes to make two of the students,  A  &  B,  pay the bill.  Starting with  A  and counting  clockwise by 3s,  A  &  B  are left.  Starting with  B  and counting anti‑clockwise by 3s,  A  &  B  are left.  Then the Jew is included and starts with himself, counting anticlockwise by 3s and again  A  &  B  are left.

313:  14 counted by 10s to the last man.

314:  21 counted by 8s to the last man.

Cassell's.  1881.  P. 103: To reward the favourites, and show no favouritism.  = Manson, 1911, p. 256.  15 & 15 counted by 9s.

Henri Gaidoz, Israël Lévi & René Basset.  Le jeu de Saint‑Pierre _ Amusement arithmétique.  This is a series of five notes in Mélusine 3 (1886‑87).

Gaidoz.  Part I.  Col. 273‑274.  Gives classical version with St. Peter, 15 Christians & 15 Jews counted by 9s.  He then gives two versions from Ceylon.  One version is called  Yonmaruma _ The massacre of the Moors _  and has 15 Portuguese & 15 Moors with a Singhalese verse mnemonic.  The second version involves the Portuguese siege of Kandy in 1821, again  15 & 15  by 9s, but different versions have the Portuguese winning or losing.  These versions come from:  The Orientalist 2 (1885) 177, ??NYS.  The editor of The Orientalist added a version learned from an Irish soldier with the vowel‑mnemonic:  From number's aid and art, Never will fame depart.  Gaidoz says he cannot venture a source for the puzzle.

Gaidoz.  Part II.  Col. 307‑308.  Comments on correspondence generated by Part I which provided:  'Populeam virgam mater regina ferebat';  the version with the Virgin instead of St. Peter;  a version with negroes and whites and a negro captain;  a version with French and English;  references to Josephus, Bachet and Ozanam.

Lévi.  Part III.  Col. 332.  Says ibn Ezra's "Tahboula" (Stratagem), c1150, is devoted to this game.  Cites  Schwenter (1623);  Steinschneider's 1880 article discussed above at Ezra;  Steinschneider's Catalog librorum hebr. Biblioth. Bodleianae, col. 687 _ all ??NYS.  Previously Steinschneider opined the game derived from Jahia ibn al‑Batrik's Secret of Secrets (8C), but Lévi says that that is a different amusement involving 9.

Gaidoz.  Part IV.  Col. 429.  Cites:  Le Manuel des Sorciers, Paris, 2nd ed., 1802, p. 70 for a version with French and English.  ??NYS, but see the 1825 ed above.

Basset.  Part V.  Col. 528.  Describes el-Qalyubi, c1650? _ cf above.

Robert Harrison.  UK Patent 15,105 _ An Improved Puzzle or Game.  Applied 25 Sep 1889;  accepted 2 Nov 1889.  2pp + 1p diagrams.  12 whites and 12 blacks on a boat with a lifeboat that will hold 12, counted by 6s, called  The Captain's Dilemma.

É. Ducret.  Récréations Mathématiques.  Op. cit. in 4.A.1.  1892? 

Pp. 105-106: Une dame pas contente.  13 counted by 2s to last person.

Pp. 118-119: Stratagéme de Joséphe.  41 counted by 3s to last two, claimed to be the method used by Josephus.

Pp. 120-121: Les marauders punis.  15 & 15 counted by 9s.  Officers and soldiers to be executed.

Pp. 121-122: Les naufrages.  Same numbers, with Turks and Christians on a boat.  Mort, tu ne failliras pas, En me livrant au trépas.

P. 122: Les Élections perfectionnées.  36 counted by 10s _ want the first six chosen.

Hoffmann.  1893.  Chap. 4.

No. 54: Tenth man out, pp. 156 & 210.  15  whites and  15  blacks on a ship, counted by  10s,  but first  15  get to go into the lifeboats.

No. 55: Ninth man out, pp. 156‑157 & 210‑211.  Same, counted by  9s.  Hoffmann cites Bachet and gives a Latin mnemonic.

É. Lucas.  Problem 32.  Intermed. des Math. 1 (1894) 9.  n²  persons, counted by  n  until  n‑1  are left.  "Problème dit de Caligula".

E. Cesarò.  Solution to 32.  Ibid., pp. 30‑31.

J. Franel.  Deuxième réponse [to Problem 32].  Ibid., p. 31.  Cites:  Busche, Comptes Rendus 103, pp. 118, ??NYS.

Adrien Akar.  Troisième réponse [to Problem 32].  Ibid., pp. 189‑190.

E. Lemoine.  Problem 330.  Ibid, pp. 184‑185.  Asks for last man of  n  counted by  p.

Adrien Akar;  H. Delannoy;  J. Franel;  C. Moreau.  Independent solvers of Lemoine's problem.  Ibid., 2 (1895) 120‑122  &  229‑230.  Akar refers to Josephus, Bachet, etc.  Moreau has the clearest form of the recurrence.

Brandreth Puzzle Book.  Brandreth's Pills (The Porous Plaster Co., NY), nd [1895].  P. 7: The tenth man out.  Almost identical to Hoffmann, No. 54.  No solution.

Lucas.  L'Arithmétique Amusante.  1895.  Pp. 12-18.

Le stratagème de Josèphe, pp. 12-17.  Prob. VI.  15 Christians and 15 Turks, counted by 9s.  Vowel mnemonics:  Mort, tu ne falliras pas, En me livrant le trépas!;  Populeam virgam mater Regina ferebat.  Discusses and quotes Bachet's 1624 Préface which gives the Josephus story and the idea of counting  41  by  3s.

Prob. VII: Le procédé de Caligula, pp. 17-18.  6  and  30  counted by  10s  so as to count the  6  first.

H. Schubert.  Zwölf Geduldspiele.  1895.  P. 125.  ??NYS _ cited by Ahrens; Mathematische Spiele; Encyklopadie article, op. cit. in 3.B; 1904.

E. Busche.  Ueber die Schubert'sche Lösung eines Bachet'schen Problems.  Math. Annalen 47 (1896) 105‑112.

P. G. Tait.  On the generalization of Josephus' problem.  Proc. Roy. Soc. Edin. 22 (1898) 165‑168.  = Collected Scientific Papers, vol. II, pp. 432‑435.  Says the Josephus passage is "very obscure, ..., but it obviously suggests deliberate fraud of some kind on Josephus' part."  Develops a way of computing the last man.

Les Bourgeois Punis.  Puzzle from c1900, shown in S&B, p. 133.  8 and 2 counted by ?? to leave the 2 at the end.

Dudeney.  A batch of puzzles.  Royal Magazine 1:3 (Jan 1899)  &  1:4 (Feb 1899) 368-372.  The prisoners of Omdurman.  8 Europeans followed by 8 Abyssinians in a ring.  Start counting with the first European.  Determine the counting-out number to eliminate the Abyssinians in sequence.  Doing it in reverse sequence works for any multiple of  16, 15, 14, ..., 9.  The LCM is  720720.  But doing it in forward sequence can be done with  360361.  Since the pattern is symmetric in the two types of people, a change of initial position, but keeping the same direction, will count out the others first.

Dudeney.  "The Captain" puzzle corner.  The Captain 3:2 (May 1900) 97 & 179  &  3:4 (Jul 1900) 303.  The "blacks" and "whites" puzzle  or  The twelve schoolboys.  6 consecutive "blacks" and 6 consecutive "whites" in a circle.  What is the smallest number to count out by which will count out the "whites" first?  You can start anywhere and in any direction.  Answer is  322  and one starts counting on the fourth "white" in the direction of counting.

Ahrens.  Mathematische Spiele.  Encyklopadie article, op. cit. in 3.B.  1904.  Pp. 1088‑1089 discusses Schubert's work and its later developments.

Benson.  1904.  The black and white puzzle, pp. 225‑226.  As in Hoffmann, no. 54, but first 15 get thrown overboard.  Solution is linear rather than circular.

Dudeney.  Tit‑Bits (14 Oct  &  28 Oct 1905).  ??NYS _ described by Ball; MRE, 5th ed., 1911, pp. 25‑27.  5 and 5 arranged so that one method eliminates one group while another method eliminates the other group.  Determine the two starting points and counts.  Ball doesn't give these values, but seems to imply that both counts go in the same direction, and this is the case in the examples given below.  Ball asks if the starting points can ever be the same for two groups of  C?  He gives solutions for  C = 2  (counted by  4 & 3),  3 (counted by  7 & 8),  4 (counted by  9 & 5).  He believes this question is new.  Note on p. 27 gives the solution for  C = 5,  but with different starting points.  See MRE, 10th ed., 1920, for a general solution with the same starting points.  See:  Kanchusen, 1727;  Dudeney, 1899;  Shaw, 1944?

Pearson.  1907.  Part II, no. 62, pp. 126 & 203.  15 Christians, including St. Peter, who does the counting, and 15 Jews, counted by 9s.

Ball.  MRE, 5th ed.  1911.  See under Dudeney, 1905.

Loyd.  Cyclopedia.  1914.  Christians and Turks, pp. 198 & 365.  = MPSL2, prob. 42, pp. 30‑31 & 134.  Like Dudeney's 1905 version with a different arrangement of 5 and 5.

Williams.  Home Entertainments.  1914.  A decimation problem, pp. 122-124.  15 whites & 15 blacks counted by 10s.  Half have to go over because of shortage of provisions.  Simple circular picture with man counting in middle.

Ball.  MRE, 10th ed., 1920, pp. 26-27.  See under Dudeney, 1905, for the previous version.  Incorporates the solution for the case  C = 5  into the text and adds a general solution due to Swinden.

See:  Will Blyth; Money Magic; 1926 for a related problem.

Collins.  Book of Puzzles.  1927.  Sailors don't care puzzle, pp. 70-71.  15 whites & 15 blacks counted by  10s.  Captain has to throw half over because of shortage of provisions.  Diagram of 15 circles in a row above a picture with 15 circles in a row below, but normally numbered _ it would seem natural in this problem to have the lower row numbered backward to simulate a circle.

William P. Keasby.  The Big Trick and Puzzle Book.  Whitman Publishing, Racine, Wisconsin, 1929.  Counting out problem, pp. 71 & 95.  15 & 15  shown in a circle with the starting point and direction given.  Determine the counting number.

Rudin.  1936.  Nos. 102-103, pp. 37-38 & 98.

No. 102.  13 counted by 9s until last man.

No. 103.  17 and 15 counted by 12s to eliminate the 15 first.

Ern Shaw.  The Pocket Brains Trust _ No. 2.  Op. cit. in 5.E.  c1944.  Prob. 50: Poser with pennies.  Pattern of  5  Hs  and  5  Ts  given _ determine the count to count the  Hs  first, which turns out to be  11.  Though not mentioned, the pattern of  Hs  is equivalent to that for  Ts,  so one can count out the  Ts  first by starting at a different point in the opposite direction.  The pattern is the same as Dudeney (1905).

Le Rob Alasdair MacFhraing (= Robert A. Rankin, who tells me that the Gaelic particle  'Le'  means  'by'  and is not part of the name).  Àireamh muinntir Fhinn is Dhubhain, agus sgeul Iosephuis is an dà fhichead Iudhaich (The numbering of Fionn's men and Dubhan's men, and the story of Josephus and the forty Jews) (in Scots Gaelic with English summary).  Proc. Roy. Irish Acad. Sect. A. 52:7 (1948) 87-93.  A more detailed description than in the Summary appears in Rankin's review in:  Math. Reviews 10, #509b (= A99-5 in:  Reviews in Number Theory).  I had assumed that this was a development from Murphy's article, but Rankin writes that he had not heard of Murphy's article until I wrote for a reprint of Rankin's article in 1991.  He gives a Scottish Gaelic version which is clearly a variant of those studied by Murphy.  He then studies the problem of determining the last man, citing Tait.  For counting out by 2s, the rule is simple.  [There is a story that the the number of mathematicians fluent in Scots Gaelic is so small and the author's name is so obscured that the journal sent the paper to Rankin to referee, not knowing he was actually the author.  The story continues that the referee made a number of suggestions for improvement which the author gratefully accepted.  However, Rankin told me that he was not the referee.  But he did review it for Math. Reviews!]

Joseph Leeming.  Games with Playing Cards Plus Tricks and Stunts.  Op. cit. in 6.BE, 1949.  ??NYS _ but two abridged versions have appeared which contain the material _ see 6.BE.

                    24 Stunts with Cards, 7th & 19th stunts.  A surprising card deal  &  All in order.  Dover: pp. 124  &  131.  Gramercy: pp. 11  &  18.  Both stunts involve dealing cards by putting one out face up, then the next is put at the bottom of the deck, then the next is dealt face up, ....  This process is the same as counting out by  2s.  The object is to produce the cards in a particular order.  The first has 8 cards and wants an alternation of face cards and non-face cards.  The second has the 13 cards of a suit and wants them produced in order.  This is the only example that I can recall of the use of the Josephus idea as a card trick, though other forms of counting out are common, e.g. by counting  1, 2, 3, ...,  or by spelling the words  one, two, three, ....

N. S. Mendelsohn, proposer;  Roger Lessard, solver.  Problem E 898 _ Discarding cards.  AMM 57 (1950) 34 & 488-489.  Basically counting out by  2s.  Determine the position of the last card discarded.  No mention of Josephus, though the editor asks what happens if every  r-th card is discarded and gets the recurrence  f(N) º r + f(N-1) (mod N).

W. J. Robinson.  Note 2876:  The Josephus problem.  MG 44 (No. 347) (Feb 1960) 47‑52.  Analyses what sequences of persons can be removed by varying the count.  Applies to Dudeney's problem.

D. St. P. Barnard.  Fifty Observer Brain‑Twisters.  Faber, 1962.  (= A Book of Mathematical and Reasoning Problems.  Van Nostrand, 1962.)  Prob. 36: Circle of fate, pp. 41‑42, 64‑65 & 95.  Princess counts out from 17 suitors by 3s.  She sees that her favourite will be the next one out, so she reverses direction and then the favourite is the survivor.

F. Jakóbczyk.  On the generalized Josephus problem.  Glasgow Math. J. 14 (1973) 168-173.  Gives a method of determining when the  i-th man is removed and which is the  k-th to be removed.  Somewhat similar to Rankin's method.

Israel N. Herstein & Irving Kaplansky.  Matters Mathematical.  1974;  slightly revised 2nd ed., Chelsea, NY, 1978.  Chap. 3, section 5: The Josephus permutation, pp. 121-128.  They study the permutation where  f(i) = number of  i-th man eliminated, but restrict to the case where one counts by  2s,  which has considerable structure.  Gives a substantial bibliography, mostly included here.

Sandy L. Zabell.  Letter [on the history of the Josephus problem].  Fibonacci Quarterly 14 (1976) 48 & 51.  Sketches the history.

I. M. Richards.  The Josephus problem.  MS 24 (1991/92) 97-104.  Studies the case of counting out by 3s.  Shows the 'Tait numbers', i.e.  n  such that  L(n) = 1 or 2,  are given by  [η(3/2)ⁱ + 1/3],  where  η = 1.216703...,  and obtains a formula for  L(n).  Presumably this could be extended to the general case??

Michael Dean.  Josephus and the Mamako-date san (Scheme to benefit the step-children).  International Netsuke Society Journal 17:2 (Summer 1997) 41-53.  There are inro boxes from late 17C Japan which have pictures of the 15 children and 15 stepchildren  problem.  These initially mystified the art historians, but eventually they discovered the Josephus problem and its Japanese forms, but only as far back as Bachet.  Dean gives a brief history for the benefit of art collectors, with references to a number of Japanese sources (some of which I have not seen) _ see above at 1767, 1795, 1818 _ and some photos of the inro boxes (including a fine late 17C example from the collection of Michael and Hiroko Dean) and other material. 

Ian M. Richards.  The Josephus Problem and Ahrens arrays.  MS 31:2 (1998/9) 30-33.  He has finally obtained a copy of Ahrens' work, but from the first edition, and states the result clearly.  For  n  persons, labelled  1, 2, ..., n,  counted out by  k,  if we want to locate the  e-th man counted out, form a sequence starting with  1 + k(n-e)  and then form each next term by multiplying a term by  k/(k-1)  and rounding the result up to an integer.  (I.e.  x_n+1 = éx_n * k/(k-1)ù.)  Then the position number of the  e-th person eliminated is the difference between  kn + 1  and the largest term in the sequence less than  kn + 1.  The sequence is giving the points where  L(n, k)  is zero in some sense.  Note that when  e = n,  so we are looking for the last person, then the sequence starts at  1,  which is because we start counting with the first person as one.  kn + 1  is the total amount counted in counting  n  people by  k,  For other values of  e,  the change of the starting point of the sequence compensates for the fact that one only counts  k(n-e) + 1  to eliminate the  e-th person.  Ahrens then examined the sequences obtained, with rational multipliers, and found some nice properties which Richards states.  Richards generalises to arbitrary multipliers and finds connections with Beatty sequences,  ëaⁿû.

Ian M. Richards.  Towards an analytic solution of the Josephus problem.  Unpublished preprint sent to me on 21 Mar 1999, 12pp.  (Available from the author, 3 Empress Avenue, Penzance, Cornwall, TR18 2UQ.)  Gets formulae for the case  k = 4  which give the result with a maximum error of ±1.

 

          7.C.    EGYPTIAN FRACTIONS

 

          The basic problem is to represent a given fraction as a sum of fractions with unit numerators and distinct denominators, as done by the Egyptians.

 

Papyrus Rhind, c‑1650.  A. B. Chace, ed.  (1927‑29);  c= NCTM, 1978.  Pp. 21‑22, 50‑51.

Fibonacci.  1202.  Pp. 77‑83: ... de disgregatione partium in singulis partibus.  He clearly has the idea of taking the smallest  n  such that  1/n £ a/b,  but he doesn't prove that this gives a finite sequence.

J. J. Sylvester.  On a point in the theory of vulgar fractions.  Amer. J. Math. 3 (1880) 332‑335 & 388‑389.

M. N. Bleicher.  A new algorithm for the expansion of Egyptian fractions.  J. Number Theory 4 (1972) 342‑382.  The Introduction, pp. 342‑344, outlines the history.  Pp. 381‑382 give 41 references.

E. J. Barbeau.  Expressing one as a sum of distinct reciprocals.  CM 3:7 (1977) 178‑181.  Bibliography of 20 items.

Paul J. Campbell.  A "practical" approach to Egyptian fractions.  JRM 10 (1977-78) 81-86.  Discusses Fibonacci & Sylvester's methods, etc.  22 references.

Charles S. Rees.  Egyptian fractions.  Math. Chronicle 10 (1981) 13‑30.  Survey with 47 references.

 

          7.D.   THE FIRST DIGIT PROBLEM

 

S. Newcomb.  Note on the frequency of use of the different digits in natural numbers.  Amer. J. Math. 4 (1881) 39‑40.  Obtains the law by simply considering logarithms.

F. Benford.  The law of anomalous numbers.  Proc. Amer. Phil Soc. 78 (1938) 551‑572.

E. H. Neville.  Note 2540:  On even distribution of numbers.  MG 39 (No. 329) (Sep 1955) 224‑225.  Says the problem is not precisely defined.  (Not cited in Raimi.)

R. A. Fairthorne.  Note 2541:  On digital distribution.  Ibid, p. 225.  Cites earlier results (see Raimi) and says the law is "a consequence of the way we talk about [numbers]."  (Not cited in Raimi.)

R. A. Raimi.  The first digit problem.  AMM 83 (1976) 521‑538.  Extensive survey and references.

G. T. Q. Hoare & E. E. Wright.  Note 70.5:  The distribution of first significant digits.  MG 70 (No. 451) (Mar 1986) 34‑37.  Generates numbers as ratios of reals uniformly distributed on  (0, 1).  Finds explicit and surprisingly simple probabilities for initial digits of these numbers, which are reasonably close to Benford's probabilities.

Peter R. Turner.  The distribution of l.s.d. and its implications for computer design.  MG 71 (No. 455) (Mar 1987) 26‑31.  l.s.d. = leading significant digit.  Cites some recent articles.

 

          7.E.    MONKEY AND COCONUTS PROBLEMS

 

          Most of these problems are determinate.  Mahavira gives two indeterminate problems, but the next are in Ozanam, with the classic version of the problem first reappearing in Clark, 1904, and Pearson, 1907, qv.

          NOTATION.  The classic coconuts problem has the following recurrence for the number of coconuts remaining: 

                    A_i+1  =  (n-1)/n [A_i - 1], 

i.e. each sailor removes  1  (given to the monkey) and  1/n  of the rest.  There are two common endings of the problem.

Ending 0  _  the  n‑th  man leaves a multiple of  n,  so the monkey doesn't get a final coconut.  See:  Mahavira: 131, 132;  Williams;  Moritz;  Meynell;  Leeming.

Ending 1  _  the  n‑th  man leaves one more than a multiple of  n,  so the monkey gets another coconut.  See:  Pearson; Roray; Collins; Kraitchik; Phillips; Home Book; Leeming; Devi; Allen.

One can extend this to  Ending E  _  the  n-th  man leaves a number  º E (mod n).

Other indeterminate versions:  Ozanam; Dudeney; Weber (Dirac); Rudin.

For the solution with  -(n-1)  coconuts, see:  Roray; Weber (Dirac); Birkhoff & Mac Lane; Gardner; Pedoe, Shima & Salvatore; Singmaster.

See  Morris (1988);  Singmaster (1993)  for the alternative division form where the pile is divided equally and the monkey takes one from the remainder, i.e. each sailor takes  1/n  of the pile and then the monkey then takes  1  from the remainder, so the recurrence is 

                    A_i+1  =  (n‑1)A_i/n - 1.  This is similar to the form of recurrence occurring in the determinate versions of the problem, where division takes place first and then some more is included.  Comparing this with the standard form, we see that the difference in the forms can be expressed as the number of coconuts  (mod n)  at each stage.  In the classic form, each  A_i º 1 (mod n),  and in Morris's form, each  A_i º 0 (mod n), so we can conveniently name these  Form 1  and  Form 0.  Unless specified, all examples have  Form 1.

It is easy to generalize to giving  c  coconuts to the monkey at each stage, in either Form, but only Kircher; Pedoe, Shima & Salvatore; Singmaster consider this.

Only Kircher considers giving variable amounts to the monkey and he even permits negative values,  e.g. if the monkey is adding coconuts to the pile!

Birkhoff & Mac Lane; Herwitz; Pedoe, Shima & Salvatore consider a variation where no ending is specified except that there is an integral number left after the  n-th division.  A discussion of this version has now been added to Singmaster.

Jackson gives a simple form with no monkey.  Edwards gives a form where the monkey only gets a coconut at the end.

See Tropfke 582.  See also 7.S.1.

 

Hermelink, op. cit. in 3.A, says there are Egyptian versions, presumably meaning some of the simpler determinate types of heap or 'aha' problems in the Rhind Papyrus.

Old Babylonian tablet YBC 4652.  c-1700?.  Transcribed, translated and commented on in:  O. Neugebauer & A. Sachs; Mathematical Cuneiform Texts; American Oriental Society and American Schools of Oriental Research, New Haven, 1945, pp. 100-103, plate 13 & photo on plate 39.  This has fragments of 22 simple problems, of which six can be restored.  The authors say the dating of the tablets discussed in the book is quite uncertain, only stating "they are to be dated to the centuries around 1700 B.C."

No. 7 is reconstructed as:  I found a stone, but did not weigh it; after I added one-seventh and added one-eleventh, I weighed it: 1 ma-na.  What was the original weight of the stone?  In modern notation, this is:  x + x/7  +  (x + x/7) / 11  =  1,  or simply:  x (8/7) (12/11) = 1  which is a simple 'aha' problem. 

No. 8 leads to  x - x/7  +  (x - x/7) / 11  =  1.

No. 9 leads to  x - x/7  +  (x - x/7) / 11  -  [x - x/7 + (x - x/7)/11] / 13  =  1.

No. 19 leads to  6x + 2  +  (6x + 2)·24/21  =  1.

No. 20 leads to  8x + 3  +  (8x + 3)·21/39  =  1.

No. 21 leads tp  x - x/6  +  (x - x/6) / 24  =  1.

Old Babylonian tablet YBC 4669.  c-1700?.  Neugebauer and Sachs continue on p. 103 with a new analysis of this table which Neugebauer had previously treated in Mathematische Keilschrift-texte III, op. cit. in 6.BF.2, p. 27.  It leads to  (2/3) (2/3) x + 10 = x/2.

Chiu Chang Suan Ching (Jiu Zhang Suan Shu).  c‑150. 

Chap. VI, prob. 27, p. 69.  Man carrying rice through customs pays  1/3,  then  1/5,  then  1/7  and has  5  left. 

Chap. VI, prob. 28, pp. 69‑70.  Man pays  1/2, 1/3, 1/4, 1/5, 1/6,  making  1  paid out.

Chap. VII, prob. 20, pp. 79‑80.  Man  gains  30%  and sends home  14000;  then  gains  30%  and sends  13000;  then  30%  and  12000;  then  30%  and  11000;  then  30%  and  10000;  leaving  0.  Capital was  30468  84876/371293.  (English in Lam & Shen, HM 16 (1989) 113.)

Zhang Qiujian (= Chang Chhiu‑Chien  = Chang Ch'iu Chien  = Zhang Yo Chien).  Zhang Qiujian Suan Jing (= Chang Chhiu‑Chien Suan Ching) (Zhang Qiujian's Mathematical Manual).  468.  ??NYS.  Chap. II, no. 17.  Man  gains  40%  and withdraws  16000;  then gains  40%  and withdraws  17000;  then gains  40%  and withdraws  18000;  then gains  40%  and withdraws  19000;  then gains  40%  and withdraws  2000;  leaving  0.  Capital was  35326  5918/16807.  (English in Lam & Shen, HM 16 (1989) 117.)

Anania Schirakatzi (= Ananias of Shirak).  Arithmetical problems.  c640.  Translated by:  P. Sahak Kokian as:  Des Anania von Schirak arithmetische Aufgaben; Zeitschrift für d. deutschösterr. Gymnasien 69 (1919) 112-117.  Kokian cites several versions and editions of this Armenian MS as well as some studies on Ananias, but I haven't been able to determine just where Shirak was.  The title varies on the different MSS and Kokian heads the text with one version translated into German:  Des Anania Vardapet Schirakuni Frage und Auflösung [Questions and Solutions of the Priest Ananias of Shirak.]  There are 24 problems, mostly of the 'aha' or 'heap' type.  Only the numerical solutions are given _ no methods are given.  There are several confusing errors which may be misprints or may be errors in the MS, but Kokian says nothing about them.  One problem seems to have omitted an essential datum of the number of grains of barley in a 'kaith'.  I cannot reconcile one solution with its problem (see 7.H).

Prob. 11.  Spend  5/6  thrice, leaving  11.  Answer:  2376.

Prob. 13.  Spend  3/4  thrice, leaving  5.  Answer:  320.

Prob. 19.  (Double and give away  25)  thrice to leave zero.  Answer:  21_.  Kokian notes that this and prob. 22 are the earliest occurrences of fraction signs in Armenian.  Hermelink, op. cit. in 3.A, points out that here the doubling is done by God in response to prayer in churches _ then the Arabic world converts the churches to mosques, and then the West reverts to churches, while in the Renaissance, the doubling is by winning at gambling.  In fact, during the Renaissance, it often was by profit from trade.

Prob. 21.  Give away  1/2,  then  1/7,  then  1/8,  then  1/14,  then  1/13,  then  1/9,  then  1/16,  then  1/20,  leaving  570.  Answer:  2240.

Papyrus of Akhmim.  c7C.  Jules Baillet, ed.  Le Papyrus Mathématique d'Akhmîm.  Mémoires publiés par les membres de la Mission Archéologique Français au Caire, vol. IX, part 1, (1892) 1‑89.  Brief discussion of the following problems on pp. 58-59.

Prob. 13, p. 70.  Take  1/13th,  then  1/17th  of the rest, leaving  150.  Answer:  172 + 1/2 + 1/8 + 1/48 + 1/96.  (Also given in HGM II 544.  Kaye I 48, op. cit. under Bakhshali MS, discusses the Akhmim problem and says both Heath and Cantor give misleading references, but I don't see what he means.)

Prob. 17, p. 72.  Take  1/17th,  then  1/19th  of the rest, leaving  200.  Answer:  224 + 1/4 + 1/18.

Bakhshali MS.  c7C. 

                    In:  G. R. Kaye, The Bakhsh_li manuscript.  J. Asiatic Soc. Bengal (2) 8:9 (Sep 1912) 349‑361.  P. 358: Sutra 25: example _ merchant pays customs of  1/3,  1/4,  1/5  and finds he has paid  24.  = Kaye III 205, f. 14r.

                    Hoernle, 1888, op. cit. under Bakhshali MS, p. 277 gives the above and the following.  Merchant gains  1/3,  1/4,  1/5,  1/6  and finds he has gained  40.  (Kaye III 205, f. 14r gives this in less detail and it is not clear if Hoernle's statement is what is intended.)  Merchant loses  1/3,  1/4,  1/5  for a total loss of  27.  (Kaye III 205, f. 14v says the remainder is  27  but gives the original amount as  45,  so he seems to have loss and remainder interchanged.).  Merchant loses  1/3,  1/4,  1/5  leaving  20 (can't find in Kaye III ??).

                    Kaye I 48, section 89, says there are 17 examples of this general form, some with the initial value given and the final result wanted, others with the final result given and the initial value wanted.  Gives the first example above and two others with the same rates and a payment of  280  (Kaye III 165, ff. 52r-52v) or a result of  2x - 32,  where  x  is the initial value (Kaye III 207, f. 15r).  Kaye III 204, f. 13v:  start with  60, lose  1/2,  gain  1/3,  lose  1/4,  gain  1/5.  Kaye III 208, f. 16r:  give  2/3,  then  2/5,  then  2/7,  then  2/9,  leaving  3.  How much was given?

                    See also Datta, op. cit. under Bakhshali MS, pp. 44 & 52‑53.  He says the Akhmim problems give the remainder, while the Bakhshali MS and Mahavira problems give the amount paid _ but above we have seen both kinds.  Datta, p. 46, says (Kaye III 184,) f. 70v has a badly damaged problem about a king who gives away  1/2,  1/3  and  1/4  of his money, making  65  given away.  Datta says that the king had only  60  to start!!  But if this is a problem of the type being treated here, then the fractions are applied to the amount left after the previous stage and the king would have  1/4  of his original amount left and he must have had  86 1/3  to start.

Mahavira.  850.  Chap. III, v. 129-140, pp. 67-69 are simple problems of this general type, involving sums of numbers diminished by fractions _ I give just some examples.  Chap. VI,  v. 112, 114, 130, 131, 132, pp. 116 & 123‑125.

  Chap. III.

133.  x(3/4)(4/5)(5/6) + y(1/2)(5/6)(4/5) + z(3/5)(3/4)(5/6)  =  1/2.  This reduces to:  x/2 + y/3 + 3z/8  =  1/2  and he arbitrarily picks two of the values, getting:  1/3, 1/4, 2/3.

134.  x(1/2)(5/6)(4/5)(7/8)(6/7)  =  1/6.

  Chap. VI.

112.  Double and subtract  5,  triple and subtract  5,  ...,  quintuple and subtract  5,  leaving  0  (Datta & Singh I 234, note this gives  43/12  flowers and hence replace  5  by  60  to give  43). 

114.  Less regular problem, leaving  0.

130.  Gives a general technique.

131.  Two sons and mangoes _  (subtract  1  and halve)  twice,  leaving some even number  _  i.e.  Ending 0  with  2  men.  Cf Pearson, 1907.

132.  Man placing flowers in a temple  _  (subtract  1  and delete  _)  thrice,  leaving some multiple of  3  _  i.e.  Ending 0  with  3  men.  Cf Pearson, 1907.

There are some simpler problems in Chap. IV, v. 29‑32, pp. 74‑75.

Chaturveda.  860.  There are some simple examples on pp. 282‑283 of Colebrooke.

Sridhara.  c900.  V. 74(i), ex. 97, pp. 59‑60 & 96.  Give away  1/2,  then  2/3,  then  3/4,  then  4/5,  leaving  3.

Tabari.  Mift_h al-mu‘_mal_t.  c1075.  ??NYS.

Pp. 177f. & 128, no. 28 & 45.  Tropfke 585 says these are business trips. 

P. 127, no. 44.  Hermelink, op. cit. in 3.A, says this is Fibonacci's seven gate problem of p. 278, with oranges instead of apples.  Tropfke 585 says it is a problem with porters at an orange orchard.

Abraham.  Liber augmenti et diminutionis.  Translated from Arabic in 12C (Tropfke 662 says early 14C).  Given in:  G. Libri; Histoire des Sciences Mathématiques en Italie; vol. 1, pp. 304-376, Paris, 1838.  ??NYS _ cited by Hermelink, op. cit. in 3.A. 

Bhaskara II.  Bijaganita.  1150.  Chap. 4, v. 114.  In Colebrooke, pp. 196‑197.  (Lose  10,  double,  lose 20)  thrice to triple original.

Fibonacci.  1202.  Pp. 258‑267, 278, 313‑318 & 329 gives many versions!  Chap. 12, part 6, pp. 258‑267: De viagiorum propositionibus, atque eorum similium is devoted to such problems.

P. 258.  (Double & spend  12)  thrice to leave  0  or  9.  Answers:  10½,  11_.  H&S 60 gives English.

P. 259.  Start with  10½,  (double and spend  x)  thrice to leave  0.  H&S 60 gives English.

P. 259.  Same, starting with  11_  and leaving  9.

P. 259.  (Triple and spend  18)  four times to leave  0.  Answer:  8 8/9.

P. 260.  Start with  8 8/9,  (triple and spend  x)  four times to leave  0.

Pp. 260‑261.  (Triple and spend  18)  four times to leave  12,  or the original amount,  or original amount + 20.

P. 261.  Three voyages, making profits of  1/2,  1/4,  1/6  and spending  15  each time to leave final profit of  1/2.  Answer:  24 6/7.  Same, with initial amount  24 6/7,  find the common expenditure.  Same, with 'leave final profit  1/2'  replaced by 'leave  21'.

Pp. 261‑266.  Many variations.

Pp. 266‑267.  Start with  13,  (double and spend  14)  X  times to leave  0.  H&S 60 gives English.  He gets  X = 3¾  voyages, by linear interpolation between  3  and  4.  Exact answer is  log₂ 14 = 3.80735.

P. 278.  De illo qui intravit in viridario pro pomis collegendis.    Man collects apples in a garden with  7  gates.  (Subtract half and one more)  seven times to leave  1.  H&S 60 and Sanford 221 give English.  Answer:  382.

Pp. 313‑316.  Man starts with  100  and spends  1/10  twelve times.  This is not strictly of the type we are looking at, but it is notable that he computes  100 (.9)¹²  using a form of decimal fraction, getting  28.2429536481.  See 7.L for related problems.

Pp. 316‑318.  Exit from a city with  10  gates.  He pays  2/3  of his money and  2/3  more, then  1/i  of his money and  1/i  more for  i = 2, ..., 10,  leaving  1.

P. 329.  Same as on p. 258, done by false position.

P. 329.  Start with  12,  (double and spend  x)  thrice to leave  0.

Abbot Albert.  c1240.  Prob. 8, p. 334.  (Double and subtract  1)  thrice, leaving  0.

Chu Shih‑Chieh (= Zhu Shijie).  Ssu Yuan Yü Chien (= Siyuan Yujian) (Precious Mirror of the four Elements  = Jade Mirror of the Four Unknowns).  1303.  Questions in Verse, prob. 4.  ??NYS.  English in Li & Du,  p. 179.  (Double and drink  19)  four times to leave  0.

BR.  c1305.

No. 89, pp. 108‑109.  (Double and spend  35)  thrice leaving  0.

No. 119, pp. 134‑135.  (Double and spend  40)  thrice leaving  0.

Munich 14684.  14C.

Prob. V, p. 78.  (Double and subtract  2)  some times to leave  0  _ determines initial values for various numbers of times as  2(1 - 1/2ⁿ).  The text seems to also consider  (Double and subtract  5).

Prob. VI, p. 78.  (Halve and subtract  1)  thrice to leave  3.

Prob. XXXIV, p. 84.  (Double and subtract  100)  thrice, then  (double and subtract  50)  thrice, leaving  0.

Gherardi.  Libro di Ragioni.  1327. 

Pp. 47‑48.  Man gathering apples.  Four porters _  ith takes half plus  5 ‑ i,  leaving  1.

P. 100.  Man makes  12d  on his first trip.  He earns at the same rate on his second trip and then has  100d.  This leads to a quadratic and he finds the positive solution.  See Van Egmond, op. cit. in Common References, pp. 168, 177 & 185 for Italian, English and algebraic versions and some corrections.

Gherardi.  Liber habaci.  1327?  Pp. 144‑145.  Three porters _  ith takes half plus  i,  leaving none.

Paolo Dagomari (dell'Abbaco).  Trattato d'Abbaco, d'Astronomia, e di segreti naturali e medicinali.  c1339.  MS in Plimpton collection, Columbia Univ.  ??NYS.  (Rara, 435‑440.  On p. 438 is a reproduction showing a garden with three gates and guards.)

Dell'Abbaco.  c1370. 

Prob. 47, p. 44 with plate on p.45.  (Halve and subtract  2)  thrice to leave  7.

Prob. 71, pp. 65‑67 with plate on p. 66.  (Lose  _  and  6  more)  thrice to leave  24.  (The illustrations are very different from that in Rara (see previous entry).  Rara does not show enough text to see if the numbers used are the same as here, though the wording is clearly different.)  I have a colour slide of this.

Lucca 1754.  c1390. 

Ff. 26r‑27r, pp. 64‑65.  Multiply by  6/5  and spend  12,  multiply by  5/3  and spend  17,  double and spend  20,  leaving  0.

F. 59v, p. 135.  (Double and spend  12)  thrice to leave  3.

Bartoli.  Memoriale.  c1420.  Prob. 9, f. 75v (= Sesiano pp. 138 & 148).  Man going into a garden to get apples.  Gives  3/4  plus  3  more;  2/3  plus  2  more;  1/2  plus  1  more;  to leave  1.

Provençale Arithmétique.  Written (or more likely copied) at Pamiers, c1430.  MS in Bibliothèque Nationale, Paris, fonds français, nouvelle acquisition 4140.  Previously in the collections of Colbert (no. 5194) and the King (no. 7937).  Partially transcribed/translated and annotated by Jacques Sesiano; Une Arithmétique médiévale en langue provençale; Centaurus 27 (1984) 26-75.  The problems are not numbered, so I will give the folios and the pages in Sesiano.  However the indications of the original folios have not come through on a few pages of my copy and I then only give Sesiano's page.

P. 58.  Man doubles his money and spends  1,  triples and spends  2,  quadruples and spends  2,  leaving him with  3.

F. 113v-114r, p. 60.  (Sell  1/2  and one (or  1/2 ??)  more)  three times to leave  3.  The author gives a general solution as starting with the final result, (adding the extra number and double)  three times to get the original number.

AR.  c1450.  No. 185 & 187, pp. 87‑88, 173‑174 & 220.

185  = Fibonacci, p. 258.

187.  Double and spend  6,  double and spend  12,  double and spend  15,  leaving the initial amount.

Muscarello.  1478. 

Ff. 78v-79r, pp. 194-196.  Lose  1/2  and  6 - i  more,  for  i = 1, 2, 3, 4, 5,  leaving  1.

Ff. 84r-85r, pp. 201-204.  Merchant starts with  79  and makes profits of  17%,  19%,  21%,  23%  at four fairs.

Chuquet.  1484. 

Prob. 30.  (Double and subtract  12)  thrice, leaving  0.  English in FHM 206.

Prob. 31-33 are generalized versions.  E.g. Prob. 31 is  double and spend  5,  triple and spend  9,  quadruple and spend  12  to leave  8.

Prob. 95, English in FHM 219.  Merchant makes a profit of  1/3  and  i  more on his  i-th journey.  He makes as many journeys as he has money to start with.  When does he have  15?  This gives a messy equation:  (4/3)^x  =  3 - 9/(x+12).  Chuquet uses some interpolation to estimate  X = Ö(50 16297/16384) - 4 13/128  [FHM misprints this]  =  3.03949414, but I get  3.045827298.  Chuquet says ordinary interpolation is not valid.

Calandri.  Arimethrica.  1491. 

F. 66v.  (Double and spend 2) thrice leaving  0.

F. 74r.  Double and then gain  50%  giving  1000.  Woodcut of merchant on horse.

Pacioli.  Summa.  1494.

F. 105v, prob. 20.  (Give half and one more)  thrice leaving  1.  (See also H&S 58.)

F. 105v, prob. 22.  (Double and spend  12)  thrice leaving  0.

F. 187r, prob. 8.  Start with  x  and double  x  times to get  30.  This gives us  x 2^x = 30,  whose answer is  3.21988....  He interpolates both factors linearly on the third day, getting  (3+y)(8+8y) = 30,  so  3+y = 1 + Ö(19/4) = 3.17945.... 

He approaches the following problems similarly.

Ff. 187r-187v, prob. 9.  Start with  x  and make  25%  on each of  x  trips to make  40%  overall.  This gives  x (1.25)^x = 1.4 x.

F. 187v, prob. 10.  Leads to  x (1.2)^x = x².

F. 187v, prob. 11.  Leads to  x (1.4)^x = 6x.

F. 188r, prob. 13.  Start with  13,  (double and spend  14)  x  times to leave  0.  He observes that each iteration doubles the distance from  14,  so the problem leads to  2^x = 14,  but again he has to interpolate on the third day. 

Calandri, Raccolta.  c1500.  Prob. 16, pp. 17‑18.  Merchant gains  _  of his money plus  i  on the  ith trip.  After three trips he has  15.

Pacioli.  De Viribus.  c1500.  Prob. 67: Un signore ch' manda un servo a coglier pome over rose in un giardino.  (Lose half and one more)  five times to leave  1.

Blasius.  1513.  Ff. F.iii.r - F.iii.v: Decimaquinta regula.  Sack of money.  First man takes half and returns  100;  second takes half and returns  50;  third takes half and returns  25;  leaving  100  in the sack.

Johannes Köbel.  Rechenbiechlein auf den linien mit Rechenpfeningen.  Augsburg, 1514.  With several variant titles, Oppenheim, 1518;  Frankfort, 1531, 1537, 1564.  1564 ed., f. 89r, ??NYS.  Lose half, gain  100,  lose half, gain  50,  lose half, gain  25  to yield  100.  (H&S 58‑59 gives German and English.)

Ghaligai.  Practica D'Arithmetica.  1521.  Prob. 29, f. 66v.  Start with  100  and have to bribe ten guards with  1/10  each time.  Computes the exact residue, i.e.  100 x .9¹⁰.  (H&S 59-60).

Tonstall.  De Arte Supputandi.  1522.

Quest. 43, p. 173.  Give half plus  i+1,  for  i = 1, 2, 3,  leaving  1.  (H&S 61 cites this to the 1529 ed., f. 103)

Quest. 44, pp. 173-174.  Give half and get back  2i  for  i = 1, 2, 3,  leaving  12.

P. 246.  (Double and spend  12)  thrice to leave  0.

Riese.  Die Coss.  1524.  Several examples _ no. 35, 53, 55, 56, 58, 59, 60, 61, 62.  I describe a few.

No. 35, p. 45.  Man stealing apples:  (give half and one more)  four times, leaving  1.

No. 53, pp. 47‑48.  (Double and spend  12)  thrice, leaving  0.

No. 55, p. 48.  (Double and spend  i)  for  i = 1, 2, 3,  leaving  10.

No. 58, p. 48.  x + (4x+1) + (3(4x+1)+3)  =  56.

No. 61, p. 48.  (Give half plus  2+2i  more)  for  i = 1, 2, 3,  leaving  0.

No. 62, p. 49.  (Give half less  2i)  for  i = 1, 2, 3,  leaving  12.

Tartaglia.  General Trattato.  1556.  Book 12, art. 34, p. 199v.  Book 16, art. 47 & 113-116, pp. 246r & 253v‑254r.  Book 17, art. 9 & 20, p. 268v & 271r.  Final remainder specified in each case.

12-34.  (Take half plus  i  more)  for  i = 1, 2, 3, 4,  leaving  1.  Cf 16-115.

16-47.  Take  1/2  and  1  more,  1/3  and  2  more,  1/4  and  4  more, leaving  26.

16-113.  (Double and subtract  20)  thrice, leaving  0  (H&S 61 gives Latin and English of this one and says it appears in van der Hoecke (1537), Stifel (quoting Cardan) (1544), Trenchant (1566) and Baker (1568) (but see below).

16-114.  (Halve and subtract  1)  thrice, leaving  1, 2, ....

16-115.  Halve and subtract  1,  then  2,  3,  4,  leaving  1.  Cf 12-34.

16-116.  Lose  1/2  and  3  more,  lose  2/3  and get back  10,  lose  3/4  and  6  more,  lose  4/5  and get back  16,  leaving  24.

17-9.  Double & spend  4,  double & spend  8,  leaving  24.

17-20.  Double and spend  18,  double and spend  24,  double and spend  36,  leaving  280.

Buteo.  Logistica.  1559. 

Prob. 6, pp. 334-335.  Lose  1/2  and  3  more,  lose  1/3  and  4  more,  lose  1/4  and  gain  1,  to leave  100.

Prob. 13, pp. 342-343.  Start with  X,  gain  40.  Make the same rate of profit twice again and then the second of these gains is  90.

Prob. 19, p. 347.  Double and spend  12,  triple and spend  15,  quadruple and spend  14,  leaving  12.

Prob. 20, pp. 347-348.  Gain  1/4  and spend  7,  gain  1/3  and spend  10,  lose  3/7  and spend  8,  leaving  0.

Prob. 21, pp. 348-350.  (Double and spend  10)  X  times to leave  0.  He makes an error at  X = 8  and deduces  X = 7.

Baker.  Well Spring of Sciences.  1562?

Prob. 7,  1580?: ff. 192r-193r;  1646: pp. 302‑304;  1670: pp. 344-345.  Lose half and gain  12,  lose half and gain  7,  lose half and gain  4,  leaving  20.

Prob. 8,  1580?: ff. 193r-193v;  1646: p. 304;  1670: p. 345.  (Double and spend  10)  thrice leaving  12.

Gori.  Libro di arimetricha.  1571.  F. 72r (pp. 77‑78).  (Lose half and one more)  four times to leave  3.

Ozanam.  1725.  Prob. 28, question 1, 1725: 211‑212.  (Give half the eggs and half an egg)  thrice.  He doesn't specify the remainder and says that  8n‑1  eggs will leave  n‑1  and that one can replace  8  by  2^k  if one does the process  k  times.  Montucla replaces this by some determinate problems _ see below.

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790), prob. XV, pp. 86-87 (1790: prob. XXVII, pp. 88-89.  Shepherd loses  (half and  1/2  more)  thrice to leave  5  (or  1)  sheep.

Les Amusemens.  1749. 

Prob. 108, p. 249.  (Double and give away  6)  thrice to leave  0.

Prob. 111, pp. 252-253.  Double and spend  20,  triple and spend  27,  double and spend 19,  leaving  250.

Prob. 118, p. 260.  (Halve and give  1/2  more)  thrice to leave  0.

Walkingame.  Tutor's Assistant.  1751. 

1777: p. 82, prob. 7;  1860: p. 111, prob. 7.  Stealing apples.  Give half and get back  10,  give half and get back  4,  give half, get back  1,  leaving  24.

1777: pp. 174-175, prob. 86; 1860: p. 183, prob. 85.  Sheep fold robbed of half its sheep and half a sheep more, thrice, leaving  20.

Euler.  Algebra.  1770.  I.IV.III: Questions for practice, no. 9, p. 204.  (Spend  100, then gain  _)  thrice to double original value.

Vyse.  Tutor's Guide.  1771? 

Prob. 18, p. 36 & Key p. 29.  Sheep fold.  (Lose half and  ½  more) thrice, leaving  20.

Prob. 31, p. 62 & Key p. 69.  (Gain  _,  less  100)  for  3¼  years to yield  £3179 11s 8d.  Solution assumes the final quarter year is the same as  (gain  1/12,  less  25),  but it is not obvious how to determine an appropriate expression for the quarterly effect.  In general, repeating  ax - b  four times gives  a⁴x - b(a⁴-1)/(a-1)  and setting this equal to  4x/3 - 100  gives  a = 1.075..,  b = 22.371.  However, the  100  is the expenses of the merchant's family and he may not be able to reduce it in one quarter.

Prob. 33, pp. 62-63 & Key p. 70.  Lose  ½,  get back  10;  lose  _, get back  2;  lose  ½,  get back 1;  leaving 12.

Prob. 2, p. 137 & Key p. 179.  (Double and spend  6)  thrice, leaving  0.  Solution by double position.

Prob. 4, p. 137 & Key p. 180.  Lose  ½,  gain  10;  lose  ½,  gain  4;  lose  ½,  gain  1;  yielding  18.  Solution by double position.

Dodson.  Math. Repository.  1775.

P. 10, quest. XXIV.  Double and spend 6; triple and spend 12;  quadruple and spend 18; leaving 30.

P. 47, quest. C.  Shepherd loses ¼ of his flock and ¼ of a sheep; then _ of his flock and _ of a sheep; then ½ of his flock and ½ of a sheep; and has 25 sheep left.

P. 48, quest CI.  Man (spends 50 and gains _ on the remainder) thrice, yielding double his original amount.

P. 49, quest. CII.  Lose ¼, win 3, lose _, win 2, lose 1/7, yielding 12.

Ozanam‑Montucla.  1778.

Prob. 15, part a, 1778: 207-208;  1803: 203.  Prob. 14, 1814: 175‑176;  1840: 91.  (Sell half the eggs and half an egg more)  thrice to leave  36.  This was one of the more popular forms of the puzzle after this time _ see: Jackson, Endless Amusement II, Nuts to Crack, Young Man's Book, Boy's Treasury, Magician's Own Book, Boy's Own Conjuring Book, Wehman, Collins, Sullivan.

Prob. 15, part b, 1778: 208-209;  1803: 203-204.  Prob. 15, 1814: 176-177;  1840: 91.  (Spend half and  1/2  more)  thrice leaving  0.  Gives the rule for the problem with more iterations.

Bonnycastle.  Algebra.  1782

P. 86, no. 20 (1815: p. 107, prob. 30).  Lose  1/4  of what he has, win  3,  lose  1/3,  win  2,  lose  1/7,  leaving  12.

Eadon.  Repository.  1794. 

P. 296, no. 9.  Man loses  1/4,  then gains  3,  then loses  1/3,  then gains  2,  then loses  1/7,  then has  12.

P. 296, no. 10.  Man (spends  50  and then gains  1/3)  thrice to double his money.

P. 297, no. 14.  Man sends out  1/3  and  25  more of his men, leaving  1/2  and  100  more.

Bonnycastle.  Algebra.  10th ed., 1815.  P. 106, no. 19.  (Double and pay  1)  four times, leaving  0.

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions.

No. 10, pp. 16 & 73.  (Give one half)  seven times, leaving  1.

No. 13, pp. 17 & 74.  (Give half plus  i)  for  i = 1, 2, 3,  leaving  1.

No. 18, pp. 18 & 75.  (Spend half plus half a guinea)  four times, leaving  0.

No. 19, pp. 18-19 & 75-76.  (Sell half, receive  10);  (sell third, receive  2);  (sell half, receive  1);  leaving  12.

No. 23, pp. 19-20 & 77.  Same as Ozanam-Montucla, prob. 15a.

Endless Amusement II.  1826?  Prob. 27, pp. 202-203.  Identical to the 1803 English of Ozanam-Montucla, prob. 15a.

Nuts to Crack II (1833), no. 126.  Same as Ozanam-Montucla, prob. 15a.

Young Man's Book.  1839.  Pp. 234-235.  Identical to the 1803 English of Ozanam-Montucla, prob. 15a.

Boy's Treasury.  1844.  Country-woman and eggs, p. 301.  Same as Ozanam-Montucla, prob. 15a.

Magician's Own Book.  1857.

Countrywoman and eggs, pp. 238-239.  Almost identical to Boy's Treasury.

The old woman and her eggs, p. 240.  (Give half and half an egg more)  thrice, leaving  1.

The apple woman, p. 252.  Sell half, gain  10,  sell third, gain  2,  sell half, gain  1, have  12  remaining.  = Book of 500 Puzzles, 1859, p. 66.  = Illustrated Boy's Own Treasury, 1860, prob. 15, pp. 428 & 431-432.

Boy's Own Conjuring Book.  1860. 

Countrywoman and eggs, p. 205.  Almost identical to Boy's Treasury.

The old woman and her eggs, p. 212.  Identical to Magician's Own Book.

The apple woman, p. 223.  Identical to Magician's Own Book.

Mittenzwey.  1879?  Prob. 138, pp. 26 & 76‑77.  Three men successively taking  1/3  of a pile of potatoes.  Remainder is  24.  Asks how to divide the remainder to make the shares equal.

William J. Milne.  The Inductive Algebra Embracing a Complete Course for Schools and Academies.  American Book Company, NY, 1881.  Pp. 138 & 332, no. 81.  (Double and lose one)  thrice to get triple original amount.

Hoffmann.  1893.  Chap. IV gives several deterministic examples:  nos. 27, 39, 46, 67, 76 (? see 7.E.1), 111, 112.

Lucas.  L'Arithmétique Amusante.  1895.  P. 184: Prob. XLII: La marchande d'Œufs.  (Sell  1/2  plus half an egg more)  n  times to leave  0.

W. W. Rouse Ball.  Elementary Algebra.  CUP, (1890), 2nd ed., 1897.  Prob. 10, p. 260 & 475.  Three Arab jugglers and their monkey, on their way to Mecca, buy a basket of dates.  Ending 1.  He gives no background to the problem nor any indication that it is novel.  The solution is just the numerical answer.  If this appeared in the first ed. of 1890, it would provide a plausible transmission route between Mahavira and the present _ Ball was already an authority on the history of mathematics and on recreations, so the translator of Mahavira might well have communicated with Ball.

Dudeney.  Problem 522.  Weekly Dispatch (8  &  22 Nov 1903) both p. 10.  Multiple of  25  eggs.  Sell half and half an egg more until all gone.

Clark.  Mental Nuts.  1904.

1904: no. 19;  1916, no. 17.  The man and his money.  (Spend  1/2  and  1/2  more)  four times to leave  0.

1904: no. 40.  The farmer and his horses.  (Pay  1,  then sell  1/2,  then pay  1  more)  four times to leave  1.

1904: no. 92;  1916, no. 8.  The man and the stores.  (Double and pay  10)  thrice to leave  0.

1904: no. 99.  Three boys and basket of apples.  Coconuts _ Ending 1  with  3  people.  (Since this is not in the 1916 ed., it is hard to know if this might have been in the 1897 ed.  In any case, this complicates the possible connection with Mahavira _ cf the discussion under Pearson, below.)

Pearson.  1907.  Part II.  Several determinate versions, and the following.

No. 29: The men, the monkey, and the mangoes, pp. 119 & 197.  Coconuts _ Ending 1  with  3  people.  Gives only one solution.

No. 94: One for the parrot, pp. 133‑134 & 210.  Coconuts _ Ending 1  with  4  boys, a bag of nuts and a parrot.  Gives only one solution.

The connection of these with Mahavira, 850, has bemused me and I conjectured that Pearson might have heard of Mahavira's work, though the translation didn't appear until 1912.  Kaye's note (see under Mahavira in the Abbreviations) shows that an advance version of the translation was produced in 1908, which makes my conjecture much more likely.  The Frontispiece of Pearson's book shows him as a Anglican(?) clergyman of about 40-50 years old, so he might have been a missionary or had Indian contacts.  HOWEVER, I have now seen Clark and this makes the connection less clear.  But I have now discovered Ball and this may make things clearer.

Wehman.  New Book of 200 Puzzles.  1908.

P. 50: The sheepfold robbery.  (Lose  1/2  and  1/2  a sheep more)  thrice leaving  2.

P. 51: The maid and her apples.  c= Magician's Own Book, p. 252.

P. 57: The countrywoman and her eggs.  Same as Ozanam-Montucla, prob. 15a.

Nelson L. Roray, proposer;  A. M. Harding, Norman Anning and the proposer, solvers.  Problem 288.  SSM 12 (1912) 235  &  520‑521.  Coconuts _ Ending 1  with  3  men.  Anning shows that the solution is  ‑2 (mod 3⁴),  but none of the solvers generalise to  n  men.

Loyd.  Newsboys puzzle.  Cyclopedia, 1914, pp. 116 & 354.  (= MPSL2, prob. 9, pp. 8 & 123.  = SLAHP: Family rivalry, pp. 51 & 103.)  Complex specification of one amount.

Loyd.  A study in hams.  Cyclopedia, 1914, pp. 268 & 375.  (= SLAHP: The Ham peddler, pp. 81 & 117.)  (Half plus half a ham more)  four times,  (half a ham plus half),  (half plus half a ham),  leaving  0.

R. L. Weber.  A Random Walk in Science.  Institute of Physics, London & Bristol, 1973.  P. 97 excerpts a Russian book on humour in physics which states that P. A. M. Dirac heard a version of the problem with three fishermen and a pile of fish, but only three divisions, at a mathematical congress while he was a student (at Cambridge?) and gave the solution,  -2.  In fact, he only came to Cambridge as a graduate student in 1923 and became a fellow in 1927, so that the story, if true and if it refers to his time at Cambridge, relates to the mid 1920s.

Ben Ames Williams.  Coconuts.  Saturday Evening Post (9 Oct 1926) 10,11,186,188.  Reprinted in:  Clifton Fadiman, ed.; The Mathematical Magpie; Simon & Schuster, NY, 1962, pp. 196-214.  Ending 0  with  5  men.

R. S. Underwood, proposer;  R. E. Moritz, solver.  Problem 3242.  AMM 34 (1927) 98 (??NX)  &  35 (1928) 47‑48.  General version of coconuts problem,  Ending 0  with  n  men.

Wood.  Oddities.  1927. 

Prob. 52: A goose problem _ not for geese to solve, pp. 43 & 44.  Sell a half and half a goose more;  sell a third and a third of a goose more;  sell a quarter and 3/4 of a goose more;  sell a fifth and a fifth of a goose more;  leaving 19.

Prob. 57: Eggs this time, p. 46.  Sell half and half an egg more;  sell a third and a third of an egg more;  sell a quarter and a quarter of an egg more;  sell a fifth and a fifth of an egg more;  leaving a multiple of 13.  Determine the least number of possible eggs.  Gives answer  719.  Complete answer is  719 (mod 780).

Collins.  Book of Puzzles.  1927.  The basket of eggs puzzle, p. 77.  Same as Ozanam-Montucla, prob. 15a.

Collins.  Fun with Figures.  1928.  The parrot talks, pp. 183-185.  Four boys and a parrot and a bag of nuts.  Ending 1  with  n = 4.  = Pearson 94.

Kraitchik.  La Mathématique des Jeux, 1930, op. cit. in 4.A.2.  P. 13.

Prob. 39.  Monkey and mangoes problem,  Ending 1  with  3  men.  = Pearson 29.  = MR, 1942, prob. 35, pp. 32-33.

Prob. 40.  (Take  _)  thrice, leaving  8.  = MR, 1942, prob. 36, p. 32.

          He asserts these are Hindu problems but gives no source.

Rudin.  1936.  No. 5, pp. 3 & 76.  Three men.  (Take away _)  thrice, then divide in thirds.  He gives only the answer  81,  though any multiple of  81  works.

Hubert Phillips.  Question Time.  Op. cit. in 5.U.  1937.  Prob. 203: Adventure island, pp. 137 & 246.  Ending 1  with  5  men and Friday instead of a monkey.

Francis & Vera Meynell.  The Week‑End Book.  Nonesuch Press, 1924 and numerous printings and editions.  I have 8th printing, 2nd ed., Mar 1925, and a 5th(?) ed., in 2 vols., Penguin, 1938.  The earlier edition has some extra text surrounding the problems, but has only 8 of the 12 problems in the Penguin ed.  This problem is not in the 2nd ed.  5th?? ed., prob. seven, p. 408: Three men and a monkey.  Ending 0,  with  3  men.  No solution.

Joseph Bowden.  Special Topics in Theoretical Arithmetic.  Published by the author, Lancaster, Pennsylvania, 1936.  The problem of the dishonest men, the monkeys and the coconuts, pp. 203-212.  ??NYS - cited by Pedoe, Shima & Salvatore.

McKay.  At Home Tonight.  1940. 

Prob. 33: The niggers and the orchard, pp. 69 & 82.  Three men and apples.  Ordinary division with the extra thrown away and Ending 1.

Prob. 35: Dividing nuts, pp. 70 & 83.  Divide nuts among  5  girls with one left over.  One girl divides hers among the rest, with one left over.  Then another girl divides hers among the rest, with one left over.

The Home Book of Quizzes, Games and Jokes.  Op. cit. in 4.B.1, 1941.  Pp. 149‑150: The bag of peanuts.  Five Italians, a bag of peanuts and a monkey.  Coconuts _ Ending 1  with  5 people.  The given answer is  3121,  but it should be  15621.  3121  would be the answer for Ending 0 _ cf Leeming, 1946.

Garrett Birkhoff & Saunders Mac Lane.  A Survey of Modern Algebra.  Macmillan, (1941, ??NYS), revised, 1953.  Prob. 11, p. 26 (p. 28 in the 4th ed. of 1977).  Usual five men and a monkey form, but no ending is specified.  That is, nothing is stated about the number left over in the morning except that it is an integer.  No answer given.  Somewhat surprisingly, the answer is the same as for the  Ending 0  problem _ see Pedoe, Shima & Salvatore.

Leeming.  1946.  Chap. 5, prob. 16, pp. 57‑58 & 177.  Five Italian organ grinders, their monkey and a pile of peanuts.  Coconuts _ Ending 1  with  5 people.  The given answer is  3121,  but it should be  15621.  3121  would be the answer for Ending 0 _ cf Home Book, 1941.

Sullivan.  Unusual.  1947.  Prob. 40: Another old problem.  Sell (half plus half an egg more)  thrice, leaving  36.  = Ozanam-Montucla, prob. 15a.

Paul S. Herwitz.  The theory of numbers.  SA 185:1 (Jul 1951) 52-55.  With letter and response in SA 185:3 (Sep 1951) 2 & 4.  Gives the Birkhoff & Mac Lane version with  n = 3  and solves it by explicitly computing the number remaining in terms of the initial number, obtaining one linear diophantine equation in two unknowns.  He states the equation for the case  m = 4,  but doesn't give the solution, and for the general case, though he doesn't sum the geometric progression that appears.  The letter requests the solution for the case  m = 5  and Herwitz outlines how to find the solution by the Euclidean algorithm, obtaining  3121.

Anon.??  Monkeys and coconuts.  Mathematics Teacher 54 (Dec 1951) 560-562.  ??NYS - cited by Pedoe, Shima & Salvatore.

Ron Edwards.  The cocoanut poker deal.  The Cardiste (Mar 1958) 5-6.  Uses the three person problem as the basis of a card trick.  He states the original problem in a novel form _ each hunter finds the pile evenly divisible by three, so the monkey doesn't get any coconuts until the morning division when he gets one.  But in the trick, Edwards uses the classic form, with a variation.  The  52  cards are dealt into  3  piles,  with one extra put in a discard pile.  The spectator places one of the end piles on the middle giving a new deck of  34.  The process is repeated twice more leaving  22,  then  14  cards.  These  14  are dealt into three piles, but now there are two extras which are discarded and then the five discards are found to be a royal flush!  (The Cardiste was a mimeographed magic magazine which ran from 1957 to 1959.  My thanks to Max Maven for remembering and finding this and sending a copy.)

M. Gardner.  SA (Apr 1958)  = 2nd Book, chap. 9.  Describes the Williams story of 1926 and says the Post received 2000 letters the first week after publication and the editor telegrammed:  "For the love of Mike, how many coconuts?  Hell popping around here."  Gardner says Williams modified the older problem of Ending 1  with  5  men.  He gives the solution of  -4,  but says he could not trace its origin.  His addendum cites Anning (1912) for this.

Roger B. Kircher.  The generalized coconut problem.  AMM 67:6 (Jun/Jul 1960) 516-519.  Generalizes by taking any number of sailors, any number of divisions and allowing the  i‑th division to discard a variable amount  V_i  before taking away  1/n  of the rest, even allowing negative  V_i,  e.g. if the monkey is adding coconuts to the pile!  Sadly, his basic recurrence equations (1) and (2) are misprinted.  He solves this by use of difference calculus.

Nathan Altshiller Court.  Mathematics in Fun and in Earnest.  Op. cit. in 5.B.  1961.  Prob. c, pp. 188-190.  Take  _  thrice, leaving  8.  How to divide the remainder to make things equal?

Philip Haber.  Peter Pauper's Puzzles & Posers.  Peter Pauper Press, Mount Vernon, NY, 1963.  Prob. 125, pp. 34 & 57.  Basket of pears divided among four people.  First gets  ¼  of the total plus  ¼  of a pear.  Second gets  _  and  _.  Third gets  ½  and  ½.  Fourth gets remainder, which is half of what the first got.

Harold H. Hart.  Grab a Pencil No. 3.  Hart Publishing, NY, 1971.  The horse trader, pp. 41 & 118.  (Pay  1,  then half, then  1  more)  thrice, leaving  1.

Birtwistle.  Math. Puzzles & Perplexities.  1971.

The Arabs, the monkey and the dates, pp. 50-52.  Four people, Ending 1.  Gets  1021.

Baling out, pp. 52, 169 & 189.  (Lose _ of a load of bales of hay and _ of a bale more) four times, leaving an integral number and no broken bales.  Solution is to start with  80  bales.

Shakuntala Devi.  Puzzles to Puzzle You.  Op. cit. in 5.D.1.  1976.  Prob. 144: The mango thieves, pp. 89 & 135.  Ending 1  with three boys, but no monkey _ the boys each eat a mango when they steal  _  and there is an extra mango when they divide in the morning.

Michael Holt.  Figure It Out _ Book One.  Dragon (Granada), London, 1978.  Problem 14 (no page number) gives a Russian version involving a man who sells his soul to the Devil.  (Double and spend  8)  thrice to leave  0.

Birtwistle.  Calculator Puzzle Book.  1978.  Prob. 27: Shipwreck, pp. 21 & 80.  Four sailors, the ship's cat and a box of biscuits, but only three of the sailors take a fourth, with one for the cat, before the final division into four, with none for the cat.  Answer is  61,  but this should be taken  (mod 256).

Dan Pedoe, Timothy Shima & Gali Salvatore.  Of coconuts and integrity.  CM 4:7 (Aug/Sep 1978) 182‑185.  General discussion of history and examples, based on Gardner (1958).  They generalise to consider giving  c  to the monkey and to having  m  divisions.  They first do Birkhoff & Mac Lane's version and  Ending 1,  effectively noting that the latter is the same as the former but with an extra division step.  I.e., the Birkhoff & Mac Lane problem has  m = n,  while  Ending 1  has  m = n+1.  Examination of the results shows that the Birkhoff & MacLane problem with odd  n  has the same answer as the  Ending 0  problem, but for even  n,  it corresponds to an  Ending 2  problem, which turns out to be one greater than the  Form, Ending = 0, 1 problem.  They separately solve the  Ending 0  problem, again with general  c.

Ben Hamilton.  Brainteasers and Mindbenders.  (1979);  Prentice‑Hall, Englewood Cliffs, NJ, 1981.  Problem for March 29, pp. 36 & 156.  i‑th customer buys  i+1  plus  1/(i+1)  of the rest.  How many customers can be served?

Scot Morris.  The Next Book of Omni Games.  Plume (New American Library), NY, 1988.  The monkey and the coconuts, pp. 30-31 & 182-183.  Sketches usual history.  He then notes that the usual process has the pile  º 1 (mod n),  so the monkey essentially gets one, then the pile is divided into  n  parts.  But one could alternatively have the pile  º 0 (mod n),  so the pile is divided into  n  parts, the sailor takes his part and then the monkey takes  1  from the remainder.  I.e. rather than removing one and  1/n  of the rest, each step removes  1/n  plus one more.  I have now termed these  Form 1  and  Form 0.  In this situation he doesn't allow the monkey to get one in the final division, i.e. he considers  Ending 0,  but  Ending 1  could be permitted, as studied by me below.

Liz Allen.  Brain Sharpeners.  Op. cit. in 5.B.  1991.  It really takes the biscuit, pp. 50-52 & 119.  Four boys and their cat dividing biscuits.  Form 1, Ending 1  with  4  boys.

David Singmaster.  Coconuts.  The history and solutions of a classic diophantine problem.  Technical Report SBU-CISM-93-02, School of Computing, Information Systems and Mathematics, South Bank Univ., Dec 1993, 21pp, Feb 1994.  Submitted to Mathematics Review (Univ. of Warwick), 1993, but the journal closed before using it.  Revised in 1996 and again as the 3rd ed. of 11 Sep 1997, 21pp.  Extensive history, based on the material in this section.  Following Morris's comment, I consider both division Forms and both Endings, giving four basic problems rather than the three that he mentions.  In the 3rd ed., I changed the terminology to Form and Ending and have converted the material in this section to conform with this.  A little reflection shows that the solution for  Form 0, Ending 0  is one less than for  Form 1, Ending 1.  Actual calculation shows that one of the four cases has the same sequence of pile sizes as another, but shifted by one stage.  When  n  is odd,  Form 0, Ending 0  is the same as  Form 0, Ending 1,  but starting one stage earlier.  When  n  is even,  Form 1, Ending 1  is the same as  Form 1, Ending 0,  but starting one stage earlier.  Although these results are easily seen from the algebraic expressions, I cannot see any intuitive reason for these last equalities.

                    Having now seen Pedoe, Shima & Salvatore, I have added two supplementary pages discussing the Birkhoff & MacLane problem and relating it to the standard versions.

 

          7.E.1. VERSIONS WITH ALL GETTING THE SAME

 

          The  i‑th child gets some linear function of  i  applied to the remainder, but all wind up with the same amount.

          See Tropfke 586.

 

Fibonacci.  1202.

P. 279.  i-th gets  i  +  1/7  of rest.  (Sanford 219 gives the English;  H&S 61‑62 gives Latin & English.)

P. 279.  i-th gets  i  +  2/11  of the rest, but he doesn't ask for the number of children.

Pp. 279‑280.  i-th gets  (3i‑1)  +  6/31  of the rest.

Pp. 280‑281.  i-th gets  (2i+1)  +  5/19  of the rest.

Maximus Planudes.  Ψηφηφoρια κατ' Ivδoυσ η Λεγoμεvη Μεγαλη (Psephephoria kat' Indous e Legomene Megale) (Arithmetic after the Indian method).  c1300.  (Greek ed. by Gerhardt, Das Rechenbuch des Maximus Planudes, Halle, 1865, ??NYS [Allard, below, pp. 20-22, says this is not very good].  German trans. by H. Waeschke, Halle, 1878, ??NYS [See HGM II 549;  not mentioned by Allard].)  Greek ed., with French translation by A. Allard; Maxime Planude _ Le Grand Calcul selon des Indiens; Travaux de la Faculté de Philosophie et Lettres de l'Univ. Cath. de Louvain _ XXVII, Louvain‑la‑Neuve, 1981. 

                    i-th gets  i  +  1/7  of the rest, pp. 191‑194 & 233‑234.  On pp. 233‑234, Allard discusses the history of the problem, citing:  Fibonacci;  BR;  Parisinus supp. gr. 387  &  Scoriolensis Φ I 16.

BR.  c1305.  No. 84, pp. 102‑105.  i-th gets  i  +  1/7  of the rest.

Gherardi.  Libro di ragioni.  1327.  Pp. 37‑38.  i-th daughter gets  i  +  1/10  of the rest.

Dell'Abbaco.  c1370. 

Prob. 168, p. 140.  i-th gets  1000 i  +  1/10  of the rest. 

Prob. 169, pp. 140‑141.  i-th gets  1/6  plus  10 i.

Lucca 1754.  c1390.  F. 82v, p. 199.  i‑th gets  i  +  1/10  of rest.  (This problem is not clearly expressed.)

Bartoli.  Memoriale.  c1420.  Prob. 9, f. 75v (= Sesiano pp. 138 & 147-148).  i-th gets  i  +  1/7  of the rest.

AR.  c1450.  No. 114, 115, 352, pp. 64‑65, 154, 173‑174 & 220.

114:  i-th gets  i  +  1/10  of rest.

115:  i-th gets  i  +  1/6  of rest.

352:  i-th gets  1/5  of remainder  +  i ‑ 1.

Muscarello.  1478.  F. 85v, pp. 204-205.  i-th gets  i  +  1/9  of the rest.

Chuquet.  1484.  Probs. 129‑141.  English of prob. 129 in FHM 224-225, with some description of the others.  i-th child gets  (a_i + b)  plus  r  of the rest;  i-th child gets  r  of amount and  a + bi  more.  Many problems have non‑integral number of children and amounts received _ e.g. prob. 133 has  2 5/6  children receiving  6 2/3,  with the  5/6  getting  5 5/9.

HB.XI.22.  1488.   Pp. 44‑45 (= Rath 247).  i-th gets  i  +  1/9  of rest.

Calandri.  Arimethrica.  1491.  F. 65r.  i-th gets  1/10  +  1000 i

Calandri, Raccolta.  c1500.  Prob. 26, pp. 25‑26.  i-th gets  1000 i  +  1/10  of the rest.

Ghaligai.  Practica D'Arithmetica.  1521. 

Prob. 24, ff. 65v-66r.  i-th gets  1000 i  +  1/7  of rest.

Prob. 25, f. 66r.  i-th gets  1/7  +  1000 i.

Cardan.  Practica Arithmetice.  1539.  Chap. 66, section 65, f. FF.ii.r (p. 155).  i-th child gets  1/7  of remainder  +  100 i

Tartaglia.  Quesiti, et Inventioni Diverse.  Venice, 1546.  Book 9, quest. 2, pp. 98r‑98v.  i-th child gets  i  +  1/8  of the rest.

Tartaglia.  General Trattato, 1556, art. 46, pp 245v‑246r.  i-th child gets  i  +  1/6  of the rest;  discusses same with  1/7  and  1/13.

Buteo.  Logistica.  1559.  Prob. 78, pp. 286-288.  i-th child gets  1/6  +  100 i.

Bachet.  Problemes.  1612.  Addl. prob. VII: Un homme venant à mourir ..., 1612: 149-154;  1624: 221-226;  1884: 158‑161.  i-th child gets  i  +  1/7  of the rest;  also  a_i  +  1/n  and  1/n  +  a_i.  Asserts some cases are impossible, contrary to Chuquet's approach.  Labosne has much revised the entire problem.

Ozanam.  1725.  Prob. 10, question 9, 1725: 67‑68.  Prob. 1, 1778: 185;  1803: 182-183;  1814: 159;  1840: 82.  i-th gets  10000 i  +  1/7  of the remainder.

Les Amusemens.  1749. 

Prob. 55, pp. 187-188.  i-th gets  1000 i  +  1/7  of the rest.

Prob. 177, p. 328.  i-th gets  1000 i  +  1/5  of the rest.

Euler.  Vollständige Anleitung zur Algebra.  (St. Petersburg, 1770) Part 2, sect. 1, chap. 3, art. 42.  (= Opera Omnia (1) 1 (1911), pp. 226‑228.  = Algebra; 1770; I.IV.III.604: Question 21, pp. 202‑203.)  i-th gets  100 i  +  1/10  of rest.

Hutton.  A Course of Mathematics.  1798?  Prob. 10,  1833: 214-217;  1857: 218-221.  Father with three sons leaves  ai + 1/n  of the remainder to the  ith and this exhausts the fortune (but they do not get equal amounts except when  n = 4).  Finds algebraic expressions for the total and each portion, e.g. the total fortune is  (6n² - 4n + 1)a/(n - 1)².

Augustus De Morgan.  Arithmetic and Algebra.  (1831 ?).  Reprinted as the second, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  Art. 114, pp. 29-30.  First gets  2  + 1/6  of rest;  second gets  3  + 1/6  of rest;  they find they got the same amount.

Louis Pierre Marie Bourdon (1779-1854).  Élémens d'Algèbre.  7th ed., Bachelier, Paris, 1834.  (The 6th ed., was 1831, but I haven't yet looked for the early editions _ ??)  Art. 48, pp. 66-71 is the same as Hutton's 1798? problem.  Art. 49, pp. 71-73 then treats the usual problem of the same form, finding  n-1  children and a fortune of  a(n-1)².

Fred Barlow.  Mental Prodigies.  Hutchinson, London, (1951), corrected reprint, 1952.  On pp. 38-41, he describes Henri Mondeaux (1826-1862), an illiterate who developed powers of mental caclulation, and was then taught by a schoolmaster, M. Jacoby.  At some time, he was asked to solve the problem of people taking  100i  + 1/13  of the rest  and he found the answer in a few seconds by taking  12 x 100  as the amount of each person and  12 x 12 x 100  as the total amount.  I wonder if he knew this type of problem beforehand??

Hoffmann.  1893.  Chap. IV, no. 76: Another eccentric testator, pp. 166 & 221‑222.  First son gets  1/6  plus  240,  second son gets  1/5  of the remainder plus  288,  ...,  fifth gets  1/2  of the remainder plus  720  and all wind up with the same amount.

Lucas.  L'Arithmétique Amusante.  1895.  Prob. XL: Le Testament du nabab, pp. 144-147.  i‑th gets  i  +  1/7  of the rest.  Gives general solution of:  i-th gets  i  +  1/n  of the rest, using a rectangular layout of markers similar to ancient Indian multiplication tables and Lucas thinks the ancient Indians must have known this problem and solution.

 

          7.F.    ILLEGAL OPERATIONS GIVING CORRECT RESULT

 

          'Two digit' refers to illegal cancellation with two digit numbers, e.g. 16/64 = 1/4, etc.

 

A. Witting.  Ernst und Scherz im Gebiete der Zahlen.  Zeitsch. math. u. naturw. Unterricht 41 (1910) 45‑50.  P. 49 gives the rule of Ahrens, below, pp. 75‑76, for the case  k = 2.  He also gives three of the two digit examples:  26/65,  16/64,  19/95  _  omitting  49/98.

Ahrens.  A&N, 1918.

Pp. 73‑74 finds the two digit solutions and some with more digits.

Pp. 75‑76 studies  (a + m/n)^1/k  =  a * (m/n)^1/k  and finds that  m = a,  n = a^k ‑ 1  works.

W. Lietzmann.  Lustiges und Merkwürdiges von Zahlen und Formen.  1922.

                    2nd ed., F. Hirt, Breslau, 1923.  Pp. 103‑104.  Gives  26/65  and  16/64  and the general rule of Ahrens, pp. 75‑76, citing Witting and Ahrens. 

                    4th ed, same publisher, 1930, p. 153, says there are more two digit examples and gives also  266/665,  etc.

                    The material is also in the 6th ed. (1943), but not in the 7th ed. (1950).

R. K. Morley, proposer;  Pincus Schub, solver.  Problem E24.  AMM 40 (1933) 111  &  425‑426.  2 digit versions.

G. [presumably the editor, Jekuthiel Ginsburg].  Curiosa 31 _ Another illegal operation.  SM 5 (1939) 176.  Cites Morley.  Refers to E. Nannei presenting several larger examples in 1935, some involving cancellation of several digits.

William R. Ransom.  Op. cit. in 6.M.  1955.  Freak cancellations, pp. 100‑102.  Finds the 2‑digit versions and give examples of several 3‑digit forms:  138/345 = 18/45,  163/326 = 1/2,  201/603 = 21/63.

B. L. Schwartz, proposer;  C. W. Trigg, solver.  Problem 434 _ Illegal cancellation.  MM 34 (1961) ??NYS  &  367‑368.  3 digit versions.

Anon.  Curiosa 122 _ A common illegal operation.  SM 12 (1946) 111.  (a² ‑ b²)/(a ‑ b)  =  a + b.

Alan Wayne, proposer;  solution ??NYS.  Problem 3568.  SSM 75:2 (No. 660) (Feb 1975) 204.  (3/2)² - (1/2)²  gives the right answer when the exponents are interpreted as multipliers.

Ben Hamilton.  Op. cit. in 7.E, 1979.  Problem for April 9, pp. 40 & 157‑158.  249/498  gives  24/48  correctly, which gives  2/8  wrongly.

R. P. Boas.  Anomalous cancellation.  In:  R. Honsberger, ed.; Mathematical Plums; MAA, 1979.  Chap. 6, pp. 113‑129.  Surveys the problem and studies the two digit case for other bases, e.g.  32/13 = 2/1  in base 4.  Cites the SM report, 1939.

R. P. Boas.  Generalizations of the  64/16  problem.  JRM 12 (1979‑80) 116‑118.  Summarises the above paper and poses problems.

 

          7.G.    INHERITANCE PROBLEMS

 

          7.G.1. HALF + THIRD + NINTH, ETC.

 

          This is usually called 'The 17 camels', etc. 

          Early versions of this problem simply divided the amount proportionally to the given numbers, regardless of whether the numbers added to one or not.  I mention a few early examples of this below.  By about the 15C, people began objecting to such proportions, though Tartaglia (qv) and others see no difficulty with the older idea.  Sanford 218‑219 says Tartaglia was among the first to suggest the 18th camel _ but I have not found this in Tartaglia so far.  H&S 87 says that the use of the 18th camel is really a modern problem.  I haven't found it occurring until late 19C, when several authors claim it is centuries old and comes from the Arabic world or India or China!  See 1872 below.  Not everyone is happy with the problem, even to this day _ see Ashley, 1997.

 

 

1/2, 1/3                                     Dell'Abbaco;  Tagliente;  Buteo

1/3, 1/4                                     Jackson

9, 8, 7                                        Papyrus of Akhmim

4/5, 3/4, 2/3                             Tartaglia 44

2/3, 1/6, 1/8                             Eperson

2/3, 1/2, 1/4                             Chuquet;  Apianus; 

1/2, 1/3, 1/4                             Bakhshali MS;  Lucca 1754;  AR 204;  Calandri;  Tagliente;  Riese;  Tartaglia 42;  Buteo;  Ozanam;  Les Amusemens;  Bullen;  Collins; 

1/2, 1/3, 1/6                             AR, 286

1/2, 1/3, 1/9                             AR, 170, 286;  Hanky Panky;  Cassell's;  Proctor;  Cole;  Lemon;  Hoffmann;  Brandreth Puzzle Book;  Loyd;  Benson;  White;  Ball‑FitzPatrick;  Dudeney;  Kraitchik;  McKay;  Sullivan;  Ashley;

1/2, 1/4, 1/5                             Lemon;  Clark;  Ernst;  King;  Foulsham

1/2, 1/4, 1/8                             Bath

2/5, 1/3, 1/4                             AR, 202;  Wagner

1/3, 1/4, 1/5                             BR;  Riese;  Tartaglia 43;  Jackson

1/4, 1/5, 1/6                             Apianus;  Leybourn

2/3, 1/2, 1/3, 1/4                     Papyrus Rhind;  Pike

1/2, 1/3, 1/4, 1/5                     Chaturveda;  Blasius

1/2, 1/3, 1/4, 1/6                     Mahavira

1/2, 1/3, 1/6, 1/19                   Parlour Pastime

1/3, 1/4, 1/5, 1/6                     Walton;  Simpson;  Dodson;  J. King

6, 5, 4, 3, 2                                Mahavira

7/2, 5/2, 15/4, 25/4, 4             Papyrus of Akhmim

1/2, 1/3, 1/4, 1/5, 1/6             Tonstall

1/3, 1/4, 1/5, 1/6, 1/7             Walkingame;  Vyse

1/3, 1/4, 1/6, 1/8, 1/9             Meyer;  Leeming

 

Papyrus Rhind, c‑1650, loc. cit. in 7.C.  Problem 63, p. 101 of vol. 1 (1927) (= p. 53 (1978)).  Divide  700  loaves in proportion  _ : ½ : _ : ¼.

Papyrus of Akhmim.  c7C.  Jules Baillet, ed.  Le Papyrus Mathématique d'Akhmîm.  Mémoires publiés par les membres de la Mission Archéologique Français au Caire, vol. IX, part 1, (1892) 1‑89.  Brief discussion of this type of problem on p. 56.  Probs.  3, 4, 10, 11, 28?, 47, 48, 49  are of this type.  I describe two examples.

Prob. 3, pp. 64-65.  Divide  1000  in proportion  3 + 1/2  :  2 + 1/2  :  3 + 1/2 + 1/4  :  6 + 1/4  :  4.

Prob. 11, pp. 68-69.  Divide  3 + 1/2 + 1/4  in proportion  7 : 8 : 9.

Bakhshali MS.  c7C.  See in 7.E, where a king gives away  ½ + _ + ¼  of his money!

Mahavira.  850.  Chap. VI, v. 80, 86, pp. 110-111.  Divide  120  in proportion  1/2 : 1/3 : 1/4 : 1/6.  Divide  480  in proportion  2 : 3 : 4 : 5 : 6.

BR.  c1305.  No. 71, pp. 94‑95.  1/3 + 1/4 + 1/5.

Dell'Abbaco.  c1370.  Prob. 122, pp. 97‑98.  Divide into  ½ + _.  He divides in the ratio  3 : 2.

Lucca 1754.  c1390.  F. 61r, p. 140.  Divide into  ½ + _ + ¼.  He divides in proportion  6 : 4 : 3.

AR.  c1450.  Probs. 170, 202-204, 207, 229-230, 286.  Pp. 81‑82, 94‑96, 106-107, 130, 160‑161, 166‑167, 211‑213.

170:  1/2 + 1/3 + 1/9.

202:  1/3 + 1/4 + 2/5.

203:  Divide 384 into  2/3 and 6 more,  3/5 and 8 more,  5/6 and 10 more,  7/8 and 6 more.  He takes a common denominator of  360  and finds  2/3  of it is  240  and then adds  6  to get  246.  Similarly, he gets  224,  310,  321  and then divides in the proportion  246 : 224 : 310 : 321.  This is actually indeterminate as it depends on the choice of common denominator.  Vogel says the problem is unclear and the solution is false and notes that dividing  387  instead of  384  would give an integral solution.  He cites a number of other occurrences of this problem _ cf. Widman below.

204:  ½ + _ + ¼.

207:  Divide 100 into  (1/3 ‑ 1/4)  +  (1/4 ‑ 1/5)  +  (1/5 ‑ 1/6).

229:  Divide 20 into  1½, 2½, 1, 1, 1, 1.  Does as  3 : 5 : 2 : 2 : 2 : 2.

230:  Divide 20 into  1½ + _, 2½ + ¼, 1, 1, 1, 1.  Does as  22 : 33 : 12 : 12 : 12 : 12.

286 discusses problems where one removes fractions and deals with the remainder.  Notes that  1/2 + 1/3 + 1/6  leaves nothing, but  1/2 + 1/3 + 1/9  leaves something.

Chuquet.  1484.  Triparty, part 1.  English in FHM 75.  "I wish to divide 100 into three parts of such proportion as are  1/2,  2/3,  1/4  ..."

Ulrich Wagner.  Untitled text known as "Das Bamberger Rechenbuch".  Heinrich Petzensteiner, Babenberg (= Bamburg), 1483.  Reproduced, with transcription and notes by Eberhard Schröder as:  Das Bamberger Rechenbuch von 1483.  Akademie‑Verlag, Berlin, DDR, 1988. 

Pp. 69‑70 & 200.  = AR, no.  202.

Pp. 71-72 & 201-202.  = AR, no. 203.

Johann Widman.  Beh_de und hubsche Rechnung auff allen kauffmanschafft.  Conrad Kacheloffen, Leipzig, 1489.  ??NYS.  (Rara 36‑40.  This is extensively described by:  J. W. L. Glaisher in Messenger of Mathematics 51 (1921‑22) 1‑148, but he gives the title as:  Beh_de und hubsche Rechenung ....)  Smith and Glaisher give  Widman,  but Knobloch (7.L.2.c) uses  Widmann  and  Behende und hubsche Rechenung ....

F. 195v (Glaisher 14-15 & 122).  = AR, no. 230.

Ff. 195v-196 (Glaisher 15 & 122).  = AR, no. 207.

F. 196v (Glaisher 18-19, 38, 45, 122, 130).  = AR, no. 203.

                    Glaisher notes that Pacioli's Summa, (see below), gives a more natural determinate interpretation for similar problems.  In this example, this would first subtract  6 + 8 + 10 + 6  from  384,  leaving  354  which would be divided in the proportion  2/3 : 3/5 : 5/6 : 7/8.  He also notes the appearance of Widman's problem and solution in Huswirt (1501) ??NYS  and of problems similar to Widman and done in the same way, in Arithmetice Lilium (a book of c1510, ??NYS) (divide  100  into  1/2 + 3,  1/3 + 2,  1/5 + 4)  and Tonstall.  Rudolff's Kunstliche Rechnung of 1526, ??NYS, does (divide into 1/2 and 6,  1/3 and 4,  1/4 less 2) in Pacioli's manner.  Cf Apainus for a similer version.  Riese's Rechenung nach der Lenge (1550?, ??NYR) does (divide  124½  into  2/3 less 12,  1/4 and 10,  5/6 less 24,  3/8 and 6,  2/5 less 7) in Pacioli's way.

Pacioli.  Summa.  1494.  These give the more natural interpretation of this type of problem.

F. 150r, prob. 3.  Divide 100 as  1/2 plus 5;  1/3 less 4.  Subtracts 1 from 100 and divides the resulting 99 in the proportion  3 : 2. 

Ff. 150r-150v, prob. 4.  Divide 100 as  1/2 plus 3;  1/3 less 5.  Divides 102 as  3 : 2.

F. 150v, prob. 5.  Divide 100 as  1/2 less 4;  1/3 less 2.  Divides 106 as  3 : 2.

F. 150v, prob. 6.  Divide 30 as  1/2 plus 2;  1/3 plus 3.  Divides 25 as  3 : 2.

F. 150v, prob. 7.  Divide 10 as  1/2 less 3;  1/3 plus 4.  Divides 9 as  3 : 2.

F. 150v, prob. 8.  Divide 1046 as  1/2 less 2;  1/3 less 1;  1/4 plus 5.  No working or answer.

Calandri.  Aritmetica.  c1500.  Ff. 93v-94r, pp. 187‑188.  Same as Lucca 1754.

Blasius.  1513.  F. F.ii.r: Prime regula.  Man leaves  6000  to be divided  1/2 + 1/3 + 1/4 + 1/5.  There is an error in the calculation.

Tagliente.  Libro de Abaco.  (1515).  1541.

Prob. 94, part 2, ff. 48v-49r.  Divide  120  into  ½ + _.

Prob. 95, ff. 48v-49v.  Divide  12  into  ½ + _ + ¼.

Riese.  Rechnung.  1522. 

1544 ed. _ pp. 81‑82;  1574 ed. _ pp. 55r‑55v.  1/3 + 1/4 + 1/5.

1544 ed. _ pp. 98‑99;  1574 ed. _ p. 66r.  Three men take  1/2 + 1/3 + 1/4  of the profits, making  50  all told.  What was the profit?  Answer:  50 x 12/13.

Tonstall.  De Arte Supputandi.  1522. 

Quest. 17, pp. 147-149.  Divide into  1/2 + 1/3 + 1/4 + 1/5 + 1/6.  Uses  4350  as common denominator.

Quest. 18, pp. 149-150.  Same as Quest. 17, done with denominator  60.

Quest. 21, pp. 151-152.  Divide into parts:  1/3 + 1/4;  1/4 + 1/5;  1/5 + 1/6.

Quest. 22, pp. 153-154.  Divide  600  into:  2/3 plus 9;  3/5 plus 8;  5/6 plus 7;  7/8 plus 6.  See:  AR;  Widman;  Pacioli for discussion of this type of problem.  Takes common denominator of  120  and then divides as  89 : 80 : 107 : 111  which is not the way I read the problem.  I would divide  570  as  80 : 72 : 100 : 105,  as done by Pacioli.

Apianus.  Kauffmanss Rechnung.  1527. 

F. H.v.r.  Divide  1300  as  1/2 plus 8,  1/3 less 5,  1/4 less 12.  Does in Pacioli's manner, dividing  1309  in the proportion  12 : 8 : 6  and then amending by  +8, -5, -12.

F. H.v.r.  Divide  58  as  1/2 + 2/3 + 1/4.  Cf Chuquet.

F. H.vii.r.  Divide  40  as  1/4 + 1/5 + 1/6.

Cardan.  Practica Arithmetice.  1539.  Chap. 66, section 91, second part, f. HH.i.v (p. 166).  First has  _  plus  7;  second has  ¼  plus  13;  third has  ½  minus  28.  How much was there?

Tartaglia.  General Trattato.  1556.  Book 12, art. 42-44, pp. 200r-201r.

Art. 42.  ½ + _ + ¼.  Long discussion of an error of Luca Pacioli and others who assert that such problems are impossible or illegal.  Tartaglia says simply to divide in the proportion  6 : 4 : 3  and can't understand why others are making such a fuss.

Art. 43.  1/3 + 1/4 + 1/5.

Art. 44.  4/5 + 3/4 + 2/3.

(Sanford 218‑219 says Tartaglia was among the first to suggest the 18th camel _ but I see nothing of this here.  H&S 87 says that this is really a modern problem in that previously property was divided in proportion to fractions, regardless of whether they summed to unity.  Tartaglia's discussion of Pacioli and others makes it clear that people were starting to object to this at this time, but examples continue and I don't see the modern version occurring until late 19C.)

Buteo.  Logistica.  1559. 

Prob. 5, pp. 203-204.  Divide 77 into  ½ + _ + ¼.

Prob. 74, pp. 283-284.  Divide 30 into  ½ + _.  Discusses the solution.

Prob. 75, pp. 284-285.  Divide 15 as  1/2 plus 2;  1/4 + 3.  He divides 10 into  ½ + _,  as in Pacioli.  See:  AR;  Widman;  Pacioli  for discussion of this type of problem.

Prob. 76, pp. 285-286.  Divide 24 as  1/3 less 7;  1/4 less 4.

Prob. 77, p. 285.  Divide 12 as  2/3 less 3;  1/6 plus 4.

Prob. 22, pp. 350-351.  Divide 60 as  1/4;  1/3 plus 4;  3/4 less 8.

Prob. 26, pp. 353-354.  Divide 30 as  1/2 plus 2;  1/3 less 3.

Prob. 28, p. 355.  Divide 224 as  1;  6/5 plus 4.

Izaak Walton.  The Compleat Angler.  (R. Marriott, London, 1653);  Everyman edition, Dent, London, 1906, et seq.  Chap. V _ The Fourth Day, pp. 101‑102.  The World's Classics, OUP, 1935, Chap. V, pp. 114-116.  Divide 20 into  1/3 + 1/4 + 1/5 + 1/6.  Leaves one left over.

Leybourn.  Pleasure with Profit.  1694.  Prob. 11, pp. 37-38.  £6000 divided  1/4 + 1/5 + 1/6.  He divides in the proportion  15 : 12 : 10.  Cf Apianus.

Ozanam.  1725.  Prob. 24, question 5, 1725: 179.  Divide  26000  into  ½ + _ + ¼.  Takes in proportion  12 : 8 : 6.

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790), prob. XIV, pp. 85-86 (1790: prob. XXIV, pp. 86-87).  Divide 20 into  1/3 + 1/4 + 1/5 + 1/6,  done by proportion.

Les Amusemens.  1749.  Prob. 52, p. 184.  Divide 78 into  ½ + _ + ¼,  by using proportions.

Walkingame.  Tutor's Assistant.  1751.  1777: p. 177, prob. 119;  1835: p. 180, prob. 58;  not in 1860.  Divide £500 into  1/3 + 1/4 + 1/5 + 1/6 + 1/7.  Gives exact answer as integer plus a fraction.  1835 reduces the fractions to lowest terms.

Vyse.  Tutor's Guide.  1771?  Prob. 6, p. 113 & key p. 151.  Same as Walkingame.  Solution gives answers rounded to farthings and never gives the exact fractions.

Dodson.  Math. Repository.  1775.  P. 32, quest. LXXVIII.  Divide 20s in proportion: 1/3, 1/4, 1/5, 1/6.

Pike.  Arithmetic.  1788. 

P. 335, no. 4.  "Being a little dipped, they agreed that A should pay  2/3,  B  1/2,  C  1/3, and  D  1/4."

P. 355, no. 39.  A, B, C do a job.  A and B do  3/11  of it,  A and C do  5/13,  B and C do  4/14.  (Also entered at 7.H.)

Samuel Bullen.  A New Compendium of Arithmetic ...  Printed for the author, London, 1789.  Chap. 38, prob. 3, p. 239.  Divide into  ½ + _ + ¼,  phrased as  2A = 3B = 4C.

John King, ed.  John King  1795  Arithmetical Book.  Published by the editor, who is the great-great-grandson of the 1795 writer, Twickenham, 1995.  The 1795 has dividing  100  into  1/3 + 1/4 + 1/5 + 1/6.  Gives usual solution.

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions, no. 16, pp. 18 & 75.  Division in the proportion of  1/3, 1/4, and 1/5,  but last person dies.  Solution indicates this as standard practice.

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Arithmetical puzzles, no. 4, p. 173 (1868: 184).  Pay  20s  with only  19s  by dividing into  1/2 + 1/3 + 1/6 + 1/19.  "This, however, is only a payment upon paper."

Hanky Panky.  1872.  A Chinese puzzle, pp. 73-74.  17  elephants to be divided  1/2 + 1/3 + 1/9.

Cassell's.  1881.  P. 102: The clever lawyer.  = Manson, 1911, p. 255.  17  horses divided  1/2 + 1/3 + 1/9.  = Rohrbough; Puzzle Craft; 1932, p. 7.

Richard A. Proctor.  Some puzzles.  Knowledge 9 (Aug 1886) 305-306.  "... the familiar puzzle [of] the farmer, ignorant of numbers, who left  17  horses to his three sons (or, equally well it may be, an Arab sheik who left  17  camels)".  Points out that if there were  35  camels, then the Cadi could also be left a camel.

E. W. Cole.  Cole's Fun Doctor.  The Funniest Book in the World.  Routledge, London  &  E. W. Cole, Melbourne,  nd [HPL gives 1886 and lists the author as  Arthur C. Cole].  P. 224: A Chinese puzzle.  17 elephants left by a Chinaman to be divided  1/2 + 1/3 + 1/9.  Says it is in the Galaxy for August, which might have been an Australian publication by Proctor, who had connections there.

Lemon.  1890.  The legacy, no. 652, pp. 81 & 121.  19  camels divided  1/2 + 1/4 + 1/5.

Don Lemon.  Everybody's Pocket Cyclopedia.  Revised 8th ed., 1890.  Op. cit. in 5.A.  P. 135, no. 2.  17  horses:  1/2 + 1/3 + 1/9.  Dervish loans them his horse.

Hoffmann.  1893.  Chap. IV, no. 11: An unmanageable legacy, pp. 147 & 191‑192.  1/2 + 1/3 + 1/9.  Answer says "this expedient is frequently employed" in "the Mahomedan Law of Inheritance".

Brandreth Puzzle Book.  Brandreth's Pills (The Porous Plaster Co., NY), nd [1895].  P. 3: An unmanageable legacy, with nice colour picture.  Identical to Hoffmann.  No solution.

Loyd.  Problem 37: A queer legacy.  Tit‑Bits 32 (5 Jun  &  3 Jul 1897) 173  &  258.  = Cyclopedia, 1914, The herd of camels, pp. 57 & 346.  17 horses in proportion  1/2 : 1/3 : 1/9.  Says the use of proportion makes the solution actually correct.

Clark.  Mental Nuts.  1904, no. 5;  1916, no. 18.  The heirs and the sheep.  1/2 + 1/4 + 1/5  of 19 sheep.

Benson.  1904.  The lawyer's puzzle, p. 225.  1/2 + 1/3 + 1/9.  There originally were 18 horses, but one died.

William F. White.  Op. cit. in 5.E.  1908.  Puzzle of the camels, p. 193.  17 camels divided  1/2 + 1/3 + 1/9.

Ball-FitzPatrick.  2nd ed., 1908‑1909.  Part 1, p. 111, footnote says the problem is Arabic.  The material is not in the 1st ed., nor in Ball, 5th ed.  A&N, pp. 84‑85, cites this but says it has been in German oral tradition for a long time.  He gives it with 17 horses.

E. Ernst.  Mathematische Unterhaltungen und Spielereien.  Vol. 2, Otto Maier, Ravensburg, 1912.  P. 15: Das geschente Weinfass.  Divide  19  in  1/2 + 1/4 + 1/5.

Dudeney.  MP.  1926.  Prob. 89: The seventeen horses, pp. 33-34 & 123-124.  = 536, prob. 172, pp. 54‑55 & 266‑267.  Discusses interpretation of proportion, as in Loyd, in detail.

King.  Best 100.  1927.  No. 21, pp. 14 & 43.  19 horses into  1/2 + 1/4 + 1/5.

Collins.  Book of Puzzles.  1927. 

The lady bookmaker's problem, pp. 72-73.  Because  1/2 + 1/3 + 1/4 = 13/12,  one can offer odds in a three horse race of  even money,  2 to 1  and  3 to 1.

The sheik and his camels, pp. 77-78.  Usual form.  Cadi loans them his camel.

M. Kraitchik.  La Mathématique des Jeux, 1930, op. cit. in 4.A.2, chap. 1, prob. 47, p. 15.  17 sheep into  1/2 + 1/3 + 1/9.  Says it is of Hindu origin.

Anon.  Foulsham's New Fun Book.  W. Foulsham, London, nd [1930s?].  Pp. 85‑86: The farmer's horses.  Identical to King, 1927.

McKay.  Party Night.  1940.  No. 35, p. 184.  "It is said that an Arab had 17 cattle." 

Sullivan.  Unusual.  1943.  Prob. 12: Will trouble.  17 horses into  1/2 + 1/3 + 1/9.

Jerome S. Meyer.  Fun for the Family.  (Greenberg Publishers, 1937);  Permabooks, NY, 1959.  No. 30: Think cow it is done, pp. 42-43 & 241.  Herd to be divided  1/3 + 1/4 + 1/6 + 1/8 + 1/9.  Neighbour loans two cows and everything divides up properly with two cows left over for the neighbour.  How many cows were there?

Leeming.  1946.  Chap. 5, prob. 24: The herd of cattle, pp. 61 & 179‑180.  Same as Meyer.

Philip E. Bath.  Fun with Figures.  Op. cit. in 5.C.  1959.  No. 66: The vicar's garden, pp. 25 & 52.  7s divided  1/2 + 1/4 + 1/8  by adding an extra shilling.  Solution doesn't seem to understand this and claims there really should be  7/8 s  left over.

D. B. Eperson.  Puzzles, pastimes and problems.  MiS 3:6 (Nov 1974) 12‑13 & 26‑27.  Prob. 6: The Shah's Rolls‑Royces.  Divide 23 Rolls‑Royces into  2/3 + 1/6 + 1/8.  The answer erroneously asserts this works for  n º ‑1 (mod 24).

David Singmaster.  A Middle Eastern muddle.  41  oil wells to be divided into  1/2 + 1/3 + 1/7.  But then I ask if there are values other than  2, 3, 7, 41  which produce such a puzzle problem.  There are  12  such quadruples.  I recall seeing this when I was a student but I haven't relocated it.  Appeared in my puzzle columns as follows.

                    "Well, well, well."  Brain Twister.  Weekend Telegraph (27 Feb 1988) xv (misprinted),  (5 Mar 1988) (corrected) xv  &  (12 Mar 1988) xv.

                    Reprinted, with no title, in:  The Daily Telegraph Braintwisters No. 1;  Pan Books, London, 1993;  with Barry R. Clarke, Rex Gooch and Angela Newing.  Prob. 25, pp. 27, 76 & 117.

                    "A Middle Eastern muddle."  The Puzzle Box.  Games & Puzzles 12 (Mar 1995) 18-19  &  13 (Apr 1995) 40.

John P. Ashley.  Arithmetickle.  Arithmetic Curiosities, Challenges, Games and Groaners for all Ages.  Keystone Agencies, Radnor, Ohio, 1997.  P. 8: Omar divides 17 horses among 3 sons.  The Answer says: "It was a great solution, but it was not correct mathematics.  The sum of the fractional parts: 1/2, 1/3 and 1/9 do not add up to 1 but to 17/18.  Therefore, each of the heirs got a bit more than the will intended."

 

          7.G.2. POSTHUMOUS TWINS, ETC.

 

          A man dies, leaving a pregnant wife and a will explaining how his estate is to be divided between the wife and a son or a daughter.  The wife produces twins, one boy and one girl.  How is the estate to be divided?

          The most common version has  son : wife  =  wife : daughter  =  2 : 1  given in the will and derives  son : wife : daughter = 4 : 2 : 1.  I will denote this as the "usual form".  Other proportions cited by Smith are:  4 : 3 : 2;   2 : 2 : 1;   9 : 6 : 4. 

          Alcuin  &  BR  proceed by dividing the estate in half first.

          See Brooks for some odd versions.

          See Tropfke 655.

 

Moritz Cantor.  Vorlesungen über Geschichte der Mathematik.  Vol. 1, 3rd ed., Teubner, Leipzig, 1907.  Pp. 561-563 sketches the history of this problem.  This is the basis of Smith's discussion below.  This asserts that the problem is based on Roman Lex Falcidia of  ‑40  which required that at least  ¼  of an estate should go to the legal heir.  He says it first appears in the works of Celsus and quotes Julianus.  He also cites Caecilius Africanus (c100) and Julius Paulus (3C).  Describes the common case and says that if the will was invalidated, then only the children would inherit.

D. E. Smith.  Op. cit. in 3.  Based on Cantor, cites Lex Falcidia, Celsus, Julianus, Africanus.  He cites 22 medieval references, including Vander Schuere and Recorde of those below.

See also:  Sanford 218‑219;  H&S 87‑88;  A&N 24‑26.

Juventius Celsus.  De istituzione uxoris et postumi et postumae.  c75.  ??NYS _ cited by Smith.  Julianus cites him.

Salvianus Julianus.  c140.  Lex 13 principio.  Digestorum lib. 28, title 2.  ??NYS _ quoted by Cantor and cited by Buteo & Smith.  Cantor says this reports a case, though the quoted text isn't very specific.  Usual form.  Julianus cites Celsus.

Caecilius Africanus.  c150.  Lex 47 §1.  Digestorum lib 28, title 5.  ??NYS _ cited by Cantor & Smith.  Cantor says this refers to a case.

Julius Paulus.  3C.  Lex 81 principio.  Digestorum lib. 28, title 5.  ??NYS _ cited by Cantor & Smith.  Cantor says this refers to a case.

Alcuin.  9C.  Prob. 35: Propositio de obitu cujusdam patrisfamilias.  Problem of posthumous twins.  Ratios are  3 : 1  for  son : mother  and  5 : 7  for  mother : daughter.  He takes half the estate and shares it  3 : 1  and then the other half is shared   5 : 7.  This gives  9 : 8 : 7.  Ahrens, A&N, p. 26, suggests  15 : 5 : 7,  which is the result of the usual Roman process.

BR.  c1305.  No. 91, pp. 110‑111.  Son : wife  =  wife : daughter  =  3 : 2.  Divides in halves, as in Alcuin, and divides each half as  3 : 2,  giving  son : wife : daughter = 3 : 5 : 2.

Gherardi.  Libro di ragioni.  1327.  P. 37.  Son : wife  =  3 : 1;  wife : daughter  =  2 : 1.  Divides as  6 : 2 : 1.

Gherardi.  Liber habaci.  1327?  P. 145.  Usual form.

Dell'Abbaco.  c1370.  Prob. 100, p. 85 with plate on p. 86.  Posthumous twins.  Usual form.  I have a colour slide of this.

Lucca 1754.  c1390. 

F. 60r, pp. 136‑137.  Posthumous triplets, 2 boys and a girl with usual ratios.  He divides in proportion  4 : 4 : 2 : 1  for  boy : boy : mother: girl.

F. 83r, pp. 200‑201.  Posthumous twins.  Usual form.

AR.  c1450.  Prob. 209, pp. 97, 176, 223.  Man has son, wife and two daughters and gives the usual ratios, hence divides in the proportion  4 : 2 : 1 : 1.

Muscarello.  1478.  Ff. 75r-76r, pp. 189-191.  Posthumous twins.  Usual form.

Wagner.  Op. cit. in 7.G.1.  1483.  Pp. 73‑75 & 202‑203.  Usual will, but wife produces a son and two daughters.  Divides as in AR.

Chuquet.  1484.  Prob. 205.  English & discussion in FHM 205.  Usual form.  FHM say it "goes back to the Roman emperor and legislator Justinian" and quote Recorde.

HB.XI.22.  1488.  P. 44 (Rath 247).  Posthumous twins.

Pacioli.  Summa.  1494. 

F. 158r, prob. 80.  Usual posthumous twins.  Then seems to assert that Nofrio Dini of Florence, a respectable merchant in Pisa, according to Giuliano Salviati, made such a will on 16 Dec 1486.  After a bequest to the church, there was an estate of 800 to be divided.  If a son was born, the mother was to get 400;  if a daughter, the mother was to get 300.  Twins were born and he says to divided as  3 : 3 : 5.  Says one can deal similarly with similar problems.  Edward A. Fennell; Figures in Proportion: Art, Science and the Business Renaissance; The contribution of Luca Pacioli to culture and commerce in the High Renaissance; Booklet for the exhibition organized by The Institute of Chartered Accountants in England and Wales, Feb 1994; p. 11 says the story was related by Onofrio Dini, who Pacioli met in Pisa.  ??clarify

F. 158v, prob. 82.  Selling a pregnant cow which bears twins.  Gives some rules which determine the relative values.

Blasius.  1513.  F. F.ii.v: Quarta regula.  Dying man with pregnant wife.  If she has a son, he gets  3/5  and the mother and the church get  1/5  each.  If she has a daughter, the daughter and the mother get  2/5  each and the church gets  1/5.  She has a son and a daughter.  He divides in proportion  3 : 2 : 2 : 1,  but gives no reason.  Offhand, I would think that  1/5  should go to the church _ since this is specified in either case _ and then the remaining  4/5  should be divided in the proportion  3 : 1 : 1,  giving overall proportions of  12 : 4 : 4 : 5.  He says one can similarly deal with two sons or two daughters or two sons and one daughter. 

Tagliente.  Libro de Abaco.  (1515).  1541.  Prob. 100, ff. 50v-51r.  Usual form.

Ghaligai.  Practica D'Arithmetica.  1521.  Prob. 23, f. 65v.  Usual posthumous twins.

Riese.  Rechnung.  1522.  1544 ed. _ p. 80;  1574 ed. _ p. 54v.  Father leaves a widow, a son and two daughters.  Divides as in AR.

Tonstall.  De Arte Supputandi.  1522.  Quest. 16, pp. 146-147.  Usual form.  Then considers 3 sons and 2 daughters!

Cardan.  Practica Arithmetice.  1539.  Chap. 66, section 87, ff. GG.vi.v - GG.vii.r (p. 164).  Posthumous twins.  Son : wife  =  4 : 1;  wife : daughter  =  2 : 1;  divides as  8 : 2 : 1

Robert Recorde.  Ground of Artes.  R. Wolfe, London, 1542;  facsimile by Da Capo Press, NY  &  Theatrum Orbis Terrarum, Amsterdam, 1969.  Material is not in the 1542 ed.  1558 ed., ??NYS, fol. X 8 is quoted by Smith, op. cit. in 3.A, p. 69.  "If some cunning lawyers had this matter in scanning, they would determine this testament to be quite voyde, and so the man to die intestate, because the testament was made unsufficient."

Giovanni Sfortunati.  Nuovo lume.  Venice, 1545.  F. 58v.  ??NYS _ described by Franci, op. cit. in 3.A, p. 38.  Franci's discussion is about several extended versions, but it seems to indicate that Sfortunati deals with a hermaphrodite.

Tartaglia.  General Trattato.  1556.  Book 12, art. 35-41, pp. 199v-200v.

Art. 35-40 give various ratios  S : M  =  son : mother  and  M : D  =  mother : daughter, then divides in the usual way to get the proportions  s : m : d  such that  s/m = S/M  and  m/d = M/D.  E.g. when  S : M = 2 : 1 = M : D,  then  s : m : d = 4 : 2 : 1.

Art. 35.  S : M = 2 : 1;   M : D = 2 : 1.

Art. 36.  S : M = 5 : 3;   M : D = 1 : 1.

Art. 37.  S : M = 2 : 1;   M : D = 1 : 1.

Art. 38.  S : M = 2 : 1;   M : D = 2 : 1,  in a different context than Art. 35.

Art. 39.  S : M = 2 : 1;   M : D = 3 : 1.

Art. 40.  S : M = 3 : 1;   M : D = 2 : 1.

Art. 41.  S : M = 2 : 1;   M : D = 2 : 1,  but quadruplets are produced _ two sons and two daughters.  He divides in proportion  4 : 4 : 2 : 1 : 1.

Buteo.  Logistica.  1559. 

Prob. 60, pp. 264-266.  Usual form.  Cites Julianus.

Prob. 12, pp. 341-342.  Selling a pregnant cow, where the value depends on the sex of the calf.  Cow + daughter is worth  40,  while  cow + son is worth 45.  This is insufficient to determine the relative values, but he then adds excessive information:  C = 3D = 2S.  The cow produces twins _ one son and one daughter.

Gori.  Libro di arimetricha.  1571.

Ff. 75r‑75v (pp. 83-84).  Usual form.

F. 75v (p. 84).  Posthumous quintuplets _ divides in same proportions, though there is some confusion in the text of the solution.

Jacob Vander Schuere.  Arithmetica, oft Reken‑const.  G. Kooman, Haarlem, 1600.  ??NYS.  [Smith, Rara, 421‑423.]  F. 98 is quoted in Smith, op. cit. in 3, p. 69, note 7.  Posthumous triplets:  boy, girl and hermaphrodite.  Divides in proportion  12 : 4 : 2 : 7 = son : wife : daughter : hermaphrodite.  Smith doesn't give the original ratios, but they were probably  son : wife = 3 : 1,  wife : daughter = 2 : 1.

Leybourn.  Pleasure with Profit.  1694.  Prob. 15, pp. 39-40.  Posthumous triplets: boy, boy, girl.  Usual ratios.  Divides  4 : 4 : 2 : 1.

Ozanam.  1725.  Prob. 24, 1725: 179.  Prob. 4, 1778: 187-188;  1803: 185-185;  1814: 160‑161;  1840: 83.  1725 gives just posthumous triplets _ two girls and a boy.  He divides  4 : 2 : 1 : 1.  Montucla does usual form, then remarks that one could have posthumous triplets, e.g. two sons and a daughter, and that he thinks that the will would be declared legally void.

Les Amusemens.  1749.  Prob. 54, p. 186.  Usual case.

Vyse.  Tutor's Guide.  1771?  Prob. 19, p. 167 & Key p. 209.  Usual case.

Dodson.  Math. Repository.  1775.  P. 13, Quest. XXXIV.  Usual case.

Charles Hutton.  A Complete Treatise on Practical Arithmetic and Book-keeping, ....  New edition, corrected and enlarged by Alexander Ingram.  [c1780?]  G. & J. Ross, Edinburgh, 1804.  [BMC earliest entry is 7th ed., 1785, then 14th ed., 1815.]  Prob. 53, p. 137.  Usual form, but states that the mother thereby loses  2400£  compared to the case of just having a girl.  What would she have got if she had only had a son?  Answer is  2100£  which assumes the usual  4 : 2 : 1  divison for the case of twins.

Edward Brooks.  The Normal Mental Arithmetic  A Thorough and Complete Course by Analysis and Induction.  Sower, Potts & Co., Philadelphia, (1858), revised, 1863.  Further revised as:  The New Normal Mental Arithmetic: A Thorough and Complete Course by Analysis and Induction;  Sower, Potts & Co., Philadelphia, 1873.  Several examples in an unusual context.

1863 _ p. 128, no. 15;  1873 _ pp. 167-168, no. 5.  Man left  $26,000  to wife, son and daughter.  If the daughter dies before coming of age, the widow gets  ¼;  if the son dies before coming of age, the widow gets  ¾;  what happens if all live?  Unusually for this book, this problem has a remark which says the divison should be in the proportion  son : wife : daughter  =  9 : 3 : 1.

1863 _ p. 128, no. 18;  1873 _ p. 168, no. 9.  Man with children abroad and wife at home.  If the son does not return, the widow gets  _;  if the daughter does not return, the widow gets  _;  both return and it is found that the son gets  $3000  more than the duaghter.  What was the estate?

1863 _ p. 128, no. 19;  1873 _ p. 168, no. 10.  A, B, C  are thinking of buying a farm.  They agree that if  A and B  buy it, then  A  pays  2/5  and if  B and C  buy it, then  B  pays  2/5.  All three buy it together and  C  is found to pay  $500  more than  A.  What was the cost?

Susan Cunnington.  The Story of Arithmetic.  Swan Sonnenschein,, London, 1904.  Prob. 11, p. 212.  Usual form.  Asserts it is a Roman problem of +300, but gives no references.

Collins.  Fun with Figures.  1928.  Then he put in his other foot, pp. 236-237.  Usual form.  He adds:  A further complication _ triplets, two boys and a girl.  "The easiest way to find out is to let the lawyers decide it, and it is the one best bet that they will get it all."

The Little Puzzle Book.  Op. cit. in 5.D.5.  1955.  Pp. 53-54: The judge's dilemma.  Ratios are  2 : 1  for son : mother  and  3 : 1  for mother : daughter.  Divides as  6 : 3 : 1.

 

          7.H.   DIVISION AND SHARING PROBLEMS  _  CISTERN PROBLEMS

 

          See Tropfke 578.

          The earliest sources in this group include what I call 'assembly problems'.  In these, there are several processes which constitute a unit of work.  The rates for the processes are given and one has to determine the number of units which can be done in a day (or how long some number of units will take).  See the Babylonian examples below and:  Chiu Chang Suan Ching;  Heron;  Metrodorus 134, 136;  Bakhshali MS;  BR 97, 98;  AR 57, 75;  Muscarello;  Borghi;  Riese;  Cardan;  Tartaglia;  Pike;  Treatise, 1850;  Chambers;  Bullen;  Pearson.  I am indebted to Eleanor Robson for the Old Babylonian examples.  She has also provided most of the references to the source material which I not yet seen.  The dating of these examples is generally pretty vague.

 

          Note.  Cistern problems with two pipes have the same form as meeting problems, cf. 10.A.

 

          NOTATION:  (a, b, c, ...)  means that different pipes, etc. can do the job in  a, b, c, ....  How long for all together?

          Negative values indicate outlets.

 

     a      b    ...

  30    20                                    Lucca 1754;  Wingate/Kersey

  20    16                                    Tartaglia

  20    15                                    P. M. Calandri

  18    12                                    Calandri, c1500

  16    10                                    De Morgan, 1831

  15    12                                    Calandri, 1491;  Tagliente;  De Morgan, 1836; 

  14    12                                    Hutton-Rutherford

  13    10                                    Vyse;  Bonnycastle;  Colenso

  12      7                                    Pike

  10      8                                    Dell'Abbaco

  10      7                                    Muscarello

  10      6                                    BR 99

    9   -12                                    Les Amusemens

    8      6                                    Calandri, c1500;  Hutton

    7      5                                    De Morgan, 1836;  Sonnenschein & Nesbit

    6      4                                    P. M. Calandri;  Calandri, 1491;  Tagliente;  Gori

    5      4                                    Dell'Abbaco

    4      3                                    Lucca 1754;  Burnaby

    4      1                                    Heron;  BR 98

    4   -11                                    Calandri, 1491;  Tonstall

    3      2                                    Gherardi

    3     -9                                    BR 70

    2     -3                                    Buteo

 

  60    30    20                            Meichsner

  55    45   -30                            Pike

  50    40   -25                            Hutton, c1780?;  Eadon;  Colenso

  24    12      8                            Buteo

  20    15    12                            Unger 520

  20    15    10                            Calandri, c1500

  18    12      6                            Calandri, c1500

  16    12    10                            Simpson

  16    12      8                            Calandri, c1500

  15    12    10                            P. M. Calandri;  Calandri, 1491 (twice)

  12    10      9                            Sonnenschein & Nesbit

  12    10      6                            Calandri, c1500

  12      9      6                            Borghi;  Ozanam

  12      8      4                            Metrodorus 131

  12      8   -10                            Unger 521

  10      9      8                            Milne

  10      8      4                            Pacioli

  10      5      4                            Pacioli;  Tartaglia

    9      7     -2                            Pike

    8      6      4                            Calandri, 1491;  Tonstall

    8      6      3                            Pacioli

    7      5      6                            King

    7      5      4                            AR 70

    7      5      3                            Gherardi

    6      5      4                            Fibonacci

    6      4      3                            Chuquet

    6      4      2                            Metrodorus 135;  Ozanam-Montucla

    6      4     -4                            Sonnenschein & Nesbit

    6      3      1                            Faulhaber

    5      4      3                            Lucca 1754;  Gori;  Les Amusemens

    5      3      2                            Calandri, c1500

    4      3      2                            Gherardi;  AR 98;  Wagner;  Faulhaber

    4      2      1                            Gori

    3  8/3 12/5                            Newton;  Dodson;  Eadon;  Colenso

    3      2      1                            Metrodorus 133;  Anania(?);  al-Karkhi;  BR 64;  AR 51, 97; 

          Calandri, 1491;  Blasius;  Tonstall;  Riese;  Vyse;  King

    3      1  2/5                            Metrodorus 132

5/3  1/2  1/3                            Chaturveda

    1  3/4  1/2                            Wingate/Kersey

    1  1/2  1/4                            AR 281;  Tonstall

1/2  1/3  1/4                            Columbia Alg.;  Pike

1/2 -10/7 -7/3                            Wingate/Kersey

 

  27    24      9      6                    Fibonacci

    6      5      4      3                    Bartoli

    6      5      3      2                    Muscarello

    6      4      3      2                    Leybourn

    4      3      2      1                    Metrodorus 130;  Fibonacci;  Tonstall

    4      3      2  1/2                    Metrodorus 7;  BR 65;  van Etten;  Wingate/Kersey; 

    4      3      2  1/4                    Ozanam

    1  1/2  1/3  1/6                    Bhaskara II

    1  1/2  1/4  1/5                    Chaturveda

1/2  1/3  1/4  1/5                    Mahavira

1/2  1/4  1/5  1/6                    Sridhara

     

    5      3  5/2      1  1/3            Chiu Chang Suan Ching

    4      3      2     -4     -6            BR 116

1/2  1/3  1/4  1/5  1/6            Columbia Alg.

1/2  1/3  1/5  1/7  1/9            BR 25

     

    6      5      4      3      2      1    Bartoli

    3      2      1 -3/4     -4     -5      Cardan

     

  12    10      8      6     -3     -4    -5       -6     Bullen;  Treatise, 1850

 

          General solution _ see:  Newton;  Simpson;  Dodson; Bonnycastle;  Hutton;  De Morgan;  Bourdon;  Young;  Milne.

 

          The earliest forms derive joint rates from individual rates.  Deriving individual rates from joint rates seems to begin in the 14C.

          NOTATION:  (A, x) in B  means the first can do it in  A  and the first and second together can do it in  B.  How long would it take the second?  For such problems, see:  BR;  Gherardi;  dell'Abbaco;  AR;  Treviso;  Chuquet;  Calandri, 1491;  Tonstall;  Gemma Frisius;  Tartaglia;  Simpson;  Euler;  Ozanam‑Montucla;  Bonnycastle;  Bullen;  Eadon;  Hutton;  Bonnycastle, 1815;  Jackson;  Nuts to Crack;  Family Friend;  Colenso;  Thomson.

 

(50, x)                in  36                Gherardi

(48, x)                in  24                Docharty

(36, x)                in  30                Gherardi

(36, x)                in  24                Docharty

(30, x)                in  12                Dodson;  Bonnycastle;  Hutton;  Nuts to Crack

(20, x)                in  14                Gemma Frisius

(20, x)                in  12                Wingate/Kersey;  Euler;  Dodson;  Pike;  Bonnycastle, 1815

(20, x)                in    8               Treviso Arith.

(18, x)                in  11                Vyse

(35/2, x)            in  40                Docharty (gives a negative x!)

(16, x)                in  10                Treatise, 1850

(15, x)                in  18                Thomson (gives a negative x!)

(15, x)                in  10                Treatise, 1850

(13, x)                in    9               AR 76

(13, x)                in    8               Pike

(12, x)                in    3               Family Friend, 1849

(10, x)                in    7               Colenso

( 9, x) in    5       Dell'Abbaco

( 8, x) in    5       Buteo;  Eadon

( 5, x) in  15/8    BR 67

( 3, x) in   4/3     BR 66

( 3, -x)                in   9/2             BR 69 (negative value!)

(-9, x) in   9/2     BR 68

 

(80, 60, x)          in  30                Tartaglia

(44, 32, x)          in  16                Eadon

(40, 30, x)          in  15                Calandri, 1491;  Tonstall;  Wingate/Kersey

(37, 23, x)          in  15                Pike

(34, 24, x)          in  12                Vyse

(17/2, 21/4, x)   in   6/5             Treatise, 1850

(8,   6, x)            in   3                Brooks

(5/2, 9/4, x)      in    1               Treatise, 1850

 

          For the general solution of:  (x, y) in A,  (y, z) in B,  (x, z) in C,  see:  Simpson;  Euler;  Ozanam-Montucla;  Bonnycastle;  Hutton;  De Morgan, 1836;  Colenso.  For examples of this form, see also:  Muscarello;  Docharty.  This is a form of the type III problem in Section 7.R.1, where the inverses of the variables are used.

 

    A      B      C

  60      4   -40                            Colenso

  30    20    15                            AR 182

  20    15    12                            Docharty

  14    12 21/2                            Colenso

  10      9      8                            Simpson;  Euler;  Dodson;  Ozanam-Montucla;  Bonnycastle; 

                                                     Hutton

    9      8      6                            Sonnenschein & Nesbit

    5      4      3                            Muscarello

 

          Vyse, Docharty and Thomson are the only examples I have seen with four people and you know how long it takes each set of three.

 

          For problems where the combinations involve one tap or worker working only part of the time that the other does, see:  Fibonacci;  Gherardi;  Chuquet;  Cardan;  Buteo;  Pike;  Jackson;  Treatise, 1850;  Colenso;  Young;  Chambers;  Brooks;  André;  Sonnenschein & Nesbit.

          For problems like  (x, x/2, x/3) in 2,  see:  di Bartolo;  Buteo.

          For problems like  (x, x‑5) in 12,  which lead to quadratic equations, see:  Di Bartolo;  Buteo;  Tate;  Briggs & Bryan.

          Sonnenschein & Nesbit has a version where pumps can work at half or full power.

          I have included a few direct rate problems as comparisons _ these usually involve money _ see:  Bakhshali;  Chaturveda;  Pike;  Chambers.

          See Clairaut for the use of this context to discuss negative solutions.

 

          See Smith, op. cit. in 3.  See also 7.E  &  H&S 69‑71.

          5.W.1 can be viewed as parodies of this problem.

 

          COMPARISON of assembly and cistern problems.  Consider the cistern-type problem  (a₁, a₂ , ...).  In the unit of time, the pipes do  1/a₁, 1/a₂, ...  of the work, so all together they do  S = Σ 1/a_i  per unit time and so the whole job takes time  1/S. 

          In an assembly-type problem, we can do  a_i  units of process  i  per unit of time.  Hence it takes  1/a_i  time to do one unit of process  i.  If each process has to be done the same number of times, then it takes  S = Σ 1/a_i  time to do a unit of work and so  1/S  units can be done in a unit of time.  In the Babylonian problems, the unit of work may require varying amounts of the different units.  If the unit of work requires  b_i  units of process  i,  then we take  S = Σ b_i/a_i.

          Hence the problems are mathematically the same, though the formulations are different.

 

YBC 7164.  Old Babylonian problem tablet at Yale, problems 6 & 7, c‑1700?  Transcribed, translated and commented on in Neugebauer & Sachs, op. cit. in 7.E, 1945, pp. 81-88 & plate 10 & photo plate 35.  On pp. 148-149, a linguistic analysis says it probably comes from Larsa, in southern Mesopotamia.

Problem 7.  A canal has to be cleaned to 3 kùš deep.  A man can clear 20 gín of silt from the top kùš in a day or he can clear 10 gín from the lower level in a day.  How much can he clear in a day?  Here  a₁ = 20,  a₂ = 10,  and we can take  b₁ = 1,  b₂ = 2,  because the lower level is twice as thick as the upper level.

Problem 6.  This is the same, but with depth  4½  kùš divided into three levels with the rate of doing the bottom level from 3 to  4½  deep being only  7½  gín per day.  So we just add  a₃ = 7½  and  b₃ = 1½  to the previous problem.

BM 85196.  Late Old Babylonian tablet in the British Museum, prob. 16, c‑1700?.  Transcribed, translated and commented on by O. Neugebauer; Mathematische Keilschrift-texte II; Springer, Berlin, 1935, pp. 45+, 49, 56+ _ ??NX.  [See 6.BF.2 for another problem from this tablet.]  But Neugebauer was not able to make sense of it until he saw the above problems, so it is reconsidered in Neugebauer & Sachs, pp. 88‑90.  Robson says it is definitely from Sippar (middle Mesopotamia) and cites Thureau-Dangin; Revue d'Assyriologie 32 (1935) 1+ for another publication of the text, ??NYS.  This problem and those of YBC 7164 are more recently discussed by Marvin A. Powell; Evidence for agriculture and waterworks in Babylonian mathematical texts; Bulletin on Sumerian Agriculture 4 (1988) 161-172, ??NYS

                    Like problem 6 above, with each level of depth 1 kùš and rates of  20, 10,  6_  gín per day.

In Spring 1994, I mentioned the assembly problems from the Chiu Chang Suan Ching (see below) in a lecture at Oxford.  Eleanor Robson told me that such problems occur in Old Babylonian times and she sent me details, including the above references, and later provided more details and references.  She described four further examples, without specific dates, and the next four examples are simplified from her letter.  The simplifications are basically to avoid use of coefficients giving the number of bricks per unit of weight, etc.

Haddad 104, c-1770.  Tablet from Tell Haddad, near Baghdad, found in the destruction layer from when Hammurabi conquered the site _ usually dated at -1762.  The tablet is in Baghdad.  See:  Farouk al-Rawi & Michael Roaf; Ten Old Babylonian mathematical problems from Tell Haddad; Sumer 43 (1984) 175-218.

Prob. ix _ Making bricks.  One man can dig  1/3  sar of earth in a day, or he can mix  1/6  sar or he can mould  1/3  sar into bricks.  If 1 sar makes  1620  bricks, how many bricks can a team of three make in a day?  For one man, we get  S = 3 + 6 + 3,  so he can process  1/12  sar per day, or  135  bricks, so three men can make  405  bricks.

Prob. x _ Carrying earth to make bricks.  Same problem as the previous, but the earth must be carried 5 nindan from the digging site to the works.  The amount one man can carry in a day is given somewhat cryptically.  The simplest interpretation is that one man can carry  1/3  sar of earth over the 5 nindan in a day, but there still are three workers in the group.  Here we get  S = 3 + 3 + 6 + 3,  so one man can process  1/15  sar per day or  108  bricks and three men make  324  bricks.

YBC 4669.  This and the following tablet are in the same hand, but have no provenance.  See Neugebauer, vol. III, pp. 28-29 & plate 3, ??NYS.  Reverse, col. 3, lines 7-17, c-1800 _ Demolishing walls.  A man can knock down  1/15  sar of wall in  1/5  of a day and he can carry away  1/12  sar in a day.  How much wall can he demolish and carry away in a day;  and what part of the day is devoted to each task?  Here  a₁ = (1/15)/(1/5) = 1/3,  so  S = 3 + 12  and he can do  1/15  sar per day.

YBC 4673.  See Neugebauer, vol. III, pp. 30 & 32 & plate 3, ??NYS.  Obverse, col. 2, lines 10-18, c-1800 _ Constructing a pile of bricks.  A man can carry  1/18  sar of earth (bricks??) in a day.  He can pile up 1 sar of bricks in  14 2/5  days.  If a sar makes  5184  bricks of this size, how many bricks can he carry and pile up in a day?

Chiu Chang Suan Ching (Jiu Zhang Suan Shu).  c‑150?  Chap. VI. 

Prob. 22, p. 67.  Man can do two processes at rates of  38  in  3  days and  76  in  2  days.  How many of both together can he do in one day?  Answer is given as  25½  but Vogel's note on the calculation shows  25_  was meant, and this is erroneous _ the correct answer is  9½.  The error arises from taking  38  and  76  as rates per day.

Prob. 23, p. 67.  Three processes at rates  50, 30, 15  per day, how many together in a day?  Correct answer,  8_,  is obtained.  (Arrow shafts, arrow feathering, arrow heading.)

Prob. 25, p. 68.  Three processes at rates  7, 3, 5  per day, how many together in a day?  Correct answer,  105/71,  is obtained.

Prob. 26, pp. 68‑69.  Cistern:  (1/3, 1, 5/2, 3, 5).  Correct answer.  Vogel says this is the first appearance of the problem.

Heron (attrib.).  c150.  Περι Μετρov (Peri Metron).  In:  J. L. Heiberg, ed.; Heronis Alexandrini Opera Quae Supersunt Omnia, vol. V; Teubner, Leipzig, 1914; reprinted 1976, pp. 176‑177.  Greek and German texts.

Problem 20: Μετρησισ χιστερvασ (Metresis xisternas) [Vermessung einer Zisterne].  (1,4)  in hours, but he computes  1 + 4 = 5  and then sets the cistern  = 12 ft and computes  12/5  as the number of hours.  See BR, c1305, prob. 98, for the explanation of this.

Problem 21: Αλλωσ η μετρεσισ (Allos e metresis) [Die Vermessung in anderer Weise].  + 1/7 ‑ 1/11,  how long to make  100?  He treats the  ‑  as  +  and gets the correct answer for that case, though Heiberg says the calculation is senseless.

Smith, History II 538, quotes from Bachet's Diophantus, implying a date of c275, citing the 1570 edition with Fermat's notes, but Smith's citation is to the part of Bachet taken from Metrodorus!  It is Art. 130 of Metrodorus.

                    Sanford 216 also cites Diophantus, but her discussion is based on Smith's AMM article (op. cit. in 3), which is the basis of the section in Smith's History containing Smith's quote.  The problem is nowhere in Heath's edition of Diophantus.

                    However, Tropfke 578 gives a reference to the Tannery edition of Diophantus, vol. 2, p. 46 _ ??NYS.

Metrodorus.  c510.  8 cistern‑type problems.

Art. 7, pp. 30‑31.  "I am a brazen lion."  (2, 3, 4, 1/2),  where 6 hours is counted as  1/2  day, i.e. a day has 12 hours.

Art. 130, pp. 96‑97.  (1, 2, 3, 4).

Art. 131, pp. 96‑97.  (4, 8, 12).

Art. 132, pp. 96‑97.  "This is Polyphemus, the brazen cyclops."  (3, 1, 2/5).

Art. 133, pp. 96‑99.  (1, 3, 2).

Art. 134, pp. 98‑99.  Three spinners can do  1, 4/3, 1/2  unit per day, how long for all three to do one unit?

Art. 135, pp. 98‑99.  "We three Loves" (or Cupids).  (2, 4, 6).

Art. 136, pp. 98‑101.  'Brickmakers.'  Three brickmakers can make  300, 200, 250  per day.  How long for all three to make 300?

Bakhshali MS.  c7C.  Kaye I 49‑52 discusses several types, e.g. first gives 5/2 dinars in 3/2 days;  next gives 7/2 in 4/3;  third gives 9/2 in 5/4;  how long for all three to give 500 dinars? (= Kaye III 192, ff. 21v-22r).  Kaye III 191 has three rates of  1/(1/3), 1/(1/2), 3/5  _ how long to give  100?  I 51 (= III 233-234, ff. 44v-44r) is an example with an income, some capital and three rates of expenditure.  On I 50 (= III 234-235, ff. 44r-43v) is an example with an income, some capital and seven rates of expenditure!

Anania Schirakatzi (= Ananias of Shirak).  Arithmetical problems.  c640.  Translated by:  P. Sahak Kokian as:  Des Anania von Schirak arithmetische Aufgaben; Zeitschrift für d. deutschösterr. Gymnasien 69 (1919) 112-117.  See 7.E for description.  Prob. 24 is a cistern with pipes:  (1, 2, 3),  but he gives the answer:  1/4 + 1/6 + 1/16 + 1/18,  which  =  77/144,  which is close to the correct answer of  6/11.  No working is shown and I am unable to see how  77/144  can arise, even allowing for a possible misprint.

H&S 70 says the cistern problem appears in Alcuin, 9C, but the only possible problem is a trivial problem (8: Propositio de cupa) which mentions a barrel.

Mahavira.  850.  Chap. VIII, v. 32, p. 266.  Cistern:  (1/2, 1/3, 1/4, 1/5).

Chaturveda.  860.  Commentary on Brahma‑sphuta‑siddhanta, chap. XII, sect. 1, art. 9.  In Colebrooke, p. 282. 

Cistern:  (1, 1/2, 1/4, 1/5).  (Datta & Singh, I, 234 and others cite this as Brahmagupta.) 

Bestowing alms:  (1/3, 1/2, 5/3).

Sridhara.  c900.  V. 69, ex. 91, pp. 55‑56 & 95.  Cistern:  (1/2, 1/4, 1/5, 1/6).

al‑Karkhi.  c1010.  Sect II, no. 15‑16, p. 83.  Cistern:  (1, 2, 3).  No. 16 asks how often the cistern will be filled in 5 days.

Bhaskara II.  Lilavati.  1150.  Chap. IV, sect. II, v. 95.  In Colebrooke, p. 42.  Cistern:  (1, 1/2, 1/3, 1/6).  (Datta & Singh, I, 234, erroneously say this is the same problem as Brahmagupta, i.e. Chaturveda.)

Fibonacci.  1202.

P. 182:  De Leone et leopardo et urso.  Lion, leopard and bear eating a sheep:  (4, 5, 6).

P. 182:  De duabus navibus ... is in 10.A.

P. 183:  cistern problems  (1, 2, 3, 4)  &  (6, 9, 24, 27).

Pp. 183‑186 _ several problems with water butts having different size openings at different heights.  E.g., pp. 183‑184 has four openings at  1/4, 2/4, 3/4, 4/4  of the way down, which could drain the whole butt in  4, 8, 12, 16  days.  How long to drain a full butt with all holes open?  Answer:  7 267/2275  days.

BR.  c1305.

No. 25, pp. 44‑45.  Ship with 5 sails:  (1/2, 1/3, 1/5, 1/7, 1/9).  Vogel says this is the first example of the formulation of a ship with sails.

No. 64, pp. 88‑89.  Cistern,  (1, 2, 3).

No. 65, pp. 88‑91.  Cistern _ 'I am a noble lion',  (2, 3, 4, 1/2).  = Metrodorus 7.

No. 66, pp. 90‑91.  Cistern,  (3, x)  in  4/3.

No. 67, pp. 90‑93.  Cistern,  (5, x)  in  15/8.

No. 68, pp. 92‑93.  Cistern,  (x,  ‑9)  in  9/2.

No. 69, pp. 92‑93.  Cistern,  (3,  ‑x)  in  9/2.

No. 70, pp. 94‑95.  Cistern,  (3, ‑9).

No. 96, pp. 114‑117.  3 cisterns of volumes  30, 60, 120  with pipes that fill them in  6, 4, 3.  Using all three pipes, how long to fill all three cisterns?

No. 97, pp. 116‑117.  Cistern,  + 1/7 ‑ 1/11  to yield  100.  This is Heron's prob. 21 and is done in the same way _ as though it were  + 1/7 + 1/11,  _ but one MS is worded so this is the correct method, as noted by Vogel.

No. 98, pp. 118‑119.  Cistern,  (1, 4).  This is Heron's prob. 20, again noted by Vogel.  The text says to set the cistern equal to  12  and then divides by  5 = 1 + 4.  Vogel notes that this does not give the time, but  12/5  is the volume delivered by the smaller pipe.

No. 99, pp. 118‑119.  Cistern,  (6, 10).

No. 116, pp. 132‑133.  Cistern,  (2, 3, 4, ‑6, ‑4).

Gherardi.  Libro di ragioni.  1327.

Pp. 44‑45:  Volare una bocte.  (3,5),  but the  5  is half‑way down the barrel while the  3  is at the bottom.

P. 45:  Ship with three sails,  (3, 5, 7).

Pp. 56‑57:  Uno chavaleri che vuole far fare uno pallagio.  Three workers,  (50, B) in 36,  (50, B, C) in 30.

Gherardi.  Liber habaci.  1327?  Pp. 143‑144: Compangnia et viaggio.  Baratti xvii.  Three workers,  (2, 3, 4).   Ship with two sails,  (2, 3).

Columbia Algorism.  c1370.

Prob. 66, pp. 87‑88.  Cask can be emptied in  1/2, 1/3, 1/4, 1/5, 1/6  of a day.  (See also Cowley 399.)

Prob. 141, p. 150.  Same with  1/2, 1/3, 1/4.

Dell'Abbaco.  c1370.

Prob. 23, p. 32.  (4, 5).

Prob. 50, p. 49 with plate on p. 50.  Ship with two sails,  (8, 10).  I have a colour slide of this.

Prob. 62, p. 59.  (9, x)  in  5.

Lucca 1754.  c1390.  F. 59r, p. 134.

Ship with 3 or 2 sails:  (3, 4, 5),  (3, 4).

Two couriers meeting:  (20, 30).

Giovanni di Bartolo.  Certi Chasi.  c1400.  Copied by Maestro Benedetto (da Firenze), in Cod. L.IV.21, Biblioteca degli Intronati di Siena, 1463.  Edited by M. Pancanti, Quaderni del Centro Studi della Matematica Medioevale, No. 3, Univ. di Siena, 1982.

Prob. 1, pp. 4‑5.  (x, x/2)  in  5.

Prob. 2‑8, pp. 5‑17 are more complex examples, often leading to quadratic equations, e.g.  (x, x‑5)  in  12.

Bartoli.  Memoriale.  c1420. 

Prob. 14, f. 76v (= Sesiano, pp. 140-142 & 149, with facsimile of the relevant part of f. 76v on p. 141.  Cask with four taps:  (3, 4, 5, 6).

Prob. 33, f. 79r (= Sesiano, pp. 146 & 150).  Six workers building a wall:  (1, 2, 3, 4, 5, 6).  He correctly finds the total rate is  137 [/ 60],  but then uses  36  instead of  60  _ "Now make  6  times  6,  which is  36,  because there are  6  workers."  Sesiano describes this as fantasy.

AR.  c1450.  Prob. 51, 57, 70, 75, 76, 97, 98, 182, 281.  Pp. 42, 44‑45, 48‑50, 58‑59, 85‑86, 128‑129, 157, 160‑161, 165‑166, 175, 211‑213, 221.

51:  cistern with three drains,  (1, 2, 3),  erroneously done _ see 97.

57:  three mills, but gives amounts each can do per day.

70:  three builders,  (7, 5, 4).

75:  three tailors, but gives amounts each can do per day.

76:  (13, B)  in  9.

97:  cistern with three drains,  (1, 2, 3),  = Metrodorus 133.

98:  ship with three sails,  (2, 3, 4).

182:  three scribes;  (A, B) in 20,  (A, C) in 30,  (B, C) in 15,  how long for each?

281:  barrel with three taps,  (1, ½, ¼).

"The Treviso Arithmetic" = Larte de labbacho (there is no actual title).  Treviso, 1478.  Translated by David Eugene Smith, with historical commentary by Frank J. Swetz, as:  Capitalism and Arithmetic; Open Court, La Salle, Illinois, (1987), improved ed., 1989.  This is discussed in:  D. E. Smith; The first printed arithmetic (Treviso, 1478); Isis 6 (1924) 310‑331.  Facsimile edition, from the copy at the Diocese of Treviso, with commentary booklet by Giuliano Romano, Libreria Canova, Treviso, nd [Swetz, p. 324, says 1969, but the commentary cites a 1976 book.]  The facsimile has taken its title from the end of the opening sentence.  Romano's commentary calls it:  L'Arte dell'Abbacho.  The text nowhere gives a publisher's name.  Smith, Rara, pp. 3-7, says it was probably Manzolo or Manzolino, while Swetz, p. 26, specifies Michael Manzolo or Manazolus.  Romano says it was published by Gerardus de Lisa.

                    F. 57r (pp. 162‑163 in Swetz).  Carpenters,  (20, x)  in  8.

Muscarello.  1478. 

F. 58r, p. 161-162.  (B, C)  in  3,  (A, C)  in  4,  (A, B)  in  5.  There are two copying errors in the MS answers.

F. 62v, pp. 169-170.  Four workers,  (2, 3, 5, 6).

Ff. 77r-77v, pp. 192-193.  Three mills can do  9, 8, 5  per day.  How long will it take them to do  6  and how much does each do?

F. 77v, p. 193.  Ship with two sails,  (7, 10).

F. 81r, p. 196.  Cask with three spouts which can let out  6, 7, 8  per hour.  How long will it take to empty a cask of  23?

P. M. Calandri.  c1480.

Pp. 90‑91:  ship with three sails  (10, 12, 15).

P. 91:  cistern  (4, 6).

P. 91:  two workers  (20, 5).

Wagner.  Op. cit. in 7.G.1.  1483.  Regel von einem Fass, pp. 114 & 224.  Cask with three taps  (2, 3, 4).

Chuquet.  1484. 

Prob. 21.  English in FHM 204.  Cistern emptying,  (3, 4, 6).

Prob. 53.  English in FHM 209-210.  The first says:  "If you help me  8  days, I will build it in  20".  The second responds:  "If you help me  10  days, I can do it in  15".  How long for each alone?

Prob. 54.  Same as prob. 53 with parameters   5, 17;  6, 24.

Borghi.  Arithmetica.  1484. 

F. 106v (1509: ff. 91r-91v).  Three mills can grind  6, 9, 11  per day.  How long to do  100?

F. 109r (1509: ff. 91v-92r).  Ship with three sails,  (6, 9, 12).  (H&S 70 gives Latin and English.)

Johann Widman.  Op. cit. in 7.G.1.  1489.  (On pp. 131-132, Glaisher mentions the following.)  Ff. 136r‑138v:  Eyn fasz mit dreyen Czapfen;  Von der Mulen;  Leb, wolff, hunt;  Schiff.  (Cistern problem;  3 mills;  lion, wolf, dog eating a sheep;  ship with 3 sails.)

Calandri.  Arimethrica.  1491.

F. 68v.  Ship with two sails.  (12, 15).  Woodcut of ship with indeterminate number of sails.

F. 69r.  Cask with two taps.  (4, 6).  Woodcut of cask with two taps.

F. 70r.  Ship with three sails.  (10, 12, 15).  Same woodcut as on f. 68v.

F. 70r.  Cask with three taps.  (4, 6, 8).  Same woodcut as on f. 69r.

F. 70v.  Three masters build a house.  (10, 12, 15).  Woodcut of two builders.  (H&S 70 gives Italian and English and says it also occurs in Treviso Arithmetic (1478) [but that has a very different type!], Pacioli, Cataneo, Tartaglia, Buteo (1559), Clavius, Tonstall.)

F. 70v.  Three masters doing a job.  (30, 40, x)  in 15. 

F. 71v.  Cistern.  (4, ‑11).  Woodcut of cistern.  (Rara, 48 is a reproduction.)

F. 72v.  Lion, leopard & wolf eating a sheep.  (1, 2, 3)  days.  Nice woodcut.  (H&S 70 gives Italian and English, says there is a remarkable picture and says it occurs in Fibonacci [again, there it occurs in a different form] and Cataneo.)

Pacioli.  Summa.  1494.  See also Buteo.

F. 99r, prob. 6.  Building a house,  (8, 10, 4).  Says one can have more builders and it is similar to dog, wolf & lion eating a sheep.

F. 99v, prob. 16.  Three mills,  (6, 8, 3)  days.

F. 99v, prob. 17(not printed).  Three mills,  (10, 5, 4)  days.

Calandri.  Aritmetica.  c1500.

F. 91r, p. 182.  Ship with 2 sails:  (12, 18).

F. 91v, p. 183.  Three men in prison:  (6, 12, 18).  (Tropfke 520 reproduces this in B&W.)

F. 93r, p. 186.  Emptying a cask:  (6, 8).

F. 95v, p. 191.  Ship with three sails:  (6, 10, 12).  (Coloured plate opp. p. 120 of the text volume.)

F. 96v, p. 193.  Emptying a cask:  (8, 12, 16).

F. 97v, p. 195.  Lion, wolf & fox eating a goat:  (2, 3, 5).  (Tropfke 581 reproduces this in B&W.)

Ff. 98v-99r, pp. 197‑198.  Furnace with 3 fires:  (10, 15, 20).

Blasius.  1513.  F. F.iii.r: Decimatertia regula.  Three rivers can water a field in  (1, 2, 3)  days.  Gets  13 1/11  hours for all three _ so he is using  24  hour days.

Tagliente.  Libro de Abaco.  (1515).  1541. 

Prob. 117, f. 58r.  Ship with two sails _  (12, 15).

Prob. 119, f. 58v.  Cask with two taps _  (4, 6).

Tonstall.  De Arte Supputandi.  1522. 

Quest. 26, pp. 157-159.  Three mills can do at rates of  18, 13, 8  per day.  How long to do 24?

Quest. 27, pp. 159-161.  Cistern,  (1, 2, 3)  and  (4, 6, 8).

Quest. 28, pp. 161-162.  Cistern,  (1/4, 1/2, 1).

Quest. 29, pp. 162-163.  Cistern,  (4, -11).

Quest. 32, pp. 164-165.  Four architects building a house,  (1, 2, 3, 4)  years.  Says it is similar to a cistern problem.

Quest. 33, p. 166.  Three architects building a house,  (30, 40, x)  in  15.

Riese.  Die Coss.  1524.

No. 117, p. 55.  Cask with three taps,  (1, 2, 3).

No. 118, p. 56.  Three windmills can grind  20, 17, 15  per day.  How long to do  24?

Cardan.  Practica Arithmetice.  1539.

Chap. 47, ff. L.iii.r - L.iii.v (pp. 70-71).  Simple example _  5  mills grind  7, 5, 3, 2, 1  per hour, how long will they take to grind  500? 

Chap. 66, section 125, ff. kk.vi.r - kk.vi.v (pp. 180-181).  Cask with four taps located at levels  1/3,  1/3 + 1/4,  1/3 + 1/4 + 1/6,  1  from the top and which empty their respective portions in  4, 3, 2, 1  hours.  How long to empty the cask with all taps?

Chap. 66, section 126 (misprinted 123), ff. kk.vi.v - kk.vii.r (p. 181).  Cistern:  (1, 2, 3, ‑4, -5, -3/4).

Gemma Frisius.  Arithmetica.  1540.  (20, x)  in  14  _ man & wife drinking a cask of wine.  ??NYS _ Latin given in H&S, p. 71.

Tartaglia.  General Trattato, 1556, art. 74, p. 248v;  art. 176‑177, p. 261v;  art. 187‑188, pp. 262r‑262v.

Art. 74:  120  per  40  and  15 ½  per  6  to do  120.

Art. 176:  (16, 20).

Art. 177:  (60, 80, x)  in  30.

Art. 187:  1 per 8,  1 per 6  and  1 per 3  to do  25.

Art. 188:  (10, 5, 4).

Buteo.  Logistica.  1559. 

Prob. 6, pp. 205-206.  Three mighty drinkers drinking an amphora of wine in  (24, 12, 8)  hours.  Cites Pacioli.  (H&S 71)

Prob. 7, pp. 206-208.  Three architects build a house:  (x, x/2, x/3)  in  2  months.  Says Pacioli gives  (x, x+6, x+8)  in  2  and solves it wrongly.

Prob. 8, pp. 208-209.  Ship with two sails,  (8, x)  in  5.

Prob. 61, pp. 266-268.  Cask with three taps  1/4, 2/3, 1  of the way down which could empty the whole cask in  (6, 3, 3)  hours.

Prob. 62, pp. 268-269.  Cistern,  (+2, -3).

Gori.  Libro di arimetricha.  1571.

F. 74v (pp. 82‑83).  Cistern empties in  (4, 6)  hours.  Ship with three sails,  (3, 4, 5)  days.

F. 77v (p.83).  Lion, bear and wolf eating a sheep,  (4, 2, 1)  hours.

Johann Faulhaber.  Arithmetischer Wegweiser ...  Ulm, 1614.  ??NYS.  A 1708 ed. is quoted in Hugo Grosse; Historische Rechenbücher des 16. und 17. Jahrhunderts; (1901);  reprinted by Sändig, Wiesbaden, 1965, p. 120. 

No. 91, p. 228:  wolf, sheepdog and dog eating a sheep,  (1, 3, 6). 

No. 92, p. 229:  ship with three sails,  (2, 3, 4).

van Etten.  1624.  Prob. 83 (76): Du Lyon de Bronze posé sur une fontaine avec cette epigraphe, pp. 94‑95 (140).  (2, 3, 4, 1/2)  = Metrodorus, art. 7.

Georg Meichsner.  Arithmetica Historica.  Hieronymus Körnlein, Rotenburg/Tauber, 1625.  No. 68, p. 209.  ??NYS.  Quoted in Hugo Grosse, op. cit. under Faulhaber, above, p. 77.  Three men with devices to pump out flooded lands in Holland,  (60, 30, 20).

Wingate/Kersey.  1678?. 

Quest. 4, pp. 476-477.  Workmen:  (20, 30).

Quest. 5, pp. 477-478.  "I am a brazen lion ..." in Latin.  (2, 3, 4, 1/2),  where 6 hours is counted as  1/2  day, i.e. a day has 12 hours.  = Metrodrus, Art. 7.

Quest. 6, pp. 478-479.  (1/2, -10/7, -7/3).

Quest. 7, pp. 479-480.  Dog, wolf, lion eating a sheep: (1, 3/4, 1/2),  but the lion has a head-start of  1/8  hour.

Quest. 12, p. 484.  (20, x)  in  12.  Man and wife driniking beer.

Quest. 13, pp. 484-485.  (30, 40, x)  in  15.  Carpenters building a house.

Leybourn.  Pleasure with Profit.  1694.  Prob. 14, p. 39.  Cistern emptying: (6, 4, 3, 2).

Isaac Newton.  Arithmetica Universalis, 1707.  ??NYS.  English version:  Universal Arithmetic, translated by Mr. Ralphson, with revisions and additions by Mr. Cunn, Colin Maclaurin, James Maguire and Theaker Wilder; W. Johnston, London, 1769.  (De Morgan, in Rara, 652‑653, says there were Latin editions of 1722, 1732, 1761 and Raphson's English editions of 1720 and 1728, ??NYS.)  Resolution of Arithmetical Questions, Problem VII, pp. 184‑185.  "The Forces of several Agents being given, to determine  x  the Time, wherein they will jointly perform a given Effect  d."  Gives general approach for three workers.  Example is  (3, 8/3, 12/5),  where the Force of the second is expressed as saying he can do the work "thrice in  8  weeks".

Ozanam.  1725.

Prob. 24, question 9, 1725: 180‑181.  Prob. 5, 1778: 188-189;  1803: 185-186;  1814: 161-162;  1840: 84.  Same as Metrodorus 7, except that a day is considered as  24  hours, so the problem is done as  (2, 3, 4, ¼).

Prob. 24, question 10, 1725: 181.  (6, 9, 12)  months to print a book.

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790).

Prob. XI, pp. 83-84 (1790: prob. XXX, pp. 91‑92).  General problems,  (a, b)  and  (a, b, c).  Example:  (10, 12, 16).

Prob. XXXI, pp. 90-92 (1790: prob. XXXVI, pp. 96‑98).  General problem:  given  (x, y)  in  a,  (x, z)  in  b,  (y, z)  in  c, determine  x, y, z.  Example with  a, b, c   =  8, 9, 10.

Alexis-Claude Clairaut.  Elémens d'Algèbre.  1746.  Vol. I, art. LVI (my source quotes from the 6th ed. of 1801).  ??NX.  He uses the context of this type of problem to study the meaning of negative solutions.  A cistern of size  a  is filled by a source running for time  b  together with another source running for time  c.  Another reservoir of size  d  is filled by the sources in times  e  and  f.  Determine the rate of each source.

Les Amusemens.  1749. 

Prob. 173, p. 321.  Cistern:  (9, -12).

Prob. 174, pp. 322-323.  Reservoir with three nymphs:  (3, 4, 5).

"By his Holiness the Pope".  The Gentleman and Lady's Palladium (1750) 22.  Qn. 11.  (??NYS, cited by E. H. Neville; Gleaning 1259: On Gleaning 1146; MG 23 (No. 254)  (May 1939) 149.  "If a Cardinal can pray a soul out of purgatory ...."  See Welch, 1833, below.

Euler.  Algebra.  1770.  I.IV.III: Questions for practice.

No. 14, pp. 204‑205.  Cistern,  (x, 20)  in  12.

No. 22, p. 205.  Same as the example in Simpson's prob. XXXI

Vyse.  Tutor's Guide.  1771? 

Prob. 61, p. 74 & Key p. 100.  Two workers,  (10, 13).

Prob. 62, pp. 74-75 & Key p. 100.  Boatbuilders,  (18, x) in 11.

Prob. 6, p. 136 & Key p. 178.  Cistern emptying,  (1, 2, 3,).  Gives volume of cistern but never uses it.  = Metrodorus 133.

Prob. 15, p. 167 & Key p. 208.  Builders,  (34, x, 24) in 12.

Prob. 5, p. 201 & Key pp. 244-245.  Trenching a field.  A, B, C  can do in  12;  B, C, D  can do in  14;  C, D, A  can do in  15;  D, A, B  can do in  18.  How long for all four together and for each one singly?  Solution in decimals..

Dodson.  Math. Repository.  1775.

P. 22, Quest. LVIII.  Man and wife drinking beer.  (30, x) in 12.

P. 23, Quest LIX.  Cistern:  (20, x) in 12.

Pp. 52-53, Quest. CV.  Workers.  A & B in 8;  A & C in 9;  B & C in 10.

P. 56, Quest 56.  (3, 8/3, 12/5),  phrased as in Newton.  Does the problem in general, then applies to the data.

Ozanam-Montucla.  1778.

Question 5, 1778: 193-194;  1803: 190-191;  1814: 165-166;  1840: omitted.  Same as Metrodorus 135.

Prob. 22, 1778: 214;  1803: 209.  Prob. 21, 1814: 181;  1840: 94.  Same as the example in Simpson XXXI.

Charles Hutton.  A Complete Treatise on Practical Arithmetic and Book-keeping.  Op. cit. in 7.G.2.  [c1780?]  1804: prob. 70, pp. 139-140.  Cistern:  (40, 50, -25).

Bonnycastle.  Algebra.  1782.

Pp. 82-83, no. 10 (1815: p. 102, no. 9).  (10, 13).  = Vyse, prob. 61.

P. 83, no. 11 (1815: pp. 102-103, no. 10).  (a, b)  done in general.

P. 86, no. 25 (1815: p. 108, no. 34).  Two drinkers:  (30, x)  in  12  days.

P. 86, no. 26 (1815: p. 108, no. 36).  Same as the example in Simpson's prob. XXXI.

P. 86, no. 27.  "If three agents,  A,  B,  and  C,  can produce the effects  a,  b,  c,  in the times  e,  f,  g,  respectively;  in what time would they jointly produce the effect  d?"

Pike.  Arithmetic.  1788. 

P. 335, no. 8.  Cistern:  (1/2, 1/4, 1/3).

P. 350, no. 14.  Merchant gaining and losing, equivalent to  (7, 9, ‑2)  _ how long to empty a full tank?

P. 350, no. 19.  Two workers:  (7, 12).

Pp. 350-351, no. 20.  Boatbuilders:  (20, x)  in  12.

P. 351, no. 21.  Two workers:  (13, x)  in  8.

P. 351, no. 22.  Three workers:  (23, 37, x)  in  15.

P. 351, no. 23.  Cistern:  (55, 45, -30).

P. 351, no. 24.  Cistern of  73,  inflow of  7/5  and outflow of  20/17  both run for two hours, then the outflow is stopped.

P. 355, no. 39.  A,  B,  C  do a job.  A  and  B  do  3/11  of it,  A  and  C  do  5/13,  B  and  C  do  4/14.  (Also entered at 7.G.1.)

Bullen.  Op. cit. in 7.G.1.  1789.  Chap. 38. 

Prob. 34, p. 243.  Mother & two daughters spin  3  lb flax in  1  day;  mother can do it in  2½  days;  elder daughter in  2¼  days;  how long does it take the younger daughter? 

Prob. 51, pp. 245‑246.  Cistern:  (6, 8, 10, 12, ‑6, ‑5, ‑4, ‑3)  _ how long to empty from full?

Eadon.  Repository.  1794. 

P. 78, no. 21.  If  3  men or  4  women can do a job in  68  days, how long will it take  2  men and  3  women?

P. 79, no. 26.  If  5  oxen or  7  colts can eat a close in  87  days, how long will it take  2  oxen and  3  colts?  Answer is  105,  which neglects growth of grass.

P. 195, no. 10.  (8, x)  in  5.

P. 195, no. 11.  (32, 44, x)  in  16.

P. 195, no. 12.  (3, 8/3, 12/5).  = Newton's example.

P. 367, no. 6.  (40, 50, -25).

John King, ed.  John King   1795   Arithmetical Book.  Published by the editor, who is the great-great-grandson of the 1795 writer, Twickenham, 1995.  P. 100:  (1, 2, 3);  (7, 5, 6).

Hutton.  A Course of Mathematics.  1798? 

Prob. 8,  1833: 212-213;  1857: 216-217.  (6, 8),  then generalises to  (a, b, c, d).

Prob. 15,  1833: 220-221;  1857: 224-225..  A  and  B  can do in  a  days;  A  and  C  in  b  days;  B  and  C  in  c  days.  How long for each singly and all three together?

Prob. 39,  1833: 223;  1857: 227.  (x, 30)  in  12.

Prob. 40,  1833: 223;  1857: 227.  Problem 15 above with numerical values:  a, b, c  =  8, 9, 10.

Bonnycastle.  Algebra.  10th ed., 1815.  P. 226, no. 6.  Cistern:  (20, x)  in  12  hours.

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions.

No. 4, pp. 15 & 71.  A  &  B  earn  40s  in  6  days;  A  &  C  earn  54s  in  9  days;  B  &  C  earn  80s  in  15  days.  What does each earn per day?

No. 11, pp. 16 & 73.  "A  in five hours a sum can count, Which  B  can in eleven;  How much more then is the amount They both can count in  7?"

No. 29, pp. 21 & 80-81.  Lion, wolf, dog eating a sheep:  (1/2, 3/4, 1),  but the lion begins  1/8  before the others.

Augustus De Morgan.  Arithmetic and Algebra.  (1831?).  Reprinted as the second, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  Arts. 3 & 112, pp. 1-2 & 28-29.  In Art. 3, he mentions the problem  (10, 16) as an example of algebraic formulation.  In Art. 112, he solves it and  (a, b).

Welch.  Improved American Arithmetic.  1833 ed.  "If a Cardinal can pray a soul out of purgatory by himself in 1 hour, a bishop in 3 hours, a Priest in 5 hours, a Friar in 7 hours, in what time can they pray out 3 souls, all praying together?"  In the 1842 ed., this was changed to steam, water, wind and horse power.  ??NYS _ quoted in Gleaning 1146, MG 21 (No. 245) (Oct 1937), 258.  See above at 1750 for an earlier version.

Nuts to Crack II (1833), no. 129.  (30, x)  in  12.  Identical to Bonnycastle, 1782, no. 25.

Louis Pierre Marie Bourdon (1779-1854).  Élémens d'Algèbre.  7th ed., Bachelier, Paris, 1834.  (The 6th ed., was 1831, but I haven't yet looked for the early editions _ ??)  Art. 57, p. 85.  General solution for  (a/b, c/d, e/f).

Augustus De Morgan.  On The Study and Difficulties of Mathematics.  First, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  P. 31.  (12, 15),  then does  (a, b).

Augustus De Morgan.  Examples of the Processes of Arithmetic and Algebra.  Third, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836. 

Prob. 16, p. 28.  Given  (5, 7),  how long will it take to do  _?

P. 91.  A & B  in  c;  A & C  in  b;  B & C  in  a.

Unger.  Arithmetische Unterhaltungen.  1838. 

Pp. 135 & 258, no. 519.  A  can do  63  in  8  days;  B  can do  37  in  6  days  and  C  can do  25  in  3  days.  How long for all three to do  268½?

Pp. 135 & 258, no. 520.  Cistern  (20, 15, 12).

Pp. 135-136 & 258, no. 521.  Cistern  (12, 8, -10).

Hutton-Rutherford.  A Course of Mathematics.  1841?  Prob. 8,  1857: 81.  Two workers:  (12, 14).

T. Tate.  Algebra Made Easy.  Op. cit. in 6.BF.3.  1848.

P. 75, no. 11.  (x + 6, x + 3/2)  in  x.

P. 76, no. 12.  (x, x+5)  in  6.

P. 85, no. 8.  A  can reap a field in  a  days.  If assisted by  B  for  b  days, then  A  only has to work  c  days.

Family Friend 1 (1849) Answers to correspondents, pp. 4 & 6.  Questions requiring answers. No. 1.  (12, x)  in  3.

Anonymous.  A Treatise on Arithmetic in Theory and Practice: for the use of The Irish National Schools.  Third edition, revised, improved and enlarged.  Published by direction of the Commissioners of National Education in Ireland, Dublin, 1850.  Many examples, of which the following are the more interesting.

Pp. 200-201, no. 133.  (17/2, 21/4, x)  in  6/5.

P. 347, no. 8.  (15, x)  in  10.

P. 347, no. 11.  5  mills grind  7, 5, 4, 3, 1  bushels per hour.  How long for all five to grind  500  bushels?

P. 356, no. 10.  A,  B  &  C  can do in  10  days;  B  &  C  can do in  16  days;  how long for  A?  (This is equivalent to  (x, 16)  in  10.)

P. 358, no. 30.  A,  with  2  days' help from  B,  can do in  12  days.  B,  with  4  days' help from  A,  can do in  8  days.  How long for both together?

P. 358, no. 31.  A  and  B  can do in  8  12-hour days.  A  can do in  12  16-hour days.  How many  14-hour days would  B  need?

P. 359, no. 36.  (5/2, 9/4, x)  in  1.

P. 359, no. 39.  (6, 8, 10, 12, -6, -5, -4, -3)  _ how long to empty from full?

John William Colenso (1814-1883).  Arithmetic Designed for the Use of Schools ....  New edition.  Longman, Brown, Green, and Longmans, 1853.  ??NX - Wallis 246 COL.  I have 1871, which seems identical for pages 1 - 144, then adds new material on decimal coinage and examination papers, then has the Answer section shifted ahead by 50pp.  The Advertisement is dated 1864, so I will date the new material as 1864?  This gives a large number of variations of the problem which I have included here as representative of mid 19C texts. 

   Miscellaneous Examples, pp. 122‑136, with answers on pp. 161-163 (1871: 211‑213).

No. 21.  (10, 13).

No. 27.  (10, B)  in  7.

No. 31.  Cistern, (40,  50, ‑25).

No. 52.  (3, 8/3, 12/5),  where  B  is determined from  "B  can do thrice as much in  8  days".  I.e. Newton's example; see also Eadon.

No. 80.  If  5  oxen or  7  horses can eat the grass of a field in  87  days, how long will it take  2  oxen and  3  horses?  (The grass is not assumed to grow.)   = Eadon, No. 26.

No. 101.  If  3  men,  5  women or  8  children can do a job in  26½  hours, how long will it take  2  men,  3  women and  4  children?

   Examination _ Paper VIII, pp. 170‑172, with answers on p. 214.  (This material is 1864?)

No. 1.  M  can do in  20  7‑hour days.  N  can do in  14  8‑hour days.  How many hours per day must they work together to do it in  10  days?

No. 2.  Cistern,  (20, 24, ‑30).  How full is it after 15?

No. 3.  Reapers,  (F, G) in  8¾,  with  3½ : F  =  5 : G.

No. 4.  (34, 38),  but second man stops  4  days before the end.

No. 5.  Cistern.  (A, B)  fills in  4;  (A, ‑C)  empties in  40;  (B, ‑C)  fills in  60.

No. 6.  4  men, working various parts of the time.

No. 7.  (A, B)  in  14,  (B, C)  in  10½,  (A, C)  in  12.

No. 8.  B = twice (A, C);  C = thrice (A, B);  (A, B, C)  in 5.

No. 9.  Three men working various parts of the time.

No. 10.  Cistern with 2 inlets and 2 outlets running various parts of the time.

John Radford Young (1799-1885).  Simple Arithmetic, Algebra, and the Elements of Euclid.  IN: The Mathematical Sciences, Orr's Circle of the Sciences series, Orr & Co., London, 1854.  [An apparently earlier version appeared as Circle of the Sciences, vol. (2 &) 3, with no TP or other details except two editorial remarks referring to Professor Young as author of the Arithmetic and Algebra sections.  I have vol. 3, which begins with the last sheet of Arithmetic and then covers the other material.  (I saw a vol. 8? on zoology? in the same bookshop.)  This version includes detailed solutions and some plates not connected with the text.  The 1854 text is identical except that the order of topics has been changed and there are some consequent changes to the text.]  This is a typical mid 19C text with a number of cistern problems, of which the more interesting are the following.

No. 16, p. 178.  A  can do a job in  10  days.  After he has worked for  4  days,  B  comes to assist and they finish it in  2  more days.  How long would  B  take by himself?

No. 4, p. 207.  A man and his wife can drink a barrel in  15  days.  After  6  days the man leaves and the woman finishes it in  30  days.  How long would it take her to drink the whole barrel by herself?

No. 10, p. 208.  Similar to the last, with two workers and numbers  16,  4,  36.

No. 12, p. 208.  General solution for  (a, b, c).

Gerardus Beekman Docharty.  A Practical and Commercial Arithmetic: ....  Harper & Brothers, NY, 1854.  Many examples on pp. 166-167, 242-243, 247, including Simpson's XXXI;  the same problem with values  12, 20, 15;  (35/2, x)  in  40  and the following.

P. 167, no. 36.  A,  B  and  C  can do in  24;  A  and  B  can do in  48;  A  and  C  can do in  36.  How long for each separately?

Pp. 247-248, no. 64.  = Vyse, prob. 5, but only asks how long for all together, and gives solution in fractions.

James B. Thomson.  Higher Arithmetic; or the Science and Application of Numbers; ....  Designed for Advanced Classes in Schools and Academies.  120th ed., Ivison, Phinney & Co, New York, (and nine copublishers), 1862.  Lots of straightforward examples and the following.

Prob. 65, p. 397 & 422.  (15, x)  in  18.  Note that  x  has a negative value, i.e. is an outlet.  See also BR & Docharty.

Prob. 93, p. 398 & 422.  = Vyse, prob. 5, with solution in fractions.

Edward Brooks.  The Normal Mental Arithmetic  A Thorough and Complete Course by Analysis and Induction.  Sower, Potts & Co., Philadelphia, (1858), revised, 1863.  Further revised as:  The New Normal Mental Arithmetic: A Thorough and Complete Course by Analysis and Induction;  Sower, Potts & Co., Philadelphia, 1873.  Lots of examples.  I mention only those of some novelty to illustrate mid/late 19C texts.

1863 _ p. 122;  1873 _ pp. 140-141;  no. 16.  (6, 8, x)  in  3.

1863 _ p. 122;  1873 _ p. 141;  no. 18.  (A, B, C)  in  4;  (A, B)  in  6;  (B, C)  in  9.

1863 - pp. 122-123;  1873 _ p. 141;  no. 19.  (A, B, C)  in  6;  (A, B)  in  8;  B  in  12.

1863 _ p. 123;  1873 - p. 141;  no. 26.  (A, B, C)  can do in  6;  (A, B)  can do in  9;  all three work for  2  days, then  C leaves _ how long for  A, B  to finish?

1863 - p. 155, no. 2;  1873 _ p. 174, no. 9.  (A, B, C)  can do in  20;  (A, B)  can do in  30;  (B, C)  can do in  40;  all three work for  5  days, then  B  leaves _ how long for  A, C  to finish?

Boy's Own Magazine 2:2 (No. 8) (Aug 1863) 183  &  2:4 (No. 11) (Nov 1863) 367.  (Reprinted as (Beeton's) Boy's Own Magazine 3:8 (Aug 1889) 351  &  3:10 (Oct 1889) 431.)  Mathematical question 87.  Complicated version, typical of its time.  Bacchus drinks from a cask for  _  of the time it would take Silenus to drink the whole cask.  Silenus then finishes it off and the total time is two hours longer than if they had drunk together.  But if they had drunk together, Bacchus would only have drunk half as much as he left for Silenus.

[Robert Chambers].  Arithmetic.  Theoretical and Practical.  New Edition.  Part of: Chambers's Educational Course _ edited by W. & R. Chambers.  William and Robert Chambers, London and Edinburgh, nd, [1870 written on fep].  [Though there is no author given, Wallis 242 CHA is the same item, attributed to Robert Chambers, with 1866 on the fly-leaf, so I will date this as 1866? _??check in BMC.]

P. 263, quest. 4.  Person doing business, equivalent to a cistern of  8000  with inlets of  1500  and  1000  per year and a drain of  3000  per year.  Wehn is he broke?  This is not really a cistern problem since the rates are given rather than the times to fill or empty, but the format is sufficiently similar that I have included it here as an example of the more straightforward rate problems.  Also, the formulation with money is not common.

P. 266, quest. 44.  Cistern  (10, 8),  but the second pipe is not turned on until the cistern is half full.

Bachet-Labosne.  Problemes.  3rd ed., 1874.  Supp. prob. V, 1884: 187-188.  Poetic and complicated form.  In  3/10  of the time that Silenus would take to drink the whole amphora, Bacchus drinks  1/4  of what he leaves for Silenus to finish.  But if they drank it all together, they would finish it in two hours less than the previous time.

M. Ph. André.  Éléments d'Arithmétique (N^o 3) a l'usage de toutes les institutions ....  3rd ed., Librairie Classique de F.-E. André-Guédon, Paris, 1876.  Several straightforward problems and the following.  Prob. 99, p. 62.  Four companies of workers can do a job in  45, 9, 27, 36  days.  How long will it take  2/5  of the first company,  3/4  of the second company,  1/2 of the third company and  1/3  of the fourth company?

Fred Burnaby.  On Horseback Through Asia Minor.  Sampson Low, et al., London, 1877.  Vol. 1, pp. 208‑210.  One man can mow in  3  days,  the other in  4.  How long together?

William J. Milne.  The Inductive Algebra ....  1881.  Op. cit. in 7.E.

No. 106, pp. 142 & 332.  Asks for general solution of  (a, b)  and two specific cases.

No. 6, pp. 162 & 334.  (8, 9, 10).

No. 2, pp. 166 & 334.  General solution of  (a, b)  and one specific case.

William Briggs & George Hartley Bryan.  The Tutorial Algebra, based on the Algebra of Radhakrishnan _ Part II _ Advanced Course.  W. B. Clive, London, (1898), 1900.  Exercises X, prob. 6, pp. 123 & 579.  Two reapers,  (x, x‑5)  in  6.

A. Sonnenschein & H. A. Nesbit.  The New Science and Art of Arithmetic For the Use of Schools.  A. & C. Black, London, 1903.  This is a typical text of its time and has a number of variations on the basic problem.

Pp. 233-234 & 479, prob. 37.  Workers _  (10, 12, 9).  How long will it take to do  2½  tasks _ how much does each person do?

Pp. 234 & 479, prob. 38.  Workers:  man, woman & boy can do in  (5, 8, 12).  Man works  1¼  days, then is joined by the woman for  1½  days, then the remainder is left to the boy.  When does he finish and how much does each do?

Pp. 234 & 479, prob. 44.  Cistern of size  600  _  (5, 7).  How much does each pipe pass?

Pp. 234 & 479, prob. 45.  Workers _  (A, B)  in  6;  (A, C)  in  8;  (B, C)  in  9.

Pp. 235 & 479, prob. 54.  Cistern _  (4, 6, -4).

Pp. 288 & 484, prob. 42.  Two pumps and they can work at half or full speed.  If  (A, B/2)  in  5  and  (A/2, B)  in  4,  determine  A,  B  and  (A,B).

Pearson.  1907.  Part II, no. 162, pp. 145 & 223.  A  brings a pint every  3  minutes,  B  a quart every  5  minutes and  C  a gallon every  7  minutes.  How long to fill a  55  gallon drum and who finishes the job?

Stephen Leacock.  A, B, and C.  IN: Literary Lapses, (1910).  The book has been frequently reprinted and the piece has been widely anthologised.  It is pp. 237-245 in my 9th English ed.

Collins.  Fun with Figures.  1928.  According to Hoyle, not arithmetic, pp. 32-33.  "[I]f your father can build a chicken coop in  7  days and your Uncle George can build it in  9  days, how long ...."  "They'd never get it done;  they'd sit down and swap stories of rum runners, and bootleggers and hijackers."

C. Dudley Langford.  Note 1558:  A graphical method of solving problems on "Rate of Work" and similar problems.  MG 25 (No. 267) (Dec 1941) 304-307.  +  Note 2110:  Addition to Note 1558:  "Rate of Work" problems.  MG 34 (No. 307) (Feb 1950) 44.  Uses a graph to show  (a, b)  problems as meeting problems.  Also solves problems  (A, x)  in  B  and  (a, -b),  the latter appearing as an overtaking problem.  The Addition gives a clearer way of viewing  (a, ‑b)  problems as overtaking problems.

David Singmaster.  How long is a brick wall? (my title is: Three bricklayers).  Weekend Telegraph (30 May 1992) xxxii  &  (7 Jun 1992) xxx.  Al and Bill can build a wall in 12 days;  Al & Charlie in 15;  Bill & Charlie in 20.  How long does it take each individually and how long does it take all three together?  This is well known, but then I ask how can you determine integer data to make all the results come out integers?  Let  A,  B,  C  denote the amounts each can build in a day.  To make all the data and results come out as integers, we have to have all of  A,  B,  C,  A+B,  A+C,  B+C,  A+B+C  be fractions with unit numerators.  To combine them easily, imagine that all these fractions have been given a common denominator  d,  so we can consider  A = a/d,  B = b/d,  etc., and we want  a,  b,  c,  a+b,  a+c,  b+c,  a+b+c  to all divide  d.  We can achieve this easily by taking any three integers  a,  b,  c,  and letting  d  be the least common multiple of  a,  b,  c,  a+b,  a+c,  b+c,  a+b+c.  Taking  a, b, c = 3, 2, 1,  we find  d = 60  and the given problem is the simplest example with distinct rates  A,  B,  C.

 

          7.H.1. WITH GROWTH  _  NEWTON'S CATTLE PROBLEM

 

Isaac Newton.  Arithmetica Universalis, 1707.  ??NYS.  English version:  Universal Arithmetic, translated by Mr. Ralphson, with revisions and additions by Mr. Cunn, Colin Maclaurin, James Maguire and Theaker Wilder; W. Johnston, London, 1769.  (De Morgan, in Rara, 652‑653, says there were Latin editions of 1722, 1732, 1761 and Raphson's English editions of 1720 and 1728, ??NYS.)  Resolution of Arithmetical Questions, Problem 11, pp. 189‑191.  (Sanford 165 quotes from the 1728 English edition and it is the same as the 1769.)  "If the Number of Oxen  a  eat up the Meadow  b  in the time  c;  and the Number of Oxen  d  eat up as good a Piece of Pasture  e  in the Time  f,  and the Grass grows uniformly;  to find how many Oxen will eat up the like Pasture  g  in the Time  h."  Gives a general solution:  (gbdfh ‑ ecagh ‑ bdcgf + ecfga)/(befh - bceh)  and an example with  a, b, c;  d, e, f;  g, h  =  12, 3_, 4;  21, 10, 9;  24, 18.  One easily finds that the rate of grass growth per unit area per unit time is   G  =  (ace - bdf)/cf(bd - ae)   and the rate of grass consumption per ox per unit time is   E  =  be(c-f)/cf(bd - ae).   There are actually three unknowns since we also don't know the initial amount of grass per unit area,  G₀.  We can either take proportions of these or we can adopt a unit of grass such that the initial amount of grass per unit area is  1.  In the second case, the basic equation   b (G₀ + Gc)  =  aEc   becomes   b (1 + cG)  =  acE.

Walkingame.  Tutor's Assistant.  1751.  1777: p. 177, prob. 121;  1860: p. 185, prob. 114.  Identical to Newton.

Eadon.  Repository.  1794. 

Pp. 208-209, no. 6.  Same as Newton.  Gives a specific solution.  "This is deemed a curious and difficult question, it was first proposed by Sir Isaac Newton, in his Universal Arithmetic, and there universally solved by an algebraic process: but I have never seen a numerical solution independent of algebra, except this of my own; ...."  He then states the general solution as  [bdfg(h-c) + aceg(f-h)]/beh(f-c). 

Pp. 209-210, no. 7.  "If  3  oxen or  5  colts can eat up  4 1/5  acres of pasture in  7  weeks, and  5  oxen and  3  colts can eat up  9  acres of like pasture in  10  weeks, the grass gowing [sic] uniformly;  how many sheep will eat up  48  acres in  20  weeks, supposing  567  sheep to eat just as much as  6  oxen and  11  colts?"  Answer is  1736.  He introduces heifers, where a heifer is  1/5  of an ox or  1/3  of a colt, which brings the problem into Newton's form with values  a, b, c;  d, e, f;  g, h  =  15, 4 1/5, 7;  34, 9, 10;  48, 20,  which gives the answer in terms of heifers, which he converts to sheep.

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Arithmetical puzzles, no. 6, p. 174 (1868: 185): Sir Isaac Newton's problem.  Gives the numerical values from Newton, but  3_  is often mis-set as  3½,  and when so done, is identical in all three editions.

M. Ph. André.  Éléments d'Arithmétique (N^o 3) a l'usage de toutes les institutions ....  3rd ed., Librairie Classique de F.-E. André-Guédon, Paris, 1876.  a, b, c;  d, e, f;  g, h  =  3, 2, 2;  2, 2, 4;  6, 6.  No answer.

G. H. Mapleton, proposer;  Charles Hammond, solver.  Arithmetical problem.  Knowledge 1 (30 Dec 1881) 191  &  (20 Jan 1882) 258, item 9.  a, b, c;  d, e, f;  g, h  =  12, 10, 16;  18, 10, 8;  40, 6.  Cites Newton, 1722 ed., p. 90.

Charles Pendlebury.  Arithmetic.  Bell, London, (1886), 30th corrected and expanded printing, 1924.  Section XLIV: Pasture with growing grass, pp. 336q-336s & answers, part II, p. xv.  (These seem to be added pages, so the date is uncertain.)  This section carefully works an example to answer different questions, then gives 12 problems.  In all cases, all the fields have the same size, which much simplifies things. 

The example problem has  a, b, c;  d, e, f;  g, h  =  70, b, 24;  30, b, 60;  b, 96. 

Prob. 12.  A cistern has a number of equal holes in its base and a pipe is adding water.  When 10 holes are open, the cistern will empty in 20 minutes.  When 8 are open, it empties in 35 minutes.  How long will it take with 12 holes open?  This is equivalent to  a, b, c;  d, e, f;  g, h  =  10, b, 20;  8, b, 35;  b, 12. 

I have read an article (in HM??) which described this as a common problem in 19C textbooks.

Walter Percy Workman.  The Tutorial Arithmetic.  University Tutorial Press, (1902);  2nd ed., 1902.  [There is a 3rd ed., (c1908), c1928, which contains a few more pages.]  Section IX [= Chap. XXXI in the 1928 ed.], examples CXLV, prob. 59‑60, pp. 430 & 544 [= 436 & 577 in the 1928 ed.]. 

Prob. 59.  a, b, c;  d, e, f;  g, h  =  15, b, 8;  9, b, 16;  b, 12  _ this has all the fields being the same.

Prob. 60 is more complex.  The field is divided into two equal halves.  50  oxen eat all of one half in  4  weeks.  Four oxen are slaughtered and the rest are put in the other half.  After  6  weeks,  3  oxen are slaughtered.  In another week, all the grass is eaten up and the oxen are sold.  After another week, the division is removed and  85  oxen are put in the field.  How long will it last them?

Wehman.  New Book of 200 Puzzles.  1908.  An ox problem, p. 55.  Gives Newton's problem with just the numercial values.

Loyd.  Cyclopedia, 1914, pp. 47 & 345.  = MPSL1, prob. 48, pp. 46 & 138‑139.  Cow, goat and goose.

Wood.  Oddities.  1927.  Prob. 71: Ox and grass, p. 55.  a, b, c;  d, e, f;  g, h  =  6, 10, 16;  18, 10, 8;  40, 6.   Gets  88,  but is confused about the growing of the grass _ Newton's formula gives  104.  Says Newton divides the oxen into those that eat the accumulated grass and those that eat the increase, but he doesn't apply this correctly.  Indeed, for this data, the grass grows at a negative rate!  Undoubtedly intended to be the data of Mapleton, 1881, for which the answer is  88.

Perelman.  MCBF.  1937.

Cows in the meadow.  Prob. 139, pp. 229-234.  Same as Pendlebury's example.

Newton's problem.  Prob. 140, pp. 234-235.  Same data as Newton.

A. I. Ostrovsky.  Oxen grazing in a field.  MG 50 (No. 371) (Feb 1966) 46‑48.  Quotes Newton and gives a graphical solution which converts this into an overtaking problem, where the grass starts growing  12  weeks before the cows are put in.

John Bull.  Grazing Oxen.  M500 165 (Dec 1998) 1-4.  He is unhappy with some of the limiting situations and proposes a different basic equation.  However, his unhappiness is really due to not understanding the basic equation properly.

 

          7.H.2. DIVISION OF CASKS

 

          NOTATION:  (a, b, c)  among  n   means to divide  a  full,  b  half‑full and  c  empty casks among  n  people so that each has an equal amount of contents and of casks.

          Dividing  kn  casks containing  1, 2, ..., kn  among  k  people so each gets the same amount of contents and of casks  _  see:  Albert;  Munich 14684;  AR;  Günther (1887);  Singmaster (1998).

          See Tropfke 659.

 

Alcuin.  9C.

Prob. 12: Propositio de quodam patrefamilias et tribus filius ejus.  (10, 10, 10)  among  3.  This has  5  solutions _ he gives just 1:  0, 10, 0;  5, 0, 5;  5, 0, 5.

Prob. 51: Proposito de vino in vasculis a quodam patre distributo.  Divide  4  casks containing  10, 20, 30, 40  among  4  _ solution involves shifting contents, so this is not really a problem of the present type.

Abbot Albert.  c1240.  Prob. 3, p. 333.  Divide  9  casks containing  1, 2, ..., 9  among three.  He gives only one of the two solutions:  1, 5, 9;  2, 6, 7;  3, 4, 8.  See Singmaster for a generalization.

BR.  c1305.  No. 40, pp. 58‑61.  300  ewes,  100  each with  1, 2, 3  lambs, to be divided among three sons so each son has the same number of ewes and lambs and no lamb is separated from its mother.  This is the same as  (100,100,100)  among  3.  He gives one solution:  0, 100, 0;  50, 0, 50;  50, 0, 50.  [There are  234  solutions!]

Munich 14684.  14C.  Prob. XXIV, f. 32r.  Same as Abbot Albert and with the same solution.

AR.  c1450.  Prob. 351, pp. 154, 182.  Same as Abbot Albert, with the same solution, but arranged in columns.  Vogel comments that this makes a 'half‑magic square' and cites Günther, 1887, as having already noted this.

Tartaglia.  General Trattato, 1556, art. 130‑131, p. 255v. 

Art. 130:  (7, 7, 7)  among  3.  Gives one of the two solutions:  3, 1, 3;  3, 1, 3;  1, 5, 1.

Art. 131:  (9, 9, 9)  among  3.  Gives one of the three solutions:  4, 1, 4;  3, 3, 3;  2, 5, 2.

Bachet.  Problemes.  1612.  Addl. prob. IX: Trois hommes ont à partager 21 tonneaux ...., 1612: 161‑164;  1624: 233-236;  1884: 168‑171.

(7, 7, 7)  among  3 _ gives all two solutions.  1612 cites Tartaglia. 

(9, 9, 9)  among  3 _ gives all three solutions.  1612 cites Tartaglia. 

(8, 8, 8)  among  4 _ Bachet erroneously does  (6, 12, 6)  among  4,  but the editor gives all four solutions of the original problem.

Labosne adds:  (5, 11, 8)  among  3  _ giving all three solutions.

          (Ahrens, A&N, 29, says that the 1st ed. also does  (5, 5, 5)  among  3.)

van Etten.  1624.  Prob. 89 (86), part IV, pp. 134‑135 (213).  (7, 7, 7)  among  3.  One solution:  3, 1, 3;  3, 1, 3;  1, 5, 1.

Ozanam.  1725.  Prob. 44, 1725: 242‑246.  Prob. 24, 1778: 182-184;  1803: 180-182;  1803: 158-159;  1814: 158-159;  1840: 81-82. 

(7, 7, 7)  among  3  _ gives both solutions.  Notes that this cannot be divided among  4  persons because  4  does not divide  21.

1725 has a very confusing attempt at  (8, 8, 8)  among  4,  which is done as though it were  (6, 12, 6)  among  4,  and he seems to think half‑empty is different than half‑full!!  1778 onwards just do  (8, 8, 8)  among  3,  giving  3  of the  4  solutions, omitting   4, 0, 4;   4, 0, 4;   0, 8, 0.

(9, 9, 9)  among  3  _ gives all  3  solutions.

Les Amusemens.  1749.  Prob. 15, p. 137: Les Tonneaux. 

(7, 7, 7)  among  3  _ gives both solutions. 

(11, 11, 11)  among  3  _ gives  2  of the  4  solutions.

Bestelmeier.  1801.  Item 717: Die sonderbare Weintheilung unter 3 Erben.  Says there are  21  casks, so presumably  (7, 7, 7)  among  3.

Jackson.  Rational Amusement.  1821.  Arithmetical Puzzles.

No. 11, pp. 3 & 53.  (7, 7, 7)  among  3.  Both solutions.

No. 51, pp. 11 & 66.  (8, 8, 8)  among  3.  Gets three of the four solutions, omitting  4, 0, 4;  4, 0, 4;  0, 8, 0.

Endless Amusement II.  1826?  Prob. 20, pp. 199-200.  (7, 7, 7)  _ two solutions.

Young Man's Book.  1839.  Pp. 239-240.  Identical to Endless Amusement II.

Walter Taylor.  The Indian Juvenile Arithmetic ....  Op. cit. in 5.B.  1849.  P. 211.  (7, 7, 7)  with both solutions.

Magician's Own Book.  1857.  The wine and the tables, p. 225.  (7, 7, 7),  (8, 8, 8),  (9, 9, 9)  among  3  _ gives two solutions for each.  = Boy's Own Conjuring Book, 1860, p. 195. 

Leske.  Illustriertes Spielbuch für Mädchen.  1864?  Prob. 593, pp. 299 & 410: Sechs Knacknüsse _ part 1.  (7, 7, 7)  _ one solution shown diagrammatically:    3, 1, 3;  3, 1, 3;  1, 5, 1.

Siegmund Günther.  Geschichte des mathematischen Unterrichts im deutschen Mittelalter bis zum Jahre 1525.  Monumenta Germaniae Paedagogica III.  1887.  Facsimile reprint by Sändig Reprint Verlag, Vaduz, Liechtenstein, 1969.  He discusses Abbot Albert and his problem on pp. 35-36, noting that the solution can be viewed as a set of lines in a magic square so that the perpendicular lines give a second solution, but that magic squares were then unknown in Europe.  He gives no other examples.

Lucas.  L'Arithmétique Amusante.  1895.  Prob. XV: Jeux de tonneau, pp. 50-51.  (7, 7, 7)  among  3.  Gives both solutions.  Notes that  half empty = half full,  so doubling gives us  empty = full!

Ahrens.  A&N.  1918.  Pp. 29‑33. 

Gives all solutions for  (n, n, n)  among  3  for  n  =  5 (1) 10;  (8, 8, 8)  among  4  &  6;  (11, 5, 8)  among  3;  (5, 11, 8)  among  3;  (4, 12, 8) among  3.

McKay.  Party Night.  1940.  No. 14, p. 178.  (7, 7, 7)  among three.  One solution.

M. Kraitchik.  Mathematical Recreations, 1943, op. cit. in 4.A.2, chap. 2, prob. 34, pp. 31‑32.  9  full,  9  three-quarter full,  9  half,  9  quarter full,  9  empty  among  5.

David Singmaster.  Triangles with integer sides and sharing barrels.  CMJ 21 (1990) 278‑285.  Shows the number of ways of sharing  (N, N, N)  among  3  is the same as the number of integer sided triangles of perimeter  N.  Shows this is the number of partitions of  N/2  or  (N‑3)/2  into  3  parts.  Finds necessary and sufficient conditions for sharing  (a, b, c)  among  k  people.

David Singmaster.  Fair division of the first  kn  integers into  k  parts.  Written in 1998.  This generalizes the problem of Abbot Albert and determines that there is a partition of the first  kn  integers into  k  sets of  n  values with each set having the same sum if and only if  n  is even  or  n > 1  and  k  is odd.  I call this a fair division of the first  kn  integers into  k  parts.  The number of such divisions is large, but might be worth examining.

 

          7.H.3. SHARING UNEQUAL RESOURCES  _  PROBLEM OF THE

                                        PANDECTS

 

          NOTATION:  (a, b, ...; c)  means persons contribute  a, b, ...   which is shared equally among themselves and an extra person who pays  c.  See Clark for a variant formulation with the same result.

          See McKay; Party Night; 1940 in 7.H.5 for a related form.

 

          INDEX  of  (a, b, ...; c)  problems with  a £ b £ ....

 

  2            3          3            Kraitchik

  2            3          5            Fibonacci;  al-Qazwînî;  Bartoli;  Calandri

  2            3        35            McKay

  3            4          4            Kraitchik

  3            4          7            Kraitchik

  3            4        14            Gherardi

  3            5          5            P. M. Calandri

  3            5          8            Jackson;  Badcock;  Magician's Own Book;  Bachet-Labosne;  Mr. X;  Pearson;  Clark;  Kraitchik;  Rohrbough;  Sullivan

  3            5        10            AR

  3            5        80            Kraitchik

  4            5          4            dell'Abbaco

  4            6        10            Mittenzwey

  5            7        12            Kraitchik

  7            8        30            Kraitchik

31          50        40            Tagliente

90         120        70            Kraitchik

 

11          14        17        42            Kraitchik

 

  2            3          6          9          4            Kraitchik

 

Fibonacci.  1202.  p. 283.  (3, 2; 5).

Qazwini  = Zakariyâ ibn Muhammad ibn Mahmûd abû Yahya al-Qazwînî.  (= al-Kazwînî  = Zakariyy_' b. Muhammad b. Mahm_d Ab_ Yahya al-Kazw_n_).  (Kitâb) ‘Ajâ’ib al-Makhlûqât wa Gharâ’ib al-Mawjûdât (= Adj_yib al-Makhl_k_t wa Gh_r_'ib al-Mawdj_d_t) ((The Book of the) Wonders of the Creation and Unique [Phenomena] of the Existence  = Prodigies of Things Created and Miraculous Aspects of Things Existing  = The Wonders of Creation and the Peculiarities of Existing Things  = The Cosmography).  c1260.  ??NYS _ the earliest dated copy, of 1458, and several others are in the Wellcome Institute.  Part 8: On the arts;  chap. 9: On reckoning.  In:  J. Ruska; Kazw_n_studien; Der Islam 4 (1913) 14‑66 & 236‑262.  German translation (Arabic omitted) of this problem on pp. 252‑253.  (3, 2; 5).  Story says one proposes a  3 : 2  split, but  4 : 1  is found to be correct.  [Qazwini also wrote a Geography, in two editions, and its titles are slightly similar to the above.  I previously had reference to the Arabic titles of the other book, but rereading of Ruska and reference to the DSB article shows the above is correct.]

Gherardi.  Libro di ragioni.  1327.  Pp. 40‑41: Chopagnia.  (3, 4; 14).

Dell'Abbaco.  c1370. Prob. 94, p. 81 with plate on p. 82.  (5, 4; 5).  I have a colour slide of this.

Bartoli.  Memoriale.  c1420.  Prob. 26, ff. 77v-78r (= Sesiano, pp. 144 & 149).  (2, 3; 5),  correctly solved.

AR.  c1450.  Prob. 212, p. 98.  (5, 3; 10)  correctly solved.  (Unusually, Vogel's notes, pp. 160‑161 & 211‑213, say nothing about this problem.)

P. M. Calandri.  c1480.  P. 106.  Part of the text is lacking, but it must be  (3, 5; 5).

Calandri.  Arimethrica.  1491.  F. 63v.  (2, 3; 5).

Tagliente.  Libro de Abaco.  (1515).  1541.  Prob. 130, ff. 60v-61r.  Philippo and Jacomo share lunch with Constanzo _  (50, 31; 40).

Ghaligai.  Practica D'Arithmetica.  1521.  Prob. 27, ff. 66r-66v.  Three men have  3, 2, 1  loaves of bread and other foods worth  8, 6, 4.  A fourth comes and shares with them, paying  9.  How much should each of the three get?  The total value of the food is  4 x 9 = 36,  so the bread is worth  36 - 8 - 6 - 4 = 18,  or  3 per loaf.  So the first should get  9,  the second  3  and the third owes them  2 !!

Cardan.  Practica Arithmetice.  1539.  Chap. 66, section 16, ff. CC.iv.v - CC.v.v (pp. 140-141).  Three men with bread, wine and fish in the amounts:  3, 4, 0;  0, 5, 6;  2, 0, 7;   which are considered of equal value.  A fourth man with one bread comes and they share the meal and the fourth man pays  5.

Tartaglia.  General Trattato.  1556.  Book 12, art. 33, pp. 199r-199v.  Three men have quails and bread.  The first has  6  quails and  2s  worth of bread;  second has  4  quails and  3s  worth of bread;  third has  2  quails and  5s  worth of bread.  They share with a fourth person who pays  8s.

Buteo.  Logistica.  1559. 

Prob. 3, pp. 201-202.  Four share their food.  First has  4  breads and  20  carrots;  second has  1  bread and  32p  of wine;  third has  7  breads and  8  carrots;  fourth has a cheese.  If all have equal value, what is the value of each item?

Prob. 4, pp. 202-203.  Three share their food with a fourth.  First has  2  breads and  7  nummos worth of fish;  second has  4  breads and  5  nummos worth of condiments;  third has  1  bread and  8  nummmos worth of wine.  Fourth pays  12  nummos for his share.

Jackson.  Rational Amusement.  1821.  Arithmetical Puzzles, no. 12, pp. 3-4 & 53-54.  (3, 5; 8).  Says it appears in an Arabian manuscript.  The man with five loaves divides  5, 3;  the other protests and divides  4, 4;  judge divides  7, 1.  Why?  = Magician's Own Book (UK version), 1871, Arithmetical paradox, pp. 28-29.

John Badcock.  Domestic Amusements, or Philosophical Recreations.  Op. cit. in 6.BH.  [1823].  Pp. 186-187, no. 255: Arithmetical paradox.  Same as Jackson, saying it appears in an Arabic manuscript.

Magician's Own Book.  1857.  The three travellers, pp. 225-226.  (3, 5; 8).  = Boy's Own Conjuring Book, 1860, pp. 195‑196.

Edward Brooks.  The Normal Mental Arithmetic  A Thorough and Complete Course by Analysis and Induction.  Sower, Potts & Co., Philadelphia, (1858), revised, 1863.  Further revised as:  The New Normal Mental Arithmetic: A Thorough and Complete Course by Analysis and Induction;  Sower, Potts & Co., Philadelphia, 1873.  Several ordinary examples and some unusual examples.

1863 _ p. 146, no. 17;  1873 _ p. 154, no. 15.  (6, 10; 16)  but the third person eats  4  more than each of the other two eat.

1863 _ p. 146, no. 19;  1873 _ p. 155, no. 17.  (5, 9; 24)  but  C  eats twice as much as  B,  who eats twice as much as  A.

Bachet-Labosne.  Problemes.  3rd ed., 1874.  Supp. prob. I: Deux Arabes allaient dîner: ..., 1884: 181.  (3, 5; 8).

Mittenzwey.  1879?  Prob. 67, pp. 16 & 63.  (6, 4; 10).

Hoffmann.  1893.  Chap IV, no. 74: The three Arabs, pp. 165 & 220.  (3, 5; 8).  First says to divide  3, 5;  second says  4, 4;  third says both are wrong.

Mr. X.  His Pages.  The Royal Magazine 10:6 (Oct 1903) 530-531.  The stranger's dinner.  (3, 5; 8).  Refers to Arabians.  Similar to Jackson.

Pearson.  1907.  Part II, no. 138, pp. 141 & 218.  (3, 5; 8).

Ball‑FitzPatrick.  Footnote cited in 7.G.1.  1908.  Says the problem is Arabic, but gives no reason.

Clark.  Mental Nuts.  1916: no. 22.  Real estate.  A invests  $5000,  B invests  $3000.  They buy three houses of equal value.  They each take one and then sell the third for  $8000.  How do they divide the money?  Answer is  $7000  and  $1000.  His  1904: no. 25;  1916: no. 39  is a atandard version of  (3, 5; 8)  with sandwiches.

Kraitchik.  La Mathématiques des Jeux.  Op. cit. in 4.A.2.  1930.  Chap. 1, pp. 7‑8: Problème des Pandects.  He gives several examples and says they come from Unterrichtsblätter für Mathematik und Naturwissenschaften 11, pp. 81‑85, ??NYS.

No. 22:  (3, 5; 8).

No. 23:  (2, 3; 3).

No. 24:  (11, 14, 17; 42).

No. 25:  (5, 7; 12).

No. 26:  (7, 8; 30).  Caius and Sempronius share  7  and  8  with Titus who paid them  14  and  16.  Sempronius protests and a judge divides it as  12  and  18.  "C'est probablement cette version qui a donné à ce problème le nom de celui de Pandectes."

No. 27:  (3, 5; 80).

No. 28:  (3, 4; 7)  _ cultivating fields.

No. 29:  (90, 120; 70)  _ digging a ditch.

No. 30:  (3, 4; 4).

No. 31:  (2, 3, 6, 9; 4)  _ heating a workshop.

Rohrbough.  Brain Resters and Testers.  c1935.  The Travelers' Dinner, pp. 23-24.  Arabs,  (3, 5,; 8).

McKay.  Party Night.  1940.  No. 25, p. 182.   A  &  B  give a party and invite  2  and  3  guests.  The party costs  35s  _ how do they divided the expense?  Initial reaction is in the ratio  2 : 3,  but it should be  3 : 4.  (Also entered in 7.H.5)

Kraitchik.  Mathematical Recreations.  Op. cit. in 4.A.2.  1943.  The problem of the Pandects, pp. 28‑29.  c= No. 26 of Math. des Jeux.

Sullivan.  Unusual.  1943.  Prob. 19: An Arab picnic.  (3, 5; 8).

 

          7.H.4. EACH DOUBLES OTHERS' MONEY TO MAKE ALL EQUAL,

                                        ETC.

 

          See 7.R for related problems.

          See Tropfke 647‑648.

 

Diophantos.  Arithmetica.  c250.  Book I.

No. 22, p. 138.  "To find three numbers such that, if each give to the next following a given fraction of itself, in order, the results after each has given and taken may be equal."  Illustrates with fractions  1/3, 1/4, 1/5. 

No. 23, pp. 138‑139, is the same with four numbers, illustrated with fractions  1/3, 1/4, 1/5, 1/6.

Mahavira.  850.  Chap. VI, v. 259‑267, pp. 160‑162.

260:  3  men, each doubles the next to make all equal.

262:  same with  5  men.

263:  each  3/2's  the next.  Also each  5/3's  the next.

266:  each doubles others.  Also each  3/2's  others.

Fibonacci.  1202.  Pp. 287‑293 gives several versions.

Pp. 287‑288.  w  5/2's  x;  x  7/3's  y;  y  9/4's  z;  z  11/5's  x  and then all are equal.  Answer:  5862, 2858, 2760, 2380  and he notes that these can be multiplied by any value.

Pp. 288‑291.  Same operations, but the results are in the proportion  5 : 4 : 3 : 2.  Answer:  22875, 10000, 8355, 7280,  and he also divides these by 5.

Pp. 291‑292.  w  5/2's  all the others, etc. with the above ratios, then all are equal.  Answer:  8436, 3288, 1440, 696.

Pp. 292‑293.  Same as the previous, but with results in proportion  5 : 4 : 3 : 2.  Answer:  29706, 11568, 498, 2256. (He erroneously has  29826  for the first value.)

Fibonacci.  Flos.  c1225.  In Picutti, pp. 326-332, numbers VIII-X.

Pp. 243-244: De quatuor hominibus qui invenerunt bizantios.  First man doubles the second's money,  then the second man triples the third's,  ...,  to make all equal.  Answer:  89, 77, 47, 27  with total of  240.

Pp. 244-245: Above continued.  First doubles the others,  second triples the others,  ...,  to make all equal.  Answer:  241, 161, 61, 17  with total  480.

Pp. 246-247: Questio similis suprascripte de tribus hominibus.  First  5/2's  the others,  second  10/3's  the others,  third  17/4's  the others to make all equal.  Answer:  1554, 738, 258  with total  2550  and he then divides through by  6.

Lucca 1754.  c1390.  Ff. 61v‑62r, p. 142.  4 & 3 men.  Each doubles the others to make all equal.  In the 4 man case, he specifies the total money is 400.

Giovanni di Bartolo.  Op. cit. in 7.H.  c1400.  Prob. 9, pp. 17‑18.  Four men, each doubles the others' money and the product of the results is 1000.  He assumes, for no clear reason, that the original amounts are proportional to  8 : 4 : 2 : 1.

AR.  c1450.  Prob. 231, pp. 107‑108 & 169‑171.  Two men, each doubles the others' money and then both have  13½.

Muscarello.  1478.  Ff. 78r-78v, pp. 193-194.  Four men, each doubles the others' money, then all are equal.  Answer:  33, 17, 9, 5.

Chuquet.  1484.  Prob. 148.  3 people.  Mentioned in passing on FHM 230.

Pacioli.  Summa.  1494. 

F. 105r, prob. 16.  First wins  1/2  of the second's;  second wins  1/3  of the third's;  third wins  1/4  of the first's to make all equal  100.  I get  (200, 400, 300)/3.  See Tonstall for corrections and intended interpretation.  Pacioli gives no working and just states answers that are printed differently in the two editions, but from Tonstall we see that they are intended to be:  55 5/9 (given as  44 4/9  and as  144 4/9),  111 1/9,  133 1/3.

F. 189r, prob. 6.  First gives  7/12  of his money to the second, who then gives  11/30  of his money to the first, when both are equal.  He gives no working and just one answer:  70/9,  658/95.  I find the general answer is  47x = 48y.

Calandri.  Aritmetica.  c1500.  Ff. 101r-102r, pp. 202‑204.  6 men.  Each doubles the others to make all equal.

Tonstall.  De Arte Supputandi.  1522.  P. 245.  First wins  1/2  of the second's;  second wins  1/3  of the third's;  third wins  1/5  of the first's to make all equal  100.  This is a correct version of Pacioli.  He gives  (500, 1000, 1200)/9  and shows the calculation which implies all three calculations are done at once _ that is, the  1/5  of the first's money is based on what he had to start, not what he has after winning from the second.

Cardan.  Practica Arithmetice.  1539.  Chap. 66, section 91, ff. GG.viii.v - HH.i.v (pp. 165‑166).  A situation somewhat similar to 7.P.7, where the first two take money leaving the third with  5.  Friend says the first is to give  10  and  _  of what he has left to the second and the second is then to give  7  and  ¼  of what he has left to the third to make their amounts proportional to  (3, 2, 1).  Answer is  x = 172,  y = 39  and the total sum is  216.

Tartaglia.  General Trattato, 1556, art. 11, 18, 37, 38, pp. 240v, 241v, 244v‑245r.  2 and 3 people versions.

Buteo.  Logistica.  1559.  Prob. 63, pp. 270-273.  Three players _ first wins  1/2  of second's,  second wins  1/3  of third's,  third wins  1/4  of what the first had originally,  and all wind up with  100.  That is, we have  x + y/2 ‑ x/4  =  y ‑ y/2 + z/3  =  z ‑ z/3 + x/4  =  100.

Bachet.  Problemes.  1612.  Addl. prob. VIII, 1612: 154-160;  1624: 226‑233;  1884: 162‑167.  3  people;  also with tripling.  Labosne adds the general case.

van Etten.  1624.  Prob. 57 (52), pp. 52‑53 (78).  3 people version used as a kind of divination.

Ozanam.  1694.  Prob. 26, 1696: 81;  1708: 72.  Prob. 47, 1725: 253.  Prob. 17, 1778: 209;  1803: 204.  Prob. 16, 1814: 177;  1840: 91.  Three person version, resulting in each having 8, used as a kind of divination.

Euler.  Algebra.  1770.  I.IV,IV.616: Question 4, pp. 211‑212.  Three players, all winding up with 24.

Hutton.  A Course of Mathematics.  1798?  Prob. 42,  1833: 223;  1857: 227.  Five players, all ending up with £32.

Thomas Grainger Hall.  The Elements of Algebra: Chiefly Intended for Schools, and the Junior Classes in Colleges.  Second Edition: Altered and Enlarged.  John W. Parker, London, 1846.  Pp. 130-131, ex. 16.  Each  3/2's  others, with total given as 162.  Cf Mahavira 266.

G. Ainsworth & J. Yeats.  A Treatise on the Elements of Algebra.  H. Ingram, London, 1854.  Exercise XXXVIII, pp. 81‑83 & 178.

No. 14: A  doubles  B,  B  doubles  A,  A  doubles  B,  to make both have  80.  How much originally?  Answer is  100, 50  instead of  110, 50.

No. 15: same as no. 14, with one more stage to make both equal  n.

No. 28: usual problem with four gamblers, ending with  64.

No. 29: usual problem with  7  baskets of apples, all ending with  128.

No. 30: usual problem with  n  persons, all ending with  a.  Solution is badly misprinted _ the  i-th should start with an amount  a(2^n-i + 1)/n×2ⁿ.

Boy's [Own] Magazine 2:5 (No. 11) (Nov 1863) 459 [answer would be in Jan 1864, ??NYS].  (Reprinted as (Beeton's) Boy's Own Magazine 3:11 (Nov 1889) 479.   Mathematical question 137.  Three boys playing.  Each pays the winner half of what he has.  Each one wins once and then they have  30d, 60d, 120d.  How much did they have to start?  [In fact, they had the same amounts.  In general, if they start with  1, 2, 4,  then the amounts after the first, second, third wins are:  4, 1, 2;  2, 4, 1;  1, 2, 4.]

Mittenzwey.  1879?  Prob. 131, pp. 25 & 75.  = Ainsworth & Yeats, no. 14, except he doesn't say what the final result is.  Solution is  110, 50.

Lucas.  L'Arithmétique Amusante.  1895.  Prob. XLIII: Les jouers, pp. 184-185.  Three players winding up with 12 each.  Does general solution for  n  players winding up with  a  each.

Workman.  Op. cit. in 7.H.1.  1902.  Section IX (= Chap. XXXI in c1928 ed.), examples CXLV, prob. 45, pp. 428 & 544 (434 & 577 in c1928 ed.).  Version with five people.

Hermann Schubert.  Beispiel‑Sammlung zur Arithmetik und Algebra.  3rd ed., Göschen, Berlin, 1913.  Section 17, no. 107, pp. 66 & 140.  Three people.  Each gives away half his money to be equally shared by the others, and then they all have 8.  Solution:  4, 7, 13.

Loyd.  Cyclopedia.  1914.  Sam Loyd's Mystery Puzzle, pp. 226 & 369.  (= MPSL2, prob. 85, pp. 61 & 148.  = SLAHP: An initiation fee, pp. 63 & 109.)  3 people version, resulting in the first person having lost 100.

Collins.  Book of Puzzles.  1927.  The five gamblers puzzle, p. 75.  They all end up with $32.

 

          7.H.5. SHARING COST OF STAIRS, ETC.

 

          See Tropfke 529.  The simplest form is given by Sridhara, BR, dell'Abbaco, Gori, von Schinnern.

 

Mahavira.  850.  Chap. VI, v. 226‑232, pp. 151‑153.

227.  Porter carrying  32  jack‑fruits over distance  1  will receive  7½  of the fruits.  He breaks down at distance  ½.  How much is he due?  Rule says  x/32*1 = (7½ ‑ x)/(32 ‑ x)(1 ‑ ½),  i.e. the wages per fruit‑mile should be the same for both parts of the journey.  (Properly, this problem leads to an exponential.)

229.  Porter carrying  24  jack‑fruits for distance  5  will earn  9  of them.  Two porters share the work, the first earns  6  and the second earns  3.  How far did the first one carry the fruits?

232.  Twenty men are to carry a palanquin a distance of  2  for wages of  720.  But two men drop off after distance  ½;  three more drop off after another  ½  and five more drop off after half the remaining distance.  How much does each earn?  He says each quarter of the distance is worth  180  and divides this equally among the carriers for each quarter.

Sridhara.  c900. 

V. 67(ii), ex. 84‑85, pp. 53 & 95.  Porter carrying  200  palas of oil for  5  panas wages.  But the bottle leaks and only  20  palas remain at the end.  How much should he be paid?  Rule says to pay  (20 + 180/2)/200  of the wages.

V. 68, ex. 86‑90, pp. 53‑55 & 95.

  Ex. 86‑87.  Four men watch a dance for  ¼, ½, ¾  and all of a day.  The dancers' fee is  96.  How much should each pay?  He charges  24/4  per watcher for the first quarter,  24/3  per watcher for the second quarter,  ..., giving payments  6, 14, 26, 50.

  Ex. 88.  Ten men are to carry a palaquin a distance of  3  for wages  100.  Two men drop off after distance  1  and another three after a total distance of  2.  How much does each earn?  Divides as in Mahavira's 232.

  Ex. 89.  Five chanters perform  1, 2, 3, 4, 5  chants for a fee of  300.  How much does each earn?  Each chant earns  60,  divided among the chanters of it.

V. 70‑71, ex. 92‑94, pp. 56‑58 & 96.

  Ex. 92.  Porter carrying  24  jack‑fruits for distance  5  will earn  9.  What does he earn for carrying distance  2?

  Ex. 93‑94.  Porter carrying  24  jack‑fruits for distance  5  will earn  9  of them.  Two porters split the carrying, the first earns  4  and the second earns  5.  How far did they carry?  Solutions of these are based on the rule in Mahavira 227.

BR.  c1305.  No. 36, pp. 54‑57.  Divide oil among  12  lamps which are to burn  1, 2, ..., 12  hours.

Dell'Abbaco.  c1370.  Prob. 92, pp. 78‑81.  House rented to  1  person the first month,  who shares with a  2nd the  2nd month,  who share with a  3rd the  3rd month,  ...,  who share with a  12th the  12th month.  How much does each pay?  He says many obtain  12/78,  11/78,  ...,  1/78,  but that it is more correct if the first pays  (1 + 1/2 + 1/3 + ... + 1/12) * 1/12,  the second pays  (1/2 + 1/3 + ... + 1/12) * 1/12,  ...,  the  12th pays  1/12 * 1/12.

Gori.  Libro di arimetricha.  1571.  F. 73r (p. 80).  Same as Prob. 92 of dell'Abbaco, but with only the first solution.

Bullen.  Op. cit. in 7.G.1.  1789.  Chap. 38, prob. 32, p. 243.  Four men hire a coach to go  130  miles (misprinted as  100).  After  40  miles, two more men join.  How much does each pay?

Clemens Rudolph Ritter von Schinnern.  Ein Dutzend mathematischer Betrachtungen.  Geistinger, Vienna, 1826, pp. 14‑16.  Discusses general problem of sharing a cost of  n  for lighting a  x  floor staircase.  Does the case  n = 48,  x = 4,  getting  3, 7, 13, 25,  which are the same proportions as Sridhara, ex. 86‑87.

Dana P. Colburn.  Arithmetic and Its Applications.  H, Cowperthwait & Co., Philadelphia, 1856.  Miscellaneous Examples No. 66, p. 365.  A and B hire a horse and carriage for $7 to go 42 miles and return.  After 12 miles, C joins them, and after 24 miles, D joins them.  How much should each pay?  No solution is given, but there is a Note after the question saying there are two common ways to do the allocation.  First is in proportion to the miles travelled, so  A : B : C : D = 42 : 42 : 30 : 18.  Second is to divide the cost of each section among the number of riders, which here gives  A : B : C : D = 29 : 29 : 17 : 9

Clark.  Mental Nuts.  1904: no. 71;  1916: no. 14.  The livery team.  I hire a livery team for $4 to go to the next city, 12 miles away.  At the crossroads 6 miles away, I pick up a rider to the city who then rides back to the crossroads.  How much should he pay?  Answer is $1.

Adams.  Puzzle Book.  1939.  Prob. C.135: Answer quickly!, pp. 158 & 187.  Man hires car to go to theatre.  He picks up and drops off a friend who lives half way to the theatre.  How do they divide the fare?  Answer is  3 : 1.

Depew.  Cokesbury Game Book.  1939.  Passenger, p. 219.  Similar to Adams.

McKay.  At Home Tonight.  1940.  Prob. 13: Sharing the cost, pp. 65 & 79.  Similar to Adams.

McKay.  Party Night.  1940.  No. 25, p. 182.   A  &  B  give a party and invite  2  and  3  guests.  The party costs  35s  _ how do they divided the expense?  Initial reaction is in the ratio  2 : 3,  but it should be  3 : 4.  (Also entered in 7.H.3.)

 

          7.H.6. SHARING A GRINDSTONE

 

          New section.  I have seen other examples.  A grindstone of radius  R  is to be shared between two (or  k)  buyers _ one grinding until his share is used.  An inner circle of radius  r  is unuseable.  The first man should grind to radius  x  where  x² - r²  =  (R² - r²)/k   or  x²  =  [(k-1)R² + r²]/k.  It is straightforward to adapt this to the case when the buyers contribute unequal amounts to the purchase price.

 

Carlile.  Collection.  1793.  Prob. XXIV, p. 16.  Three men buy a grindstone of radius  20  for  20s.  They pay  9s,  6s,  5s  respectively.  How much should each man get to grind?  He makes no allowance for wastage.

Hutton-Rutherford.  A Course of Mathematics.  1841?  Prob. 22,  1857: 558.  Share a grindstone among seven people.  2R = 60".  He takes  r = 0.

Dana P. Colburn.  Arithmetic and Its Applications.  H, Cowperthwait & Co., Philadelphia, 1856.  Miscellaneous Examples No. 43, p. 362.  Four men sharing a grindstone 4 ft in diameter.  No indication of inner wastage.  No solution given.

Collins.  Fun with Figures.  1928.  A grindstone dispute, p. 22.  2R = 5 ft 6 in;  2r = 18 in.

 

          7.H.7. DIGGING PART OF A WELL.

 

          Recently separated from 7.H.5.

          See Tropfke 529.

          The 'digging a well' problem has a contract to dig a well  a  deep for payment  b,  but the digging stops at  c.  How much should be paid?  The value of  b  is not always given and then only the ratio of the part payment to the total payment is sought.

          NOTATION _ this problem is denoted  (a, b; c).

If the difficulty is proportional to depth, then integration yields that the payment should be proportional to  (a/c)².  A common medieval approach is use the proportion  1 + ... + c  :  1 + ... + a.  We let  T_a  =  1 + ... + a,  so this proportion is  T_c : T_a.

          P. M. Calandri; Calandri, 1491, are the only cases where  c > a.

          Pacioli begins the inverse problem _ if the contract for depth  a  is worth  b  and work stops at  x  such that the value of the dug hole is  d,  what is  x?  Denote this situation as  (a, b; x)  worth  d.  See:  Pacioli;  Calandri, Raccolta;  Buteo.

          Ozanam, Vyse and Jackson are the only ones to consider use of other arithmetic progressions. 

          Berloquin gives a simple argument that work is proportional to  a²/2.

 

In Neugebauer & Sachs, op. cit. in 7.E, the problems discussed on pp. 81-91 involve digging out ditches and the cost or difficulty of digging increases with the depth, but none of these are like the problem considered here, though Tropfke 529 notes a resemblance.

Tabari.  Mift_h al-mu‘_mal_t.  c1075.  ??NYS - quoted and discussed by Tropfke 529. 

P. 96, part III, No. 17.  Calculation of ditchwork.  (15, 30; 10).  Tropfke doesn't give a solution but says it is similar to the following.

P. 227, part VI, no. 61.  "Apportionment of wells and cisterns.  In order to dig a well, the earth must be lifted out.  For the first ell, the earth comes up one ell; for the second ell, the earth comes up two ells; for the third, three ells; etc. untilt he end.  For this calculation, one must use the series of of natural numbers.  So we take 1 as first term, to which 2 as second term gives 3, to which 3 as third term gives 6 as sum, etc.  In this way, we do each time, and that is the 'bast' (literally 'extension').  Example: the well depth is 10 ells; what is the 'bast'?  1 + 2 = 3,  3 + 3 = 6,  ...,  45 + 10 = 55.  That is the 'bast' of a well 10 ells deep."  So he divides in the standard ratio  T₁₀/T₁₅ = 55/120 = 1/3 + 1/8.  "We multiply that by 30; this yields  13 [+] 1/2 [+] 1/4.  That is the payment for a well of 10 ells, when the other well costs 30 dirhems."  Tropfke says the problem is reminiscent of a Babylonian one - cf above.

Qazwini  = al‑Qazwînî.  Loc. cit. in 7.H.3.  c1260.  P. 253.  German translation only _ man contracts to dig a well  10  ells deep for  10  dirhems.  He stops at  9  ells, so we have  (10, 10; 9).  Man asks for  9  dirhems, but an expert says only  8  and somewhat more.

BR.  c1305.  No. 22, pp. 40‑43.  Man contracts to dig  10 x 10 x 10  cistern but only does  5 x 5 x 5.  Text gives him  1/8  of the value.

Dell'Abbaco.  c1370.  Prob. 102, p. 87 with plate on p. 88.  Man contracts to dig a well 20 deep and stops at  14,  i.e.  (20, b; 14).  Author divides in ratio  T₁₄ : T₂₀ = 1 : 2.  But he says he doesn't think this is a correct method, though he doesn't know a better one.  I have a colour slide of this.

Lucca 1754.  c1390.  F. 64v, p. 152.  Man digging a well,  (10, b; 8).  He divides in ratio   T₈ : T₁₀ = 36 : 55.

Muscarello.  1478.  Ff. 66v-67r, pp. 176-177.  Man to dig a hole but hits water and has to stop,  (10, b; 7).  Divides in the ratio   T₇ : T₁₀ = 28 : 55.

P. M. Calandri.  c1480.  Pp. 115‑116.  If a well  12  deep is worth  12,  how much is a well  14 deep worth?  This is  (12, 12; 14).  He takes values proportional to T_d.

Calandri.  Arimethrica.  1491.  F. 65v.  (12, 12; 16).

Pacioli.  Summa.  1494. 

F. 40v, prob. 8.  Dig a well,  (11, 11; 6).  Divides as  T₆ : T₁₁ = 7 : 22,  so the partial well is worth  7/2. 

F. 40v, prob. 9.  Dig a well,  (11, 11; x)  worth  7/2.

Part 2, f. 55v, prob. 38.  Dig a well,  (10, 10; 6).  Divides as  T₆ : T₁₀ = 21 : 55.

Part 2, f. 55v, prob. 39.  Dig a well,  (10, 10; x)  worth  4.  He notes  4 : 10 = 22 : 55,  so we want  n(n+1)/2 = 22,  which would give  n = (-1 + Ö177)/2 = 6.15207....  He interpolates as  6  days plus  1/7  of the sixth day, i.e.  n = 6.14286....

Calandri.  Raccolta.  c1500.  Prob. 38, pp. 33‑34.  If a well  24  deep is worth  24,  how deep a well is worth  40?  I.e.  (24, 24; x)  worth  40.  He takes values proportional to  T_d.

Tagliente.  Libro de Abaco.  (1515).  1541.  Prob. 110, ff. 55r-55v.  Dig a well,  (9, 24; 5).  Divides in ratio  T₅ : T₉ = 15 : 45.

Apianus.  Kauffmanss Rechnung.  1527.  F. D.vi.v.  Mason to build a tower  100  high  for  100.  He falls ill after  84.  Divides in ratio  T₈₄ : T₁₀₀.

Cardan.  Practica Arithmetice.  1539.  Chap. 66.

Section 8, ff. CC.i.r - CC.i.v (p. 138).  Men dig a well,  (34, 60; 20).  Divides in ratio  T₂₀ : T₃₄ = 6 : 17.

Section 10, ff. CC.i.v ‑ CC.ii.v (p. 138).  Man building a wall  5  high at prices  10, 20, 40, 80, 160  per unit of height.  He stops at height  2½.  Takes half a unit of height as having  Ö2  times the value of the previous half-unit, so the interval from  2  to  3  high worth  40  divides into two halves worth  x  and  xÖ2,  giving  x = 40/(1+Ö2).

Buteo.  Logistica.  1559. 

Prob. 35, pp. 238-240.  Dig a well,  (100, 50; 50).  Divides as  T₅₀ : T₁₀₀ = 1275 : 5050.  Also does  (200, 50; 100).  His depths are initially in cubits, but are converted to 'semipedes' _  120  cubits is  50  semipedes.

Prob. 36, pp. 240-241.  Dig a well,  (100, 50; x)  worth  28 22/101.

Ozanam.  1694, 1725.

Prob. 7, question 1, 1696: 30;  1708: 27.  Prob. 10, question 1, 1725: 60-61.  Prob. 2, 1778: 65;  1803: 67-68;  1814: 60;  1840: 32-33.  Man digging a well  20  feet deep, to receive  3, 5, 7, ...,  for each successive foot.  [This is not really in this section, but is included because later ed. use it as the basis of the next problem.]

Prob. 51, question 1, 1725: 256‑257.  Prob. 3, 1778: 66-67;  1803: 68-69;  1814: 61-62;  1840: 33.  Man digging a well,  (20, b; 12)  (1778 et seq change  12  to  8).  1725 divides as  T₁₂ : T₂₀ = 78 : 210.  1778 notes that the difficulty of the work increases in arithmetic progression, but that there are many such progressions.  He then posits that the first unit is worth  1/4  when the agreed payment is  20  and this gives a difference of  30/11  for the arithmetic progression.

                    If the cost per unit depth is an arithmetic progression:  A, A+D, ...,  δ = D/A  and  d  is the value of the partial well, then  d/b = (c + δc²)/(a + δa²).

Les Amusemens.  1749.  Prob. 44, p. 176.  Mason to dig a well,  (10, b; 4).  Divides as  T₄ : T₁₀ = 10 : 55.

Vyse.  Tutor's Guide.  1771?  Prob. 17, pp. 144-145 & Key p. 188.  Same as Ozanam, prob. 7, but with depth  30.

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions, no. 25, pp. 20 & 79.  Dig  20  yards for  £20.  Man falls sick after  8  yards.  "How much was then due to him, on a supposition that the labour increases in arithmetical proportion as the depth?"  I.e.  (20, 20; 8).  Solution notes that the data does not determine what the arithmetic progression is and chooses  5s  as cost of the first yard _ see Ozanam.

Unger.  Arithmetische Unterhaltungen.  1838.  Pp. 182 & 263, no. 693.  Man digging a well  49  feet deep.  First foot costs  15,  but each successive foot costs  6  more than the previous.  Find cost of last foot and total cost.  So this is really an arithmetic progression problem, but I haven't seen others of those using this context.

Pierre Berloquin.  The Garden of the Sphinx, op. cit. in 5.N.  1981.  Prob. 117: A bailout fee, pp. 66 & 166.  Man contracts to bail out a  20  yard deep well for  $400  and gives up at  10  yards.  Answer says value is proportional to the depth  d  times the average distance lifted, i.e.  d/2,  hence value is proportional to  d²/2  and this is the result that integration produces.

 

          7.I.     FOUR FOURS, ETC.

 

          Express an integer using four 4s, etc.  Cupidus Scientiae, 1881, seems to be the first to ask for solutions to a lot of the integers, rather than a few specific examples.  The next examples of the general form are Cunningham & Wiggins (1905), Pearson (1907), Ball (1911), Ball (1912).  Dawson (1916) is the first to ask for four  R's,  where  R  is indeterminate, e.g.  3  =  (R+R+R)/R.  I have included examples where a set of numbers and operations is given and one has to obtain a given value.  This overlaps a bit with 7.I.1, where the object is to find the maximum possible value, and with 7.AC.3-6, where one uses all nine or ten of the digits and I have included problems of inserting signs into  12...9  to make  100  in 7.AC.3.

 

Dilworth.  Schoolmaster's Assistant.  1743.  Part IV: Questions: A short Collection of pleasant and diverting Questions, p. 168.

Problem 4:  "Let  12  be set down in four Figures and let each Figure be the same."

Problem 9.  "Says Jack to his brother Harry, I can place four threes in such manner that they shall just make  34;  can you do so too?"

Les Amusemens.  1749.  Pp. 52-54.  Several problems leading to:  7 7/7,  33 3/3,  55 5/5,  99 9/9,  77 77/77,  2222 2222/2222,  11 1/1, etc.  See the entry in 7.AN.

Vyse.  Tutor's Guide.  1771?  Prob. 1, p. 165 & Key p. 206.  "Four Figures of nine may be so placed and disposed of as to denote and read for  100,  neither more nor less.  Pray how is that to be done?"

Pike.  Arithmetic.  1788.  P. 350, no. 16.  "Said Harry to Edmund, I can place four  1's  so that, when added, they shall make precisely  12;  Can you do so too?"

Jackson.  Rational Amusement.  1821.  Arithmetical Puzzles.

No. 2,  pp.  1 & 51.  "It is required to express  100  by four  9's."

No. 4,  pp.  2 & 51.  Use three  2s  to make  ½,  1  and  2.

No. 5,  pp.  2 & 51.  Express  12  by four equal figures.

No. 17, pp.  5 & 56.  Express  6½  by four  5s.  Answer is:  5.5 + 5/5.

No. 33, pp.  8 & 59.  Use four  2s  to make  1/8,  1/2,  2  and  8.

No. 40, pp. 10 & 62.  Use three  3s  to make  1/3,  1  and  3.

No. 42, pp. 10 & 62.  Use four  3s  to make  1/243,  1/27,  1/3,  3,  27  and  243.

No. 43, pp. 10 & 62.  Use five  3s  to make the same numbers as in no. 42.

No. 44, pp. 10 & 63.  Express  78  by six equal digits.

Endless Amusement II.  1826?  Prob. 23, p. 201.  "Put down four nines, so that they will make one hundred."

Child.  Girl's Own Book.  Arithmetical puzzles, no. 5.  1832: 170 & 179;  1833: 184 & 193;  1839: 164 & 173;  1842: 282 & 291;  1876: 231 & 244.  "Place four nines together, so as to make exactly one hundred.  In the same way, four may be made from three threes, three may be made from three twos, &c."  The 1833 solution is printed rather oddly as  199 9‑9, while the 1839 and 1842 solution is  99 9-9  and the 1876 solution is  99 9¸9. 

Nuts to Crack III (1834), no. 211.  "Write down four nines so as to make a hundred."

Family Friend 1 (1849) 150 & 178.  Problems, arithmetical puzzles, &c. _ 4.  "Put four fives in such a manner, that they shall make  6½.  _  D. F."  Answer is   5 5/5 + .5.   = The Sociable, 1858, Prob 46: A dozen quibbles, part 12: pp. 300 & 318.  = Book of 500 Puzzles, 1859, prob. 46: part 12, pp. 18 & 36.

Boy's Own Book.  To place four figures of  9  in such a manner as to represent 100.  1855: 601.

Magician's Own Book.  1857.  Quaint questions, p. 253.  [No. 4] _ "Place three sixes together, so as to make seven."  [No. 6] _ "Place four fives so as to make six and a half."  [Boy's Own Conjuring Book, 1860, pp. 224-225, has the Quaint Questions, but omits these two questions!]

Book of 500 Puzzles.  1859.

Prob 46: A dozen quibbles: part 12, pp. 18 & 36.  As in Family Friend.

Quaint questions, p. 67.  [Nos. 4 & 6] _ Identical to Magician's Own Book.

Charades, Enigmas, and Riddles.  1859?: prob. 30, pp. 60 & 64;  1862?: prob. 575, pp. 108 & 155.  "Write a Hundred with 4 nines."  (1865 has slightly different typography.)

Illustrated Boy's Own Treasury.  1860. 

Prob. 1, pp. 427 & 431.  "Put down four nines, so that they shall make one hundred."

Prob. 26, pp. 429 & 433.  "Put four fives in such a manner, that they shall make  6½."

Prob. 38, pp. 430 & 434.  "It is required to place four  2's  in such a manner as to form four numbers in geometrical progression?"  Uses four  2s  to make each of  1/8,  1/2,  2,  8.

Leske.  Illustriertes Spielbuch für Mädchen.  1864? 

Prob. 564-15, pp. 253 & 395.  Write  100  with  4  nines.  Write  1000  with no zeroes _ answer: 999 9/9.

Prob. 564-18, pp. 253 & 395.  Write  100  with no zeroes, using  4, 6 or 8 figures.  Answers:  99 3/3;  99 44/44;  99 999/999.

Magician's Own Book (UK version).  1871.  Paradoxes [no. 3], p. 37.  "With four fives make  6½? _ 5 5/5 .5."  where the  5/5  is written as a 5 over a 5 with no fraction bar.  Cf Kackson and Magician's Own Book.

Hugh Rowley.  More Puniana; or, Thoughts Wise and Other-Why's.  Chatto & Windus, London, 1875.  P. 300.  "Write down one hundred with four nines."

"Cupidus Scientiae" (possibly the editor, Richard A. Proctor).  Four fours, singular numerical relation.  Knowledge 1 (30 Dec 1881) 184, item 151.  A bit vague as to what operations are permitted, but wants four  4s  to make various values.  Says he has not been able to make  19.

H. Snell.  Singular property of number  4.  Knowledge 1 (6 Jan 1882) 209, item 178.  19 = 4! ‑ 4 - 4/4.  Editor says  4!  is not reasonable for the problem as posed.

Solutions from various contributors.  Four fours.  Knowledge 1 (13 Jan 1882) 229, item 184.  Numerous solutions for  1  through  20,  except  19.  Solutions for  19  are:  4/.4 + 4/.4;   4! - 4 - 4/4;   4/Ö.4 ‑ 4/4 ("manifestly erroneous");   (4 + 4 - .4)/.4;   (x + x ‑ .x)/.x  in general.  Four  3s  give same results as three  5s,  except for  17.

Albert Ellery Berg, ed.  Op. cit. in 4.B.1.  1883.  P. 373.  "Place three sixes together so as to make seven."

Lemon.  1890. 

Vagaries, no. 217(b), pp. 33 & 105.  Three  6s  to make 7. 

Arithmetical, no. 752, pp. 92 & 124.  = Sphinx, no. 600, pp. 81 & 118.  "Place four nines so as to make one hundred."

Berkeley & Rowland.  Card Tricks and Puzzles.  1892.  Card Puzzles No. II: A curious addition sum, p. 2.  Mentions "writing down 100 with four nines" as  99 9/9.

(Sam Loyd.)  One hundred pounds for correct answer to a puzzle.  Tit-Bits (14 Oct 1893) 25.  "Find How to Arrange the Figures  · 4 · 5 · 6 · 7 · 8 · 9 · 0 ·  in an Arithmetical Sum which Adds up the Nearest to  82."  "Mr. Loyd is confident that no one will find it out."  Indeed, Loyd will be paid £100 only if no correct answer is received.

(Sam Loyd.)  Solution of Mr. Sam Loyd's one hundred pound puzzle.  Tit-Bits (18 Nov 1893) 111.  80·5 + ·97 + ·46  =  82.  (There are points over the  5, 9, 7, 4, 6,  but my printer may not print these clearly.)  Here the mid-line dot (·) is used for a decimal point.  Because of the number of correct solutions, ten extra names were drawn from them for additional £5 prizes.  "It seems that not a single person in the whole of America has sent the correct answer when a prize was offered there, but here we have received a very large number actually correct."  [See MRE for another solution.]

Hoffmann.  1893.  Chap. IV, no. 18: Another way to make a hundred, pp. 148 & 193.  Use six  9s  to make 100.

Ball.  MRE, 3rd ed., 1896.  P. 13.  "... a question which attracted some attention in London in October, 1893, ...."  [See Loyd above.]  He says that the problem is to make  82  with the seven digits  9, 8, 7, 6, 5, 4, 0  and gives one solution as  80·69 + ·74 + ·5  (with points over the  9, 4, 5  _ there should also be points over the  6  and  7).

Mr. X.  His Pages.  The Royal Magazine 10:5 (Sep 1903) 426-427.  "How to arrange four 9's to make 100."

Ball.  MRE, 4th ed., 1905.  P. 14.  Repeats the material in the 3rd ed of 1896, again omitting two points, and adds further questions.  Use the  10  digits to total  1  _ a solution is  35/70 + 148/296  _ or to total  100  _ a solution is  50 + 49 + 1/2 + 38/76.  Use the  9  digits to make four numbers which total  100  _ a solution is  78 + 15 + ²Ö9 + ³Ö64.

A. Cunningham & T. Wiggins.  ??  Math. Quest. Educ. Times 7 (1905) 43‑46.  ??NYS _ cited in Dickson I 460, item 45d.  Expressions using four  9s  and four  4s.

Pearson.  1907. 

Part I, no. 43: The nimble nines, pp. 125 & 187.  Verse asking for three  9s  to make  16  _  solution is   96/6 !!

Part II: On all fours, p. 107.  Four fours in general, with a few examples.

Wehman.  New Book of 200 Puzzles.  1908.  P. 26.  = Magician's Own Book, no. 4.

Ball.  MRE, 5th ed., 1911.  Pp. 13-14.  Briefly restates the material in the 4th ed. as "questions which have been propounded in recent years.  ...   To the making of such questions of this kind there is no limit, but their solution involves little or no mathematical skill."

                    "Another traditional and easy recreation ....  I have never seen this recreation in print, but it seems to be an old and well-known question."  Deals just with four  4s  and says one can get up through 170.  G. N. Watson has pointed out that one can get further by using factorials and subfactorials.  (The subfactorial of  n  is  n¡ = n![1/0! ‑ 1/1! + 1/2! - 1/3! + ... ± 1/n!].)  The topic is not in earlier editions.

W. W. Rouse Ball.  Four fours.  Some arithmetical puzzles.  MG 6 (No. 98) (May 1912) 289‑290.  "An arithmetical amusement, said to have been first propounded in 1881, ...."  [This would seem to refer to Knowledge, above.]  Studies various forms of the problem.  Says it occurs in his MRE _ see above.  MRE 6th ed., 1914, p. 14, cites this article.

Ball.  MRE, 6th ed., 1914.  Pp. 13-14.  He now splits the material into three sections.

                    Empirical Problems.  Restates the material in the 5th ed. as "... numerous empirical problems, ..." and omits Loyd's problem.  "To the making of such questions there is no limit, but their solution involves little or no mathematical skill."

                    He then introduces the "Four Digits Problem".  "I suggest the following problem as being more interesting."  Using the digits  1, 2, ..., n,  express the integers from  1  up using four different digits and the operations of sum, product, positive integral power and base-10  notation (or also allowing iterated square roots and factorials).  With  n = 4,  he can get to  88  or to  264.  With  n = 5,  he can get to  231  or  790.  Using  0, 1, 2, 3,  he can get to  36  (or  40).

                    Under Four Fours Problem, he discusses what operations are permitted and says he can get to  112,  or to  877  if subfactorials are permitted (citing his MG article for this).  Mentions four 9s and four 3s problems.

Williams.  Home Entertainments.  1914.  The six 9's, p. 119.  "Express the number  100  by means of six  9's."

Thomas Rayner Dawson.  1916.  ??NYS.  Cited in:  G&PJ 3 (Jan 1988) 45  &  4 (Mar 1988) 61.  Asks for four  R's,  where  R  is indeterminate, e.g.  3  =  (R+R+R)/R.

Ball.  MRE, 7th ed., 1917.  Pp. 13-14.  The material of the first two sections is repeated, but under "Four Fours Problem", he discusses the operations in more detail.  With  +,  -,  x,  ¸,  brackets and base-10  notation, he can get to  22.  Allowing also finitely iterated square roots, he can get to  30.  Allowing also factorials, he can get to  112.  Allowing also integral indices expressible by  4s  and infinitely iterated square roots, he can get to  156.  Allowing also subfactorials, he can get to  877.  (In the 11th ed., 1939, pp. 15-16, two footnotes are added giving expressions for  22  in the first case and  99  in the third case.)  Gives some results for four  2s,  four  3s,  four  5s,  four  9s.  Mentions the general problem of  n  ds.

Smith.  Number Stories.  1919.  Pp. 112‑113 & 140‑141.  Use four  9s  to make  19,  2  and  20.

Ball.  MRE, 9th ed., 1920.  Pp. 13-14.  In the "Four Digits Problem", he considers  n = 4,  i.e. using  1, 2, 3, 4,  and discusses the operations in more detail.  Using sum, product, positive integral power and base-10  notation, he can get to  88.  Allowing also finitely iterated square roots and factorials, he can get to  264.  Allowing also negative integral indices, he can get to  276.  Allowing also fractional indices, he can get to  312.  He then mentions using  0, 1, 2, 3  or four of the five digits  1, ..., 5.

                    Under "Four Fours Problem", he repeats the material of the 7th ed., but adds some extra results so he has results for four  ds,  d = 1, 2, 3, 5, 6, 7, 8, 9.

Ball.  MRE, 10th ed., 1922.  Pp. 13-14.  In the "Four Digits Problem", he repeats the material of the 9th ed., but at the end he adds that using all of the five digits,  1, ..., 5,  he has gotten to  3832  or  4282,  depending on whether negative and fractional indices are excluded or allowed.

Dudeney.  MP.  1926.  Prob. 58: The two fours, pp. 23‑24 & 114.  = 536, prob. 109, pp. 34 & 248‑249, with extensive comments by Gardner.

King.  Best 100.  1927.

No. 44, pp. 20-21 & 48.  = Foulsham's no. 14, pp. 8 & 11.  "Can you put down four fifteens so that they come to  16,665?

No. 45, p. 21 & 49.  "Arrange the figures  1  to  7  so that they will amount to  100,  when added together."  Arrange four  9s  to make  100.  Gives two answers for the first part.

Perelman.  FFF.  1934.  1957: probs. 98, 100 & 102, pp. 137 & 143-144;  1979: probs. 101, 103 & 105, pp. 166-167 & 174-175.  = MCBF, probs. 101, 103 & 105, pp. 167 & 176‑178.

101: Two digits:  "What is the smallest integer that can be written with two digits?"  1/1 = 2/2 = ....  [Though I think  0/1  might be counted.]

103: Five 9's:  "Write 10 with five  9's.  Do it in at least two ways."

105: Four ways:  "Show four different ways of writing 100 with five identical digits."

Perelman.  MCBF.  1937.  Any number via three twos.  Prob. 202, pp. 398-399.  "A witty algebraic brain-teaser that amused the participants of a congress of physicists in Odessa."  n  =  ‑ log₂ log₂ Öⁿ2,  where  Öⁿ  means  n‑fold iterated square root.

Adams.  Puzzle Book.  1939.  Prob. B.83: Figure juggling, part 3, pp. 78 & 107.  Use a digit  8  times to make  1000.  Answer uses  8s.

Evelyn August.  The Black-Out Book.  Op. cit. in 5.X.1.  1939.  Number, please!, pp. 20 & 210. 

Use the same odd figure five times to make  14. 

Four  9s  to make  100. 

Four  5s  to make  6½.

Depew.  Cokesbury Game Book.  1939. 

Three eights, p. 216.  Use three  8s  to make  7.

Twenty-four, p. 227.  Use a digit three times to make  24.  Answers:  3³ - 3,   22 + 2.

McKay.  Party Night.  1940.  No. 9, p. 177.

(a)  Use three  9s  to make  10.  Answer:  9 9/9.

(b)  Use four  9s  to make  20.  Answer:  9 99/9.

(c)  Use three  9s  to make  100.  Answer:  99.9 (or 99.9 for clarity).

(d)  Use two  9s  to make  10.  Answer:  9/.9  or  9.9 (or 9.9).

Meyer.  Big Fun Book.  1940.  A half dozen equals 12, pp. 119 & 738.  Use six  1s  to make  12.  Answer:  11 + 11/11.

George S. Terry.  The Dozen System.  Longmans, Green & Co., NY, 1941.  ??NYS _ quoted in:  Underwood Dudley; Mathematical Cranks; MAA, 1992, p. 25.  What numbers can be expressed with four  4s  duodecimally?  About 5 dozen.  How many numbers can be expressed using each of the digits  1,  2,  3,  4  once only, again duodecimally?  About 9 dozen and nine.  Terry (or Dudley) also gives the results for decimal working as  22  and  88.

Sullivan.  Unusual.  1943.  Prob. 17: Five of a kind.  Write 100 with the same figure five times "and the usual mathematical symbols".  Says it can be done with  1s,  2s,  5s (two ways),  9s  and perhaps others.

J. A. Tierney, proposer;  Manhattan High School of Aviation Trades, D. H. Browne, H. W. Eves, solvers.  Problem E631 _ Two fours.  AMM 51 (1944) 403  &  52 (1945) 219.  Express 64 using two 4s.

Vern Hoggatt & Leo Moser, proposers and solvers.  Problem E861 _ A curious representation of integers.  AMM 56 (1945) 262  &  57 (1946) 35.  Represent any integer with  p  a's,  for any  p ³ 3  and any  a ¹ 1.  Solution for  ±n  uses  log  to base  ÖÖ...Öa,  with  n  radicals.

S. Krutman.  Curiosa 138: The problem of the four  n's.  SM 13 (1947) 47.

Sullivan.  Unusual.  1947.  Prob. 28: A problem in arithmetic.  What is the smallest number of eights which make 1000?

Anon.  The problems drive.  Eureka 13 (1950) 11 & 21.  No. 6: Use four  4s  to approximate  π.  They get  3.14159862196...

D. G. King‑Hele.  Note 2509:  The four 4's problem.  MG 39 (No. 328) (May 1955) 135.  n  =  log[{log 4}/{log Öⁿ 4}]/log Ö4  expresses any positive integer  n  in terms of three  4s.  A slight variation expresses  n  in terms of four  x's,  for any real  x ¹ 0, 1.  1  can be used by taking  x = .1.  He also expresses  n  in terms of  m  x's  for real  x ¹ 0, 1  with  m > 5  and also with  m = 5.

Philip E. Bath.  Fun with Figures.  Op. cit. in 5.C.  1959.

No. 12: One hundred every time, pp. 10 & 41.  Make an arrangement of  x's  which gives the result  100  for  x = 1, 2, ..., 9.  Answer:  xxx/x - xx/x.

No. 20: Form fours, pp. 12 & 42.  Eight  4s  to make  500.

No. 74: Signs wanted, pp. 28 & 54.  Insert signs  (+,  -,  x,  /)  into a row of four  x's  to make  10 - x,  for  x = 2, ..., 9.

J. H. Conway & M. J. T. Guy.  π  in four 4's.  Eureka 25 (1962) 18‑19.  Notes that  π  =  Ö[(‑Ö4/4)!]⁴,  if non‑integral factorials are allowed.  Shows that any real can be arbitrarily well approximated using four  4s.

D. E. Knuth.  Representing numbers using only one  4.  MM 37 (1964) 308‑310.

Gardner.  SA (Jan 1964) adapted as Magic Numbers, chap. 5.  Cites Knowledge as the origin.  Magic Numbers gives numerous other citations.

Marjorie Bicknell & Verner E. Hoggatt.  64 ways to write 64 using four  4's.  RMM 14 (Jan‑Feb 1964) 13‑15.

Jerome S. Meyer.  Arithmetricks.  Scholastic Book Services, NY, 1965.  Juggling numbers, no. 3, pp. 83 & 88.

Make  100  with four  7s.   77/.77.

Make  20  with two  3s.   3!/.3.

Make  7  with four  2s.   (2/.2)/2 + 2.

Make  37  with six  6s.   6*6 + 66/66.

Steven Everett.  Meanwhile back in the labyrinth.  Manifold 10 (Autumn 1971) 14‑16.  (= Seven Years of Manifold 1968‑1980; ed. by I. Stewart & J. Jaworski; Shiva Publishing, Cheshire, 1981, pp. 64‑65.)  n  =  ‑Ö4 * log₄ log₄ Öⁿ4,  where  log₄  means log to the base 4 and  Öⁿ  means  n‑fold iterated square root.  This is a variant of King‑Hele's form.  This article is written in a casual style and seems to indicate that this formula was devised by Niels Bohr.  He gets a form for  e,  but it uses infinitely many factorial signs!!!!...

                    Editor's note on p. 2 (not in the collection) gives an improvement due to Michael Gerzon, but it is unclear what is intended.  The note gives another method due to Professor Burgess using  sec tan^-1 Öm  =  Ö(m+1)  and  1  =  Öⁿn  which expresses  n  by one  1.

Ball.  MRE, 12th ed., 1974.  Pp. 15-17.  Under "Four fours problem", the material of the 9th ed. and the footnotes mentioned at 7th ed are repeated, but the bound for four  9s  is increased.

Bronnie Cunningham.  Funny Business.  An Amazing Collection of Odd and Curious Facts with Some Jokes and Puzzles Too.  Puffin, 1978.  Pp. 38 & 142.  Arrange three  9s  to make  20.  Answer:  (9 + 9)/.9.

Putnam.  Puzzle Fun.  1978.

Nos. 54-57: Ten is the number, pp. 10 & 35.  Express 10 using five  9s,  in four different ways.

Nos. 58-59:  3 + 3 + 3 = 30,  pp. 10 & 35.  Express 30 using three  3s,  in two differnet ways.

No. 97: Eight to one thousand, pp. 13 & 37.  Use a digit eight times to express 1000.

P. Grammer, I. McFiggans, N. Blacknell, T. Joyce, J. Anstey & A. Devonald.  Counting in fours.  MiS 9:4 (Sep 1980) 21-22.  Uses four 4s to express  1, ..., 50.  Says  51 - 100 will appear in next issue, but they weren't.

J. Bellhouse.  Four fours.  MiS 14:1 (Jan 1985) 15.  Says the promised table for  51 - 100  (see Sep 1980 above) had not appeared, so his students found their own.

Anne Williamson.  1985.  MiS 14:4 (Sep 1985) 7.  Use the four digits  1, 9, 8, 5  to express integers  1 - 100.  Unhappy with expressions for  24, 31, 65  which use !.

Ken Lister.  Letter.  MiS 15:2 (Mar 1986) 47.  Responding to Bellhouse (Jan 1985).  Corrects and improves some values, but says  71 and 73  have not been done.  Expresses  a/b,  for single digits  a, b,  by use of four  4s.

Angie Aurora.  Letter.  MiS 15:3 (May 1986) 48.  Improvements for Wiiliamson's problem _ Sep 1985 above.

Joyce Harris.  Letter:  Four fours.  MiS 15:3 (May 1986) 48.  Responding to Lister (Mar 1986), gives expressions for  71 and 73.

Bob Wasyliw.  Letter:  Four 4's _ the ultimate solution.  MiS 15:5 (1986) 39.  Adapts Everett's 1971 method to include non‑positive integers.

Simon Gray & Colin Abell.  Letters:  Four fours again.  MiS 16:2 (1987) 47.  Gray notes that  4 = Ö4 * Ö4,  so that 'four 4s' is the same as 'at most four 4s'.  He gives  π  =  Ö4 * sin^‑1(4/4)  and more complex forms.  Abell gives  π  =  ‑ Ö(‑4/4) * log (‑4/4)  [The first minus sign is ambiguous??] and  π  =  ‑ Ö(4*4) * Tan^-1(4/4)  [The minus sign is wrong.]

Tim Sole.  The Ticket to Heaven and Other Superior Puzzles.  Penguin, 1988.

   Number play (ii) ‑ (iv), pp. 15, 29 & 178‑180.

(ii).  Using  +,  ‑,  *,  /,  .,  !  and brackets,  (,  ),  one can express  1 ‑ 48  with four  3s.  Allowing also  SQR,  one can express  1 ‑ 64  with four  3s.

(iii).  Using all 7 symbols in (ii) and brackets, one can express  1 ‑ 112  with four  4s.  Gives solutions with two and three  4s.

(iv).  If we also allow  Σ(n) [=  T_n,  the  n-th triangle number], he finds solutions with one, two or three  4s.  Using  log,  all integers can be expressed with three  4s.

   Three of the best _ (iii), pp. 17 & 32.  Some solutions with one  4,  using  !,  Ö,  [n] = INT(n)  and  Σ(n).

Tony Forbes.  Fours.  M500 116 (Nov 1989) 4‑5.  Says someone (possibly Marion Stubbs?) gave a simple variation of King‑Hele's and Everett's formulae to use exactly four  4s  to yield  n.  Forbes suggests using one  4  and the three operations:  !,  Ö  and  INT.  He already gets stuck at 12.

Liz Allen.  Brain Sharpeners.  Op. cit. in 5.B.  1991.  Take two, pp. 96 & 139.  Use five  2s  to express  1, 2, ..., 26,  particularly 17 and 26.

P. H. R.  Fawlty keys.  Mathematical Pie, 129 (Summer 1993) 1023  &  Notes, p. 1.  Calculator whose only keys are  5,  7,  +,  -,  x,  ¸  and  =.  Make numbers from 0 to 20.  Solution notes you can make any number by adding enough terms of the form  5 ¸ 5  and then gives short solutions for 1 through 20.

Clifford A. Pickover.  Phi in four  4's.  Theta (Crewe) 7:2 (Autumn 1993) 5-8.  In Sep 1991 he asked for good approximations to  _  using four  4s,  either with as many symbols as you want or with each symbol used at most four times.  Says he was inspired by Conway & Guy's paper of 1962.  Brian Boutel produced  _  =  (Ö4 + Ö{4!-4})/4.  Pickover then extended the question and various solvers got  _  in five  5s,  seven  6s,  eight  8s,  nine  9s  and  2k-5  ks.

John Seldon.  Fours.  M500 136 (Jan 1994) 15‑16.  Answers Forbes' 1989 problem of expressing  1 ‑ 100  with one  4  and any number of factorials,  x!,  and integer square roots,  [Öx].

David Crawford and students.  1999  the end of an era.  MiS 28:4 (Sep 1999) 25.  Uses  1, 9, 9, 9  to make all integers from  1  to  100.  Notes that  2000, 2001, ...  are not going to be very useful for such puzzles!

 

          7.I.1.  LARGEST NUMBER USING FOUR ONES, ETC.

 

James Joyce.  Ulysses.  (Dijon, 1922);  Modern Library (Random House), NY, 1934, apparently printed 1946.  P. 684 (Gardner says the 1961 ed. has p. 699).  Bloom estimates that  9^(9^9)  would occupy "33 closely printed volumes of 1000 pages each", but he erroneously phrases the number as "the 9th power of the 9th power of 9", which is only  9⁸¹.

King.  Best 100.  1927.  No. 35, pp. 19 & 46.  Largest number using two  4s.  Gives  4⁴ = 256.

Perelman.  FFF.  1934.  Four 1's.  1957: prob. 103, pp. 137 & 144-145;  1979: prob. 106, pp. 167 & 175.  = MCBF, prob. 106, pp. 167 & 178  = MCBF, prob. 131, p. 217.  "What is the biggest number that can be written with four  1's?

Perelman.  MCBF.  1937.  Probs. 128-132, pp. 214-219.  Largest numbers with:  Three twos;  Three threes;  Three fours;  Four ones;  Four twos.

Sullivan.  Unusual.  1947.  Prob. 30: Not  999.  Largest number that can be written with three integers [sic!].  Answer:  9^(9^9).

Leroy F. Meyers.  An integer construction problem.  AMM 66:7 (Aug/Sep 1959) 556-561.  This deals with Ball's "Four Digits Pproblem" (see MRE, 6th ed., 1914 in 7.I) and generalizations.  In particular, he shows that if one uses  1, 2, 3, 4,  with operations  +,  ‑,  x  and brackets, then one can obtain precisely the following:  1, 2, ..., 28, 30, 32, 36.  In general he obtains the largest integer expressible using a given multiset of integers (i.e. one is allowed a fixed number of repeats of a value) using the operations  +,  x  and brackets.  He also shows that allowing also  -,  for both negation and subtraction, does not increase the maximum obtainable value.  He conjectures that allowing also  ¸,  for both reciprocation and division, does not increase the maximum obtainable, but Meyers has written that a student once showed him a counterxample, but he cannot remember it.  He applies his general results to show that no other values are obtainable when using  1, 2, 3, 4.

Problematical Recreations 4.  Problem 1 and its answer, pp. 3 & 36.  (This is one of a series of booklets issued by Litton Industries, Beverly Hills, California, nd [c1963], based on the series of the same name in Aviation Week and Electronic News during 1959-1971.  Unfortunately, neither the date nor location nor author is given and the booklet is unpaginated.  The answer simply states the maximum value with no argument.)  Reproduced with a proper solution in:  Angela Dunn; Mathematical Bafflers; (McGraw-Hill, 1964, ??NYS);  revised and corrected 2nd ed., Dover, 1980, pp. 119 & 132  and with just the answer in:  James F. Hurley; Litton's Problematical Recreations; Van Nostrand Reinhold, NY, 1971, chap. 7, prob. 8, pp. 238 & 329.   "What is the largest number which can be obtained as the product of positive integers which add up to 100?"  (This type of problem must be much older than this??  Meyers writes that he first encountered such problems as an undergraduate in 1947.  If one looks at maximizing the LCM instead of the product of the terms, this is the problem of finding a permutation of  100  letters with maximum order.)

Sol Golomb.  Section 13.8  The minimization of the cost of a digital device (the juke-box problem).  IN:  Ben Nobel;  Applications of undergraduate mathematics in engineering.  MAA & Macmillan, 1967, pp. 284-286.  This considers the problem of Problematical Recreations in the inverse form.  We have  r  k-state devices which allow  k^r  choices and the cost is proportional to  rk.  Minimize the ratio of cost to capacity or maximize the ratio of capacity to cost.  [Another way to express this is to ask which base is best for a computer to use?  I recall this formulation from when I was a student in the early 1960s.  The asnwer is  e,  but here only integer values are used.  One can generalise to: given a value, find the smallest sum of numbers whose product is the given value.]  The connection with juke-boxes is that they typically have two rows of  12  buttons and one has to press two buttons to make a selection of one from  144  records.  One can do much better with five rows of three buttons, but asking a customer to punch five buttons may be unreasonable, so perhaps three rows of five or six buttons might be best.

The Fortieth William Lowell Putnam Mathematical Competition, 1 Dec 1979.  Problem A-1.  Reproduced in:  Gerald L. Alexanderson, Leonard F. Klosinski & Loren C. Larson;  The William Lowell Putnam Mathematical Competition  Problems and Solutions: 1965‑1984;  pp. 33 & 109.  "Find positive integers  n  and  a₁,  a₂, ...,  a_n  such that  a₁ + a₂ + ... + a_n = 1979  and the product  a₁a₂...a_n  is as large as possible."

Cliff Pickover & Ken Shirriff.  The terrible twos problem.  Theta (Crewe) 6:2 (Autumn 1992) 3-7.  They study the problem of making numbers using just  +,  -,  x,  ^,  and  1s  (or  1s  and  2s).  For a given  n,  what is the least number of digits required?  They later permit concatenation, e.g.  11  or  12  is permitted.  They report results from various programs and mention some related problems.

Bryan Dye.  1, 2, 3, 4 _ four digits that dwarfed the universe.  Micromath 10:3 (Aut 1994) 12‑13.  Says a version appeared in SA a few years ago and is discussed in:  Clifford A. Pickover; Computers and the Imagination; Alan Sutton, 1991.  Dye's version is to make the largest number using  1,  2,  3,  4  once and the signs  +,  -,  x,  ¸,  ( ),  . (i.e. decimal point).  Exponentiation was not considered a sign and was permitted.  Pickover's version allowed only the signs  -,  ( ),  . (i.e. decimal point).  The largest value found actually fits Pickover's conditions:  .3^-(.2^-{.1^-4})  has  10⁶⁹⁹⁰  digits.  The largest number using just exponentiation was  2^(3^41))  with  10¹⁹  digits.

 

          7.J.     SALARY PUZZLE

 

          It is better to get a rise of  5  every six months than a rise of  20  every year.  The interpretation of the first phrase is somewhat ambiguous _ see Mills (1993).  If the salary is  S  every six months, the usual interpretation of the first phrase is that the half-yearly payments are:  S,  S + 5,  S + 10,  S + 15,  S + 20,  S + 25,  ...,   while the second phrase gives payments of:  S,  S,  S + 10,  S + 10,  S + 20,  S + 20,  ...,   and the former gets  5  extra every year.

 

Ball.  MRE, 3rd ed., 1896. pp. 26‑27.  £20 per year versus £5 every half year.  He says this is a question "which I have often propounded in past years."  It is not in the 1st ed.

Workman.  Op. cit. in 7.H.1.  1902.  Section IX (= Chap. XXXI in c1928 ed.), examples CXLV, prob. 16, pp. 425 & 544 (431 & 577 in c1928 ed.).  Compares rise of £15 per year every 3 years with £5 every year.  This represents a precursor of the puzzle version.

A. Sonnenschein & H. A. Nesbit.  The New Science and Art of Arithmetic For the Use of Schools.  A. & C. Black, London, 1903.  Prob. 12 & 14, pp. 341 & 489.  "A youth entered an office at the age of 15 at a salary of £40 a year, with an annual rise of £12.  ..."  "... What total sum would he have received in 30 years? and what would he have received if the increase had been at the rate of £1 per month?"

Mr. X.  His Pages.  The Royal Magazine 10:4 (Aug 1903) 336-337.  Not too obvious.  Half-yearly rise of £5 versus £20 a year.  No solution given.

Susan Cunnington.  The Story of Arithmetic.  Swan Sonnenschein,, London, 1904.  Prob. 35, p. 217.  £5 raise each six months versus £10 raise each year.

Dudeney??  Breakfast Table Problems No. 330: Smith's salary.  Daily Mail (30  &  31 Jan 1905) both p. 7.  Raise of  10  per year versus  2½  every six months.

Pearson.  1907.  Part II, no. 87, pp. 132 & 208.  As in Ball.

Loyd.  Salary puzzle.  Cyclopedia, 1914, pp. 312 & 381.  = MPSL1, prob. 84, pp. 81 & 150‑151.  = SLAHP: The stenographer's raise, pp. 60 & 108.  Raise of  100  per year versus  25  every half year.  Interpreting the raise of  25  as worth only  12.50  in a half‑year, this option loses, contrary to all other approaches.

Clark.  Mental Nuts.  1916: no. 6.  The two clerks.  $25 rise each six months versus $100 rise each year.

Dudeney.  AM.  1917.  Prob. 26: The junior clerk's puzzle, pp. 4 & 150.  Two clerks getting £50 per year with one getting a raise of £10 per year versus the other getting a raise of  £2 10s  every six months with a complicated further process of savings at different rates for five years.

Adams.  Puzzles That Everyone Can Do.  1931.  Prob. 276, pp. 103 & 171: A problem in salaries.  £20 rise every six months versus £80 rise each year.

T. O'Conor Sloane.  Rapid Arithmetic.  Quick and Special Methods in Arithmetical Calculation Together with a Collection of Puzzles and Curiosities of Numbers.  Van Nostrand, 1922.  [Combined into: T. O'Conor Sloane, J. E. Thompson & H. E. Licks; Speed and Fun With Figures; Van Nostrand, 1939.]  The two clerks, p. 168.  Raise of $50 every six months versus $200 each year.

F. & V. Meynell.  The Week‑End Book, op. cit. in 7.E.  1924.  Prob. one, p. 274 (2nd ed.), pp. 406‑407 (5th? ed.).  Raise of 20 per year versus 5 every half year.

Peano.  Giochi.  1924.  Prob. 16, p. 5.  1000 per year with rise of 20 each year versus 500 each half-year with rise of 5 each half-year.

Wood.  Oddities.  1927.  Prob. 31: A matter of incomes, p. 31.  $1000 per year with $20 per year increase versus "$5 each half year increase".

Collins.  Fun with Figures.  1928.  Do figures really lie?, pp. 35-36.  $50 every six months versus $200 per year.

R. Ripley.  Believe It Or Not!  Book 2.  (Simon & Schuster, 1931);  Pocket Books, NY, 1948.  P. 123: Figure your raise in pay.  A raise of one every day is better than a raise of 35 every week.  (Assumes a six day week.)

Phillips.  Week‑End.  1932.  Time tests of intelligence, no. 20, pp. 16 & 189‑190.  Raise of 2000 per year versus 500 half‑yearly.

Phillips.  Brush.  1936.  Prob. G.1: The two clerks, pp. 20 & 87.  Raise of 200 annually versus 50 half‑yearly.

McKay.  At Home Tonight.  1940.  Prob. 4: A choice of rises, pp. 63 & 76.  £5 per half-year versus £15 per year.  Solution is unclear and seems to be wrong.  "£5 each six months is £5 in the first half-year and £10 in the second _ that is, £15 per year.  But the man who gets £5 per six months gets £5 in the first year, and of course he keeps this advantage year by year."  I get that the first case is ahead by  £5n  in the  n-th year.

Sullivan.  Unusual.  1943.  Prob. 5: Raising the raise question.  Raise of $20 per year versus $5 every half year.

Birtwistle.  Calculator Puzzle Book.  1978.  Prob. 56: The interview, pp. 39 & 99-100.  £3050  yearly plus  £100  each year versus  £1500  half-yearly plus  £50  each half-year.

Stuart E. Mills.  Dollars and sense.  CMJ 24 (1993) 446-448.  Raise of $1000 per year versus $300 every half year.  Discusses various interpretations of the second phrase and gives some recent references.

John P. Ashley.  Arithmetickle.  Arithmetic Curiosities, Challenges, Games and Groaners for all Ages.  Keystone Agencies, Radnor, Ohio, 1997.  P. 80: The best job offer.  $20,000 per year plus a $500 raise every six months versus $20,000 per year plus a raise of $1000 each year.  He says the six-monthly payments of the first are  10,000,  10,500,  11,000, 11,500,  ...  and the payments for the second are  10,000,  10,000,  11,000,  11,000,  ....  But the second is a raise of $2,000 per year!

 

          7.K.    CONGRUENCES

 

          7.K.1. CASTING OUT NINES

 

          See Smith, History II 151‑154 for a detailed discussion.  He says it appears in al‑Khowarizmi and al‑Karkhi and that it is generally assumed to come from India, but his earliest Indian source is Lilivati, 1150.  G. R. Kaye;  References to Indian mathematics in certain Mediæval works;  J. Asiatic Society of Bengal (NS) 7:11 (Dec 1911) 801-816 notes the appearances in al-Khwarizmi, Avicenna and Maximus Planudes [Arithmetic after the Indian method; c1300; op. cit. in 7.E.1] but asserts it does not occur in early Indian sources _ but cf Aryabhata II, 950.

          Dickson I, chap. XII, pp. 337‑346, especially p. 337, gives a concise history.  He says al‑Karkhi was the first to use a  (mod 11)  check.

          See Tropfke, pp. 165-167.

          I have recently realised that certain puzzle problems should be listed here, but so far I have only noted Peano.

 

St. Hippolytus.  Κατα Πασωv Αιρεσεωv Ελεγχoσ (??= Philosphumena) (= Refutatio Omnium Haeresium  = Refutation of all Heresies).  c200.  Part iv, c. 14.  ??NYS.  Discusses adding up digits corresponding to letters  (mod 9)  and mentions considering it  (mod 7).  (HGM I 115‑117)  See also:  Smith II 152,  Dickson I 337  and Saidan (below), p. 472.  Hippolytus doesn't use the method to check any arithmetic.  (St. Hippolytus may be the only antipope to be counted a saint!  A reference says he was Bishop of Portus and the MS was discovered at Mt. Athos in 1842.)

Iamblichus.  On Nicomachus's Introduction to Arithmetic.  c325.  ??NYS.  In:  SIHGM, vol. 1, pp. 108‑109.  Special case.  (See also HGM I 114‑117.)

Muhammad ibn Mûsâ al‑Khwârizmî.  c820.  Untitled Latin MS of 13C known as Algorismus or Arithmetic, Cambridge Univ. Lib. MS Ii.6.5.  Facsimile ed., with transcription and commentary by Kurt Vogel as:  Mohammed ibn Musa Alchwarizmi's Algorismus, Das früheste Lehrbuch zum Rechnen mit indischen Ziffern; Otto Zeller Verlagsbuchhandlungen, Aalen, 1963.  English translation by John N. Crossley & Alan S. Henry as:  Thus spake al‑Khw_rizm_: A translation of the text of Cambridge University Library Ms.Ii.vi.5; HM 17 (1990) 103‑131.  [Crossley & Henry name this author  Ab_ Ja‘far Muhammad ibn M_s_ al‑Khw_rizm_,  but I have seen no other authority giving  Ab_ Ja‘far  _ several give  Ab_ ‘Abdall_h  and Rosen's translation The Algebra of Mohammed ben Musa specifically says our author "must therefore be distinguished from  Abu Jafar Mohammed ben Musa,  likewise a mathematician and astronomer, who flourished under the Caliph Al Motaded" (c900).  F. 108r  = Vogel p. 25  = Crossley & Henry p. 117.  Describes casting out  9s  in doubling and in multiplication.

Aryabhata II.  Mahâ‑siddhânta.  950.  Edited by M. S. Dvivedi, Braj Bhushan Das & Co., Benares, 1910.  English Introduction, pp. 21‑23;  Sanskrit text, p. 245.  Casting out  9s  for multiplication, division, squaring, cubing, and taking square and cube roots.  (Datta & Singh I 181 give the text in English.)

Abû al-Hassan Ahmad Ibn Ibrâhîm al-Uqlîdisî.  Kitâb al Fusûl fî al‑Hisâb al‑Hindî.  952/953.  MS 802, Yeni Cami, Istanbul.  Translated and annotated by A. S. Saidan as:  The Arithmetic of Al‑Uql_dis_; Reidel, 1978.  Book II, chap. 13, pp. 153‑155 and Book III, chap. 7‑8, pp. 195‑201 deal with checking by casting out  9s,  which is given only briefly, apparently being well‑known.  He applies it to division and square roots.  The method is also mentioned in Book II, chap. 2.  On pp. 468‑472, Saidan discusses the appearance of various rules in early texts.  His earliest Indian example is Lilivati, 1150, but he gives no reference.

K_shy_r ibn Labb_n  = Ab_ al-Hasan K_šy_r ibn Labb_n ibn B_šahri al-__l_.  Kit_b f_ us_l His_b al-Hind [Principles of Hindu Reckoning].  c1000.  Facsimile with translation by Martin Levey & Marvin Petruck.  Univ. of Wisconsin Press, Madison, 1965. 

First Book, Ninth Section: On arithmetic checks, f. 274a, pp. 70-71.  Brief description of casting out nines.

Second Book, Eleventh Section: On checks, ff. 280a-280b, pp. 94-97.  Brief description of casting out nines in base 60.

Introduction, pp. 32-33, discusses the above, noting that use of  9  in base  60  is unreasonable, but others also did it.  Says Sibt al-M_rid_n_ (16C) used casting out  8s  and  7s  in base  60,  but that al-K_sh_ (15C) used casting out  59s  in base  60.  K_shy_r does not state that the check proves the correctness of the result, though this was commonly believed, e.g. by Fibonacci and Sibt, though al-K_sh_ clearly discusses the question.

Ibn Sina  = Avicenna.  Treatise on Arithmetic.  c1020.  ??NYS.  Complete rules for checking operations by casting out  9s,  attributed to the Hindus, (Smith, Isis 6 (1924) 319).  (See also Cammann _ 3 (cited in 7.N);  Datta & Singh, I, 184;  and Kaye, above, who cite F. Woepcke;  Mémoire sur la propagation des chiffres indiens;  J. Asiatique (6) 1 (1863) 27-529;  p. 502,  ??NYS.)  The DSB entry indicates that the material is in Ibn Sina's Al‑Shifâ' (The Healing) and there doesn't appear to be a translation.  Suter 89 mentions some Latin translations but I'm not clear whether they are this book or a related book.

                    Saidan's discussion says Woepcke (p. 550 [sic]) construes ibn Sina as saying that the method is Indian, but this is a contentious interpretation.  Kaye, above, says Woepcke is wrong.  Smith, History II 151, says the expression "has been variously interpreted".

Bhaskara II.  Lilivati.  1150.  Smith, History II 152, cites this in Taylor's edition, p. 7, but the method is not in Colebrooke and neither Dickson nor Datta & Singh cite it, so perhaps it is an addition in the text Taylor used??

Fibonacci.  1202.  Pp. 8, 20, 39, 45 uses checks  (mod 7, 9 and 11).  On p. 8, he says that if the 'proof' is right, then the calculation is correct _ see comments at K_shy_r above.

Maximus Planudes.  Ψηφηφoρια κατ' Ivδoυσ η Λεγoμεvη Μεγαλη (Psephephoria kat' Indous e Legomene Megale) (Arithmetic after the Indian method).  c1300.  (Greek ed. by Gerhardt, Das Rechenbuch des Maximus Planudes, Halle, 1865, ??NYS [Allard, below, pp. 20-22, says this is not very good].  German trans. by H. Waeschke, Halle, 1878, ??NYS [See HGM II 549;  not mentioned by Allard].)  Greek ed., with French translation by A. Allard; Maxime Planude _ Le Grand Calcul selon des Indiens; Travaux de la Faculté de Philosophie et Lettres de l'Univ. Cath. de Louvain _ XXVII, Louvain‑la‑Neuve, 1981.  Proofs by casting out 9s are given in the material on the operations of arithmetic.

N_r_yana Pandita.  Ganita‑kaumud_.  1356.  Edited by P. Dvivedi, Indian Press, Benares, 1942.  Introduction in English, p. xv, discusses the material.  Allows any modulus.  (English in Datta & Singh I 183.)

The Treviso Arithmetic = Larte de labbacho.  Op. cit. in 7.H.  1478.  F. 4v onward (p. 46 in Swetz) uses casting out  9s  as a check on many examples.  Swetz (p. 189) refers to Avicenna and the Hindus.  On f. 10v (Swetz p. 59), the anonymous author says that proving a subtraction by addition "is more rapid and also more certain than the proofs by  9s"  and he makes similar statements regarding multiplication and division.

                    On p. 323 of his Isis article, Smith says the author "gives a proof by casting out sevens".  This would be on or near f. 17v (Swetz p. 73).  I can find nothing of the sort _ the author has an example of multiplication by  7,  but he checks it by casting out  9s.

Borghi.  Arithmetica.  1484.  Ff. 8r-9r (1509: ff. 9r-9v).  Casting out  7s  and  9s.  This is applied over the next few sections, but I don't see any indication that casting out  9s  is not a certain test.  However he uses casting out  7s  more often than  9s  which may indicate that he was aware that  7s  is a more secure test than  9s.

Chuquet.  1484.  Triparty, part 1.  English in FHM 41-42.  "There are several kinds of proofs such as the proof by  9,  by  8,  by  7,  and so on by other individual figures down to  2,  ....  Of these only the proof by  9,  because it is easy to do, and the proof by  7,  because it is even more certain than that by  9  are treated here."  He then notes that these proofs are not always certain.

Pacioli.  Summa.  1494.  Ff. 20v-23v.  Discusses casting out  9s  and  7s  and notes that these tests are not sufficient.

Apianus.  Kauffmanss Rechnung.  1527.  Gives numerous examples of testing by  9s,  and also by  8s,  7s  and  6s,  in his sections on the four arithmetic operations and also under arithmetic progressions.

Robert Recorde.  Ground of Artes.  1542.  Op. cit. in 7.G.2.  Discusses the proof by nines in his chapters on:  addition, ff. D.i.r - D.ii.r;  subtraction, ff. F.iii.r - F.iiii.r;  multiplication, ff. G.vi.r - G.vi.v;  and division, ff. H.iii.v - H.iiii.v.

Hutton.  A Course of Mathematics.  1798?  1833 & 1857: 6-12.  In his discussion of the basic arithmetic operations, we find on p. 7 under  To Prove Addition,  "Then, if the excess of 9's in this sum, ...., be equal to the excess of 9's in the total sum ..., the work is right."  A footnote explains the idea and is less clear as to the direction of implication being asserted: "it is plain that this last excess must be equal to the excess of 9's contained in the total sum".  The note concludes: "This rule was first given by Dr Wallis in his Arithmetic, published in the year 1657."  However, Hutton does not mention the rule under subtraction and under multiplication on pp. 10-11, he says the "remainders must be equal when the work is right."  All in all, it seems that he is surprisingly unclear for his time.

Peano.  Giochi.  1924.  Prob. 50, p. 13.  Take  x,  form  10x  and subtract  x.  Cancel a non-zero figure from the result and tell me the other figures.  I will tell what number you cancelled.

 

          7.L.    GEOMETRIC PROGRESSIONS

 

          See Tropfke 628.

          These also occur in 7.M, 7.S.1 and 10.A. 

          I am starting to include early problems which involve interpolation in a geometric series here _ these were normally solved by linear interpolation.  From about 1400, such problems arise in compound interest but I will omit most such problems.  See Chuquet here and  Chiu Chang Suan Ching  &  Cardan in 10.A.

 

H. V. Hilprecht.  Mathematical, Metrological and Chronological Tablets from the Temple Library of Nippur.  Univ. of Pennsylvania, Philadelphia, 1906.  Pp. 13, 28‑34, 62, 69, pl. 15, PL. IX, are about a tablet which has a geometric progression from c‑2300.  The progression is double:  a_n = 125 * 2ⁿ  and  60⁴/a_n  for  n = 0, 1, ..., 7.  There is no summation.

Tablet K 90 of the British Museum.  A moon tablet deciphered by Hincks containing  5, 10, 20, 40, 80  followed by  80, 96, 112, 128, ..., 240.  Described in The Literary Gazette (5 Aug 1854) _ ??NYS.  Cited and described in:  Nicomachus of Gerasa: Introduction to Arithmetic; Translated by Martin Luther D'Ooge, with notes by Frank Egleston Robbins and Louis Charles Karpinski; Macmillan, London, 1926; p. 12.

Euclid.  IX: 35, 36.  This gives the general rule for the sum of a geometric progression.

Zhang Qiujian .  Zhang Qiujian Suan Jing.  Op. cit. in 7.E.  468,  ??NYS.  Mikami 42 gives:  "A horse, halving its speed every day, runs 700 miles in 7 days.  What are his daily journeys?" _ i.e.  x*(1 + 1/2 + ... + 1/64)  =  700.  Solved by adding up.

Mahavira.  850.

Chap. IV, v. 28, p. 74:  x ‑ x/2 ‑ x/4 ‑ ... ‑ x/256  =  32.

Chap. VI, v. 314, pp. 175‑176:  Let  a_i+1 = r*a_i + c.  He sums such terms.

Fibonacci.  1202.  Pp. 313‑316.  Man has  100  and gives away  1/10  of his wealth  12  times.  This has been described under 7.E.

Columbia Algorism.  c1370.  Prob. 63, pp. 84‑85.  Same as the Fibonacci,  but he converts to pounds, shillings and pence!

Lucca 1754.  c1390.  F. 10v, pp. 36‑37.  Computes  2⁴⁰  &  2¹⁰⁰  by repeated squaring.

Chuquet.  1484.  He gives a number of such problems _ see also 7.E.

Prob. 96, English in FHM 219.  Cask of  9½  drains so first barrel takes  1  hour,  second barrel takes  2  hours,  third barrel takes  4  hours,  ....  How long to empty?  I.e. he wants the sum of  9½  terms of a geometric progression.  He gets the correct answer of  2^9.5 - 1  hours.

Prob. 97, English in FHM 219.  Man travels  1, 3, 9, ...  leagues per day.  How far has he travelled in  5½  days?  He gets the correct answer of  (3^5.5-1)/2  days.

Ghaligai.  Practica D'Arithmetica.  1521.  Prob. 29, f. 66v.  Same as Fibonacci, pp. 313-316.  (H&S 59-60.)

Buteo.  Logistica.  1559.

Prob. 69, pp. 276-278.  1 + .9 + (.9)² + ... + (.9)^x  =  7.5.  The phrasing of the problem is unclear, but this is what he considers.  He interpolates linearly between  12  and  13,  getting  12.164705107,  while the exact answer is  (log .25/log .9) ‑ 1  =  12.15762696.

Prob. 84, pp. 294-296.  Relates  2ⁱ  and  5ⁱ  for  i = 1, ..., 7.  He determines  2^3.5  as  Ö128  which he estimates as  11_.  Similarly he estimates  5^3.5  as  279.

Jacques Chauvet Champenois.  Les Institutions de L'Arithmetique.  Hierosme de Marnef, Paris, 1578, p. 70.  ??NYS.  Problem of tailor and robe involving  4888  divided by  2  twenty times.  (French quoted in H&S 14‑15.)

van Etten.  1624.  Prob. 87 (84): Des Progressions & de la prodigieuse multiplication des animaux, des Plantes, des fruicts de l'or & de l'argent quand on va tousjours augmentant par certaine proportion,  pp. 111‑118 (177‑183).  Numerous examples including horseshoe problem and chessboard problem, with ratios 1000, 4, 50.  Henrion's Notte, p. 38, observes that there are many arithmetical errors which the reader can easily correct.  In part X: Multiplication des Hommes, he considers one of the children of Noah, says a generation takes 30 years and that, when augmented to the seventh, one family can easily produce 800,000 souls.  The 1674 English ed. has: "... if we take but one of the Children of Noah, and suppose that a new Generation of People begin at every 30 years, and that it be continued to the Seventh Generation, which is 200 years; ... then of one only Family there would be produced 111000 Souls, 305 to begin the World: ... which number springing onely from a simple production of one yearly ...."

Leybourn.  Pleasure with Profit.  1694.  Chap. VI, pp. 24-28: Of the Increase of Swine, Corn, Sheep, &c.  Examples with ratios 4, 40, 2, 1000, 2, mostly taken from van Etten.  Then art. VI: Of Men, discusses the repopulation of the world from Noah's children: "... if we take but one of the Children of Noah, and suppose that a New Generation of People begin at every 30 years, and that it be continued to the seventh Generation, which is 210 years; ... then, of one ony family there would be produced 111305, that is, One hundred and eleven thousand, three hundred and five Souls to begin the World ....  ... such a number arising only from a simple production of only One yearly ...."  I cannot work out how 111305 arises _ the fact that he spells it out makes it unlikely to be a misprint.

Ozanam.  1694.  Prob. 8, 1696: 33-35;  1708: 29-32.  Prob. 11, 1725: 68-75.  Section II, 1778: 68-74;  1803: 70-76;  1814: ??NYS;  1840: 34-36.  A discussion of geometric progression and a mention of  1, 2, 4, ....  1778 et seq also mention 1, 3, 9, ....

Ozanam.  1725.  Prob. 11, questions 6 & 7, 1725: 79‑82.  Prob. 3, parts 1-3, 1778: 80-82;  1803: 82-84;  1814: 72-75;  1840: 38-39.  Examples of population growth in Biblical and biological contexts.  In 1725, he has ratios of 2, 50, 3, 4, 1000, The examples vary a bit between 1725 and 1778.

Walkingame.  Tutor's Assistant.  1751.  The section Geometrical Progression gives several problems with powers of  2  and the following less common types. 

Prob. 5, 1777: p. 95;  1835: p. 103;  1860: p. 123.  Find  1 + 4 + 16 + ... + 4¹¹  farthings.

Prob. 8, 1777: p. 96;  1835: p. 104;  1860: p. 123.  Find  2 + 6 + 18 + ... + 2 x 3²¹.  If these are pins, worth 100 to the farthing, what is the value?

Vyse.  Tutor's Guide.  1771?    The section Geometrical Progression, pp. 146-151 & Key pp. 190-192, gives several examples with doublings and triplings as well as examples with ratios of  3/2  and  10.  There is a major error in the solution of prob. 7, to find  2 + 6 + 18 + ... + 2 x 3¹⁹.

Pike.  Arithmetic.  1788.  Pp. 237-239.  Numerous fairly standard examples, mostly doubling, but with examples of powers of  3  and of  10  and the following.

Pp. 239-240, no. 8.  One farthing placed at  6%  compound interest in year  0  is worth what after  1784  years?  And supposing a cubic inch of gold is worth  £53 2s 8d,  how much gold does this make?  This is very close to  2¹⁵⁰  farthings and makes about  4 x 10¹⁴  solid gold spheres the size of the earth!

Eadon.  Repository.  1794.  P. 241, ex. 3.  Doubling  20  times from a farthing.

John King, ed.  John King  1795  Arithmetical Book.  Published by the editor, who is the great-great-grandson of the 1795 writer, Twickenham, 1995.  P. 100.  10 + 10² + ... + 10¹¹  grains of wheat, converted to bushels and value at  4s  per bushel.

(Beeton's) Boy's Own Magazine 3:6 (Jun 1889) 255  &  3:8 (Aug 1889) 351.  (This is undoubtedly reprinted from Boy's Own Magazine 1 (1863).)  Mathematical question 59.  Seller of  12  acres asks  1  farthing for the first acre,  4  for the second acre,  16 for the third acre, ....  Buyer offers  £100  for the first acre,  £150 for the second acre,  £200  for the third acre, ....  What is the difference in the prices asked and offered?  Also entered in 7.AF.

 

          7.L.1. 1 + 7 + 49 + ...  &  ST. IVES

 

          See Tropfke 629.

 

Papyrus Rhind, c‑1650, loc. cit. in 7.C.  Problem 79, p. 112 of vol. 1 (1927) (= p. 59 (1978)).  7 + 49 + 343 + 2401 + 16807.  (Sanford 210 and H&S 55 give Peet's English.)  Houses, cats, mice, ears of spelt, hekats.

L. Rodet.  Les prétendus problèmes d'algèbre du manuel du calculateur Égyptien (Papyrus Rhind).  J. Asiatique 18 (1881) 390‑459.  Appendice, pp. 450‑454.  Discusses this problem and its appearance in Fibonacci (below).

F. Cajori.  History of Mathematics.  2nd ed., Macmillan, 1919;  Chelsea, 1980.  P. 90 gives the legend that Buddha was once asked to compute  7¹⁷.

Shakuntala Devi.  The Book of Numbers.  Orient Paperbacks (Vision Books), Delhi, 1984.  This gives more details of the Buddha story, saying it occurs in the Lalitavistara and Buddha finds the number of atoms (of which there are seven to a grain of dust) in a mile, obtaining a number of 'about 50 digits'.  Note:  7⁵⁸  =  1.04 x 10⁴⁹.

Alcuin.  9C.  Prob. 41: Propositio de sode et scrofa.  This has sows which produce  7  piglets, but this results in a GP of ratio  8.

Fibonacci.  1202. 

Pp. 311‑312.  7 + ... + 117649.  Old women, mules, sacks, loaves, knives, sheathes.  (English in:  N. L. Biggs; The roots of combinatorics; HM 6 (1979) 109‑136 (on p. 110) and Sanford 210.  I have slides of this from L.IV.20 & 21.  It is on f. 147r of L.IV.20 and f. 225r of L.IV.21.)  

P. 312.  100 + 10000 + ... + 10⁸  _ branches, nests, eggs, birds.

Munich 14684.  14C.  Prob. XXXI, pp. 83‑84.  Refers to  7 + 49 + ... + 117649.

AR.  On p. 227, Vogel refers to an example in CLM 4390 which has not been published.

Peter van Halle.  MS. 3552 in Royal Library Brussels, beginning "Dit woort Arithmetica coomt uuter griexer spraeken ...."  1568.  F. 23v.  "There were 5 women and each woman had 5 bags but in each bag were 5 cats and each cat had 5 kittens question how many feet are there to jump with?"  Copy of original Dutch text and English translation provided by Marjolein Kool, who notes that van Halle only counts the feet on the kittens.

Josse Verniers.  MS. 684 in University Library of Ghent, beginning "Numeration heet tellen ende leert hoemen die ghetalen uutghespreken ende schrijven ...."  1584.  F. 7v.  "Item there is a house with 14 rooms and in each room are 14 beds and each bed lay 14 soldiers and each soldier has 14 pistols and in each pistol are 3 bullets Question when they fire how many men, pistols and bullets are there"  Copy of original Dutch text and English translation provided by Marjolein Kool.

Harley MS 7316, in the BM.  c1730.  ??NYS _ quoted in:  Iona & Peter Opie; The Oxford Dictionary of Nursery Rhymes; OUP, (1951);  2nd ed., 1952, No. 462, p. 377.  The Opies give the usual version with  7s,  but their notes quote Harley MS 7316 as:  "As I went to St. Ives I met Nine Wives And every Wife had nine Sacs And every Sac had nine Cats And every Cat had Nine Kittens."  The Opies' notes also cite Mother Goose's Quarto (Boston, USA, c1825), a German version with  9s  and a Pennsylvania Dutch version with  7s.

Child.  Girl's Own Book.  1842: Enigma 35, pp. 233-234;  1876: Enigma 27, pp. 196-197.  "As I was going to St. Ives,  I chanced to meet with nine old wives:  Each wife had nine sacks,  Each sack had nine cats,  Each cat had nine kits;  Kits, cats, sacks and wives,  Tell me how many were going to St. Ives?"  Answer is  "Only myself.  As I met all the others, they of course were coming from St. Ives."  The 1876 has a few punctuation changes.

Halliwell, James Orchard.  Popular Rhymes & Nursery Tales of England.  John Russell Smith, London, 1849.  Variously reprinted _ my copy is Bodley Head, London, 1970.  P. 19 refers to "As I was going to St. Ives" in MS. Harl. 7316 of early 18C, but doesn't give any more details.

Kamp.  Op. cit. in 5.B.  1877.  No. 20, p. 327.  12  women, each with  12  sticks, each with  12  strings, each with  12  bags, each with  12  boxes, each with  12  shillings.  How many shillings?

Mittenzwey.  1879?  prob. 24, pp. 9 & 57.  Man going to Stötteritz meets  9  old women, each with  9  sacks, each with  9  cats, each with  9  kittens.  How many were going to Stötteritz?  Answer is one.

Fireside Amusements _ A Book of Indoor Games.  Chambers, London & Edinburgh, nd [probably the 1890 ed, but the material may date back to the 1848 ed., ??NYS].  P. 102, no. 34:  "As I was going to St Ives, I chanced to meet with nine old wives;  Each wife had nine sacks, Each sack had nine cats, Each cat had nine kits.  Kits, cats, sacks and wives, Tell me how many were going to St Ives?"  Answer:  one.

Mary & Herbert Knapp.  One Potato, Two Potato ...  The Secret Education of American Children.  Norton, NY, 1976.  Pp. 107-108 gives a modern New York City version:  "There once was a man going to St. Ives Place.  He had seven wives;  each wife had seven sacks;  each sack had seven cats;  each cat had seven kits.  How many altogether were going to St. Ives Street?  One."  St. Ives has become attached to a location in New York!

Colin Gumbrell.  Puzzler's A to Z.  Puffin, 1989.  Pp. 9 & 119: As I was going ... 

          "As I was going to St Ives,   I met a man with seven wives;   And every wife had seven sons;   But they were not the only ones,   For every son had seven sisters!   Bewildered by so many misters   And by so many misses too,   I quickly cried: 'Bonjour!  Adieu!'   And hurried to another street,   Away from all their trampling feet.   Now, here's the point that puzzles me yet:   Just how many people had I met?"  Answer is 106, or 64 if there are just seven girls who are half-sisters to all the 49 sons.

The Michelin Green Guide to Brittany (3rd ed., Michelin et Cie, Clermont‑Ferrand, 1995, pp. 178 & 237-238) describes the Breton St. Yves, the original of the Cornish St. Ives.  St. Yves (Yves Helori (1253‑1303)) was once parish priest at Louannec, Côtes-d'Armor, where a chasuble of his is preserved in the church.  His tomb is in the Cathedral of St. Tugdual in Tréguier, Côtes-d'Armor.  He worked as a lawyer and is the patron saint of lawyers.  His head is in a reliquary in the Treasury.  He was born in the nearby village of Minihy-Tréguier and his will is preserved in the Chapel there.  The Chapel cemetery contains a monument known as the tomb of St. Yves, but this is unlikely.  Attending his festival, known as a 'pardon', is locally known as 'going to St. Yves' _ !!

 

          7.L.2. 1 + 2 + 4 + ...

 

Chiu Chang Suan Ching.  c‑150?  Chap. III, prob. 4, pp. 28‑29.  Weaver weaves  a (1 + 2 + 4 + 8 + 16),  making  5  in all.  (English in Needham, pp. 137‑138.  Needham says this problem also occurs in Sun Tzu (presumably the work cited in 7.P.2, 4C), ??NYS.)

Alcuin.  9C.  Prob. 13: Propostio de rege et de ejus exercitu.  1 + 1 + 2 + 4 + ... + 2²⁹  =  2³⁰.  Calculations are suppressed in the Alcuin text, but given in the Bede.  Murray 167 wonders if there is any connection between this and the Chessboard Problem (7.L.2.a).

Bhaskara II.  Lilavati.  1150.  Chap. V, sect. II, v. 128.  In Colebrooke, pp. 55‑56.  2 + 4 + ... + 2³⁰.

Leybourn.  Pleasure with Profit.  1694.  See in 7.L.

Ozanam.  1725.  Prob. 11, question 5, 1725: 78‑79.  1 + 2 + ... + 2³¹.

Walkingame.  Tutor's Assistant.  1751.  The section Geometrical Progression gives several problems with straightforward doublings _ see 7.L and 7.L.2.b for some more interesting examples.

Vyse.  Tutor's Guide.  1771?  Same note as for Walkingame.

Eadon.  Repository.  1794.  P. 241, ex. 3.  Doubling 20 times from a farthing.

John King, ed.  John King  1795  Arithmetical Book.  Published by the editor, who is the great-great-grandson of the 1795 writer, Twickenham, 1995.  P. 115.  For  20  horses, is starting with a farthing and doubling up through the  19th  horse, with the  20th  free, more or less expensive than  £20  per horse? 

Boy's Own Book.

Curious calculation.  1868: 433.  1 + 2 + ... + 2⁵¹  pins would weigh  628,292,358  tons and require  27,924  ships as large as the Great Eastern to carry them.

Arithmetical [sic] progression.  1868: 433.  1 + 2 + ... + 2⁹⁹  farthings.  Answer is wrong.

Ripley's Believe It or Not, 4th Series, 1957.  P. 15 asserts that the Count de Bouteville directed that his widow, age 20, should receive one gold piece during the first year of widowhood, the amount to be doubled each successive year she remained unmarried.  She survived 69 years without marrying!  Ripley says the Count 'never suspected the cumulative powers of arithmetical [sic!] progression'.

 

          7.L.2.a.         CHESSBOARD PROBLEM

 

          See Tropfke 630.  See 5.F.1 for more details of books on the history of chess.

 

Perelman.  FFF.  1934.  1957: prob. 52, pp. 74-80;  1979: prob. 55, pp. 92-98.  = MCBF, prob. 55, pp. 90-98.  This describes a Roman version where the general Terentius can take  1  coin the first day,  2  the second day,  4  the third day,  ..., until he can't carry any more, which occurs on the  18th  day.  A footnote says this is a translation "from a Latin manuscript in the keeping of a private library in England."  ??

Murray mentions the problem on pp. 51‑52, 155, 167, 182 and discusses it in detail in his Chapter XII: The Invention of Chess in Muslim Legend, pp. 207‑219.  He discusses various versions of the invention of chess, some of which include the doubling reward.  He describes the doubling legends in the following.

al‑Ya‘qûbî (c875).

al-Mas‘udi (943).

Firdawsî's Shâhnâma (1011).

Kitâb ash‑shatranj [= AH] (1141) as AH f. 3b (= Abû Zakarîyâ [= H] f. 6a).

BM MS Arab. Add. 7515 (Rich) [= BM] (c1200?).

von Eschenbach (c1220).

BM Cotton Lib. MS Cleopatra, B.ix [= Cott.] (13C).

ibn Khallikan (1256).

Dante (1321).

Shihâbaddîn at‑Tilimsâni [= Man.] (c1370), which gives five versions.

Kajînâ [= Y] (16C?).

          Murray 218 mentions two treatises on the problem:

Al‑Missisî.  Tad‘îf buyût ash‑shatranj.  9C or 10C.

Al‑Akfânî.  Tad‘îf ‘adad ruq‘a ash‑shatranj.  c1340.

                    On p. 217, Murray gives 10 variant spellings of Sissa and feels that Bland's connection of the name with Xerxes is right.

                    On p. 218, he says the reward of corn would cover England to a depth of  38.4  feet.

 

Murray 218.  "This calculation is undoubtedly of Indian origin, the early Indian mathematicians being notoriously given to long‑winded calculations of the character."  He suggests the problem may be older than chess itself.

Al‑Ya‘qûbî.  Ta’rîkh.  c875.  Ed. by Houtsma, Leyden, 1883, i, 99‑105.  ??NYS.  Cited by Murray 208 & 212.  "Give me a gift in grains of corn upon the squares of the chessboard.  On the first square one grain (on the second two), on the third square double of that on the second, and continue in the same way until the last square." [Quoted from Murray 213.]

al-Mas‘udi (= Mas'udi  = Maçoudi).  Murûj adh‑dhahab [Meadows of gold].  943.  Translated by:  C. Barbier de Meynard & P. de Courteille as:  Les Prairies d'Or; Imprimerie Impériale, Paris, 1861.  Vol. 1, Chap. VII, pp. 159‑161.  "The Indians ascribe a mysterious interpretation to the doubling of the squares of the chessboard;  they establish a connexion between the First Cause which soars above the spheres and on which everything depends, and the sum of the square of its squares.  This number equals  18,446,744,073,709,551,615 ...."  [Quoted from Murray 210.  The French ed. has two typographical errors in the number.]  No mention of the Sessa legend.

Muhammad ibn Ahmed Abû’l-Rîhân el-Bîrûnî  (= al‑Bîrûnî  = al-Biruni).  Kitâb al‑âtâr al‑bâqîya ‘an al‑qurûn al‑halîya  (= al‑Âthâr al‑bâqiya min al‑qurûn al khâliya   = Athâr‑ul‑bákiya)  (The Chronology of Ancient Nations).  1000.  Arabic (and/or German??) ed. by E. Sachau, Leipzig, 1876 (or 1878??), pp. 138‑139.  ??NYS.  English translation by E. Sachau, William H. Allen & Co., London, 1879, pp. 134‑136.  An earlier version is:  E. Sachau; Algebraisches über das Schach bei Bîrûnî; Zeitschr. Deut. Morgenländischen Ges. 29 (1876) 148‑156, esp. 151‑155.  Wieber, pp. 113‑115, gives another version of the same text.  Computes  1 + 2 + 4 + ... + 2⁶³  as  2⁶⁴ ‑ 1  by repeated squaring.  Doesn't mention Sessa.  He shows the total is  2,305  mountains.  "But these are (numerical) notions that the earth does not contain."  Murray 218 gives: "which is more than the world contains." but I'm not sure if al-Biruni means the mountains or the numbers are more than earth can contain.

BM MS Arab. Add. 7515 (Rich).  Arabic MS with the spurious title  "Kitâb ash‑Shatranj al Basrî",  perhaps c1200.  Copied in 1257.  Described by Bland and Forbes, loc. cit. in 5.F.1 under Persian MS 211, and by Murray on p. 173.  Murray denotes it BM.

                    Bland, p. 26, says p. 6 of the MS gives the story of  Súsah ben Dáhir  and the reward.  Bland, p. 62, says the various forms of the name Sissah are corruptions of Xerxes.  Forbes, pp. 74‑76, does not mention the story or the reward.

                    Murray 217 says all the Arabic MSS include the reward problem as part of one of their stories of the invention of chess, but on pp. 173 & 211‑219, he doesn't mention the story in this MS specifically.  However, on p. 173, he notes that the spurious first page gives the calculation in 15C Arabic and again in Turkish.

Fibonacci.  1202.  Pp. 309‑310.  He induces the repeated squaring process and gets  2⁶⁴ ‑ 1.  He computes the equivalent number of shiploads of grain _ there is a typographic error in his result.

Wolfram von Eschenbach.  Willehalm.  c1220.  Ed. by Lachmann, p. 151, ??NYS _ quoted by Murray 755.  "Ir hers mich bevilte, der Zende ûz zwispilte ame schâchzabel ieslîch velt mit cardamôm."

Murray 755 gives several other medieval European references.

(Al-Kâdi Shemseddîn Ahmed) Ibn Khallikan.  Entry for:  Abû Bakr as‑Sûli.  In:  Kitab wafayât al‑a‘yân.  1256.  Translated by MacGuckin de Slane as:  Biographical Dictionary; (London, 1843‑1871;) corrected reprint, Paris, 1868.  Vol. III, p. 69‑73.  Sissah ibn Dâhir,  King Shihrâm  and the chessboard on pp. 69‑71.  An interpolation(?) mentions King Balhait.

BM Cotton Lib. MS Cleopatra, B.ix.  c1275.  Anglo‑French MS of c1275, described by Murray 579‑580, where it is denoted Cott.  No. 18, f. 10a, gives doubling.

Dante.  Divina Commedia: Paradiso XXVIII.92.  1321.  "Ed eran tante che'l numero loro Piu che'l doppiar degli scacchi s'imila."  [Quoted from Murray 755.]

Thomas Hyde.  Mandragorias seu Historia Shahiludii, ....  (= Vol. 1 of De Ludis Orientalibus, see  4.B.5 for vol. 2.)  From the Sheldonian Theatre (i.e. OUP), Oxford, 1694.  Prolegomena curiosa.  The initial material and the Prolegomena are unpaged but the folios of the Prolegomena are marked (a), (a 1), ....  The material is on  (d 1).r - (d 4).v, which are pages 25-32 if one starts counting from the beginning of the Prolegomena. 

                    He mentions Wallis (see below) and arithmetic (sic!) progressions and says the story is given in  al-Safadi  (Salâhaddîn as‑Safadî  = al-Sâphadi  = AlSáphadi) (d. 1363) in his  Lâmiyato ’l Agjam  (printed various ways in the text).  This must be his  Sharh Lâmîyat al‑‘Ajam  of c1350.  Hyde gives some Arabic text and a Latin translation.  Wallis gives the full Arabic text and translation.  This refers to Ibn Khallikan.  In his calculation, he uses various measures until he takes  2³⁹  grains as a granary, then  1024  granaries (=  2⁴⁹  grains) as a city, so the amount on the  64th square is  16384  (=  2¹⁴)  cities, “but you know there are not so many cities in the whole world".  He then gives  2⁶⁴ - 1  correctly and converts it into cubic miles, but seems off by a factor of ten _ see Wallis, below, who gives details of the units and calculations involved, noting that al‑Safadi is finding the edge (= height) of a square pyramid of the volume of the pile of wheat.  Hyde then adds a fragment from a Persian MS,  Mu’gjizât,  which gives the story with drachmas instead of grains of wheat, but the calculations are partly illegible.  In his main text, pp. 31-52 are on the invention of the game and he gives various stories, but doesn't mention the reward.

Shihâbaddîn Abû’l‑‘Abbâs Ahmad ibn Yahya ibn Abî Hajala at‑Tilimsâni alH‑anbalî.  Kitâb ’anmûdhaj al‑qitâl fi la‘b ash‑shatranj  (Book of the examples of warfare in the game of chess).  c1370.  Copied by  Muhammed ibn ‘Ali ibn Muhammed al‑Arzagî  in 1446.

                    This is the second of Dr. Lee's MSS, described in 5.F.1, denoted Man. by Murray.  Described by Bland and Forbes, loc. cit. in 5.F.1 under Persian MS 211, and by Murray 175-177 (as Man) & 207‑219.  Gives five versions of the chessboard story.  The first is that of ibn Khallikan;  others come from  ar-Râghib's  Muhâdarât al-Udabâ’;  from  Qutbaddîn Muhammad ibn ‘Abdalqâdir's  Durrat al-Mudî’a  and from  al-Akfânî.  One calculates in lunar years and another version calculates in miles!!

Columbia Algorism.  c1370.  Prob. 88, pp. 106‑107.  Chessboard.  Uses repeated squaring.  Copying error in the final value.

Persian MS 211.  Op. cit. in 5.F.1.  c1400.  Bland, loc. cit., p. 14, mentions "the well known story of the reward asked in grain".  Forbes' pp. 64‑66 is a translation of the episode of the Indian King Kaid's reward to Sassa.  On p. 65, Forbes mentions various interpretations of the total.

AR.  c1450.  Prob. 319, pp. 141, 180, 227.  Chessboard, with very vague story.

Benedetto da Firenze.  Italian MS, c1460, in Plimpton Collection Columbia Univ.  ??NYS.  Chessboard.  (Rara, 464‑465.)

Pacioli.  Summa.  1494.  Ff. 43r-43v, prob. 28.  First mentions  1, 2, 6, 18, 54, ..., where each cell has double the previous total.  Then does usual chessboard problem, but with no story.  Computes by repeated squaring.  Converts to castles of grain.  Shows how to do  1 + 2 + 6 + 18 + ...  for  64  cells and computes the result.

Muhammad ibn ‘Omar Kajînâ.  Kitab al‑munjih fî‘ilm ash‑shatranj  (A book to lead to success in the knowledge of chess).  16C?  Translated into Persian by  Muhammad ibn Husâm ad‑Daula,  copied in 1612.  Described by Bland and Forbes and more correctly by Murray on p. 179, where it is identified as MS BM Add. 16856 and denoted Y, since it was a present from Col. Wm. Yule.

                    Bland, p. 20, mentions  Sísah ben Dáhir al Hindi  and the reward claimed in grain.  "The geometrical progression of the sixty four squares ... is computed here at full length, commencing with a Dirhem on the first square, and amounting to two thousand four hundred times the size of the whole globe in gold."

                    Forbes describes this on pp. 76‑77 and in the note on p. 65, where he computed the reward to make a cube of gold about 6 miles along an edge.  He says the above Persian value is wrong somewhere, but he hasn't been able to see the original.  [I can't tell if he means the Persian or the Arabic MS.  If a dirhem was the size of an English  2p  coin or an American quarter, the reward is about  2 x 10⁴ km³,  compared to earth's volume of about 10¹² km³.  The reward would make a cube about  27 km on an edge or about 17 miles on an edge.]  Murray doesn't refer to this MS specifically.

Ian Trenchant.  L'Arithmetique.  Lyons, 1566, 1571, 1578, ...   ??NYS.  1578 ed., p. 297.  1, 3, 9, 27.  (H&S 91 gives French and English and says similar appear in Vander Hoecke (1537), Gemma Frisius (1540) and Buteo (1556).)

Clavius.  ??NYS.   Computes number of shiploads of wheat required.  (H&S 56.)

van Etten.  1624.  Prob. 87, pp. 111‑118 (not in English editions).  Includes chessboard as part XI, on p. 117.  Henrion's Notte, p. 38, observes that there are many arithmetical errors in prob. 87 which the reader can easily correct.

John Wallis.  (Mathesis Universalis.  T. Robinson, Oxford, 1657.  Chap. 31.)  = Operum Mathematicoroum.  T. Robinson, Oxford, 1657.  Part 1, chap. 31: De progressione geometrica, pp. 266-285.  This includes the story of Sessa and the Chessboard in Arabic & Latin, taken from  al-Safadi,  c1350, giving much more text than Hyde (see above) does and explaining the units and the calculation, showing that  al-Safadi's  60 miles should be about 6 miles and this is the edge and height of a square pyramid of the same volume as the wheat.  He then computes all the powers of two up to the  63rd and adds them!  John Ayrton Paris [Philosophy in Sport made Science in Earnest; (Longman, Rees, Orme, Brown, and Green; London, 1827);, 8th ed., Murray, 1857, p. 515] says Wallis got 9 English miles for the height and edge.

Euler.  Algebra.  1770.  I.III.XI: Questions for practice, no. 3, p. 170.  Payment to Sessa, converted to value.

Ozanam‑Montucla.  1778.  Prob. 3, 1778: 76-78;  1803: 78-81;  1814: 70-72;  1840: 37‑38.  Problem wants the results of doublings, with no story.  Discussion gives the story of  Sessa,  taken from  Al‑Sephadi.  Gives various descriptions of the pile of grain, citing Wallis for one of these and says it would cover three times the area of France to a depth of one foot.

Eadon.  Repository.  1794.  Pp. 369-370, no. 11.  Indian merchant selling  64  diamonds to a Persian king for grains of wheat, in verse.  Supposing a pint holds  10000  grains and a bushel of  64  pints weighs  50  pounds, how many horse loads (of a thousand pounds each) does this make?  How many ships of  100  tons capacity?

Manuel des Sorciers.  1825.  P. 84.  ??NX  Chessboard.

The Boy's Own Book.  The sovereign and the sage.  1828: 182;  1828‑2: 238;  1829 (US): 106;  1855: 393;  1868: 431.  Uses 63 doublings for no reason.

Hutton-Rutherford.  A Course of Mathematics.  1841?  Prob. 26,  1857: 82.  Reward to Sessa for inventing chess.  Takes a pint as 7680 grains and 512 pints as worth  27/6  to value the reward at  6.45 x 10¹² £.

Walter Taylor.  The Indian Juvenile Arithmetic ....  Op. cit. in 5.B.  1849.  P. 200.  King conferring reward on a general.  Computes number of seers, which contain 15,360 grains, and the value if 30 seera are worth one rupee.

Nuts to Crack XIV (1845), no. 73.  The sovereign and the sage.  Almost identical to Boy's Own Book.

Magician's Own Book.  1857.  The sovereign and the sage, pp. 242-243.  A simplified version of Ozanam-Montucla.  = Book of 500 Puzzles, 1859, pp. 56-57.  = Boy's Own Conjuring Book, 1860, p. 213.

[Chambers].  Arithmetic.  Op. cit. in 7.H.  1866?  P. 267, quest. 54.  Story of Sessa with grains of wheat.  Suppose  7680  grains make a pint and a quarter is worth  £1 7s 6d,  how much was the wheat worth?

James Cornwell & Joshua G. Fitch.  The Science of Arithmetic: ....  11th ed., Simpkin, Marshall, & Co., London, et al., 1867.  (The 1888 ed. is almost identical to this, so I suspect they are close to identical to the 2nd ed. of 1856.)  Exercises CXLIII, no. 6, pp. 299 & 371.  Chessboard problem with no story, assumes  7680  grains to a pint.

Cassell's.  1881.  P. 101: Sovereign and the sage.  Uses sage's and king's common age of 64, with no reference to chessboard.

Lucas.  L'Arithmétique Amusante.  1895.  Le grains du blé de Sessa, pp. 150-151.  Says it would take 8 times the surface of the earth to grow enough grain.

Berkeley & Rowland.  Card Tricks and Puzzles.  1892.  The chess inventor's reward, pp. 112‑114.  Assumes  7489 29/35  grains to the pound, with  112  pounds to the cwt,  20  cwt to the ton and  1024  tons to the cargo, getting  1,073,741,824  cargoes, less one grain.  The number of grains is chosen so that a ton contains exactly  2²⁴  grains of rice and the answer is  2³⁰  cargoes less one grain.

 

          7.L.2.b.        HORSESHOE NAILS PROBLEM

 

          See Tropfke 632.

 

AR.  c1450.  Prob. 274, 317, 318, 353.  Pp. 125, 140, 154, 180, 227.

274:  32  horseshoe nails.

317:  16  cow nails.

318:  32  horseshoe nails.

353:  28  nails _  text is obscure.

Riese.  Rechenung nach der lenge ....  1525.  (Loc. cit. under Riese, Die Coss.)  Prob. 32, p. 20.  32  horseshoe nails.

Christoff Rudolff.  Künstliche rechnung mit der ziffern und mit den zal pfenninge.  Vienna, 1526;  Nürnberg, 1532, 1534, et seq.  F. N.viii.v.  ??NYS.  32  horseshoe nails.  (H&S 56 gives German.)

Apianus.  Kauffmanss Rechnung.  1527.  Ff. D.vi.r - D.vi.v.  32 horseshoe nails.

Anon.  Trattato d'Aritmetica, e del Misure.  MS, c1535, in Plimpton Collection, Columbia Univ.  ??NYS.  Horseshoe problem:  1 + 2 + 4 + ... + 2²³.  (Rara, 482‑484, with reproduction on p. 484.)

Robert Recorde.  Ground of Artes.  1542.  Op. cit. in 7.G.2.  Ff. L.ii.r - L.ii.v.  24  horseshoe nails.

Buteo.  Logistica.  1559.  Prob. 34, pp. 237-238.  24  horseshoe nails.  (H&S 56.)

van Etten.  1624.  Prob. 87, pp. 111‑118 (not in English editions).  Includes  24  horseshoe nails problem as part VII on p. 115.  Henrion's Notte, p. 38, observes that there are many arithmetical errors in prob. 87 which the reader can easily correct.

Ozanam.  1725.  Prob. 11, question 4, 1725: 77‑78 & 80.  Part of prob. 3, 1778: 79-80;  1803: 81;  1814: 72;  1840: 38.  24  nails _ first asks for the price of the  24th, then the total.

Dilworth.  Schoolmaster's Assistant.  1743.  P. 96, no. 1.  32  nails, starting with a farthing.

Walkingame.  Tutor's Assistant.  1751.  Geometrical Progression, prob. 6, 1777: p. 95;  1835: p. 103;  1860: p. 123.  32  nails,  one farthing for the first,  wants total, which he gives in £ s/d.

Euler.  Algebra.  1770.  I.III.XI.511, p. 166.  Horse to be sold for the value of  32  nails,  1  penny for the first, ....

Vyse.  Tutor's Guide.  1771?  Prob. 2, p. 148 & Key p. 190.  36  horseshoe nails.  Want value of last one, starting  ¼, ½, 1, ....

Bullen.  Op. cit. in 7.G.1.  1789.  Chap. 31, prob. 1, p. 215.  Same as Walkingame.

Bonnycastle.  Algebra.  10th ed., 1815.  P. 80, no. 6.  Same as Walkingame.

Manuel des Sorciers.  1825.  P. 84.  ??NX  24 horseshoe nails.

The Boy's Own Book.  The horsedealer's bargain.  1828: 182;  1828-2: 238;  1829 (US): 106;  1855: 393‑394;  1868: 431-432.  Wants value of  24th nail, starting with a farthing.  = Boy's Treasury; 1844; Amusements in arithmetic: The horse-dealer's bargain, p. 304.

Nuts to Crack XIV (1845), no. 74.  The horsedealer's bargain.  Almost identical to Boy's Own Book.

Walter Taylor.  The Indian Juvenile Arithmetic ....  Op. cit. in 5.B.  1849.  P. 199.  32  nails.

[Chambers].  Arithmetic.  Op. cit. in 7.H.  1866?  P. 268, quest. 64.  24  nails, starting with a farthing.  Finds total.

Cassell's.  1881.  P. 101: The horse‑dealer's bargain.  24  nails, unclear, but uses  2²³  farthings as the answer.

 

          7.L.2.c.         USE OF 1, 2, 4, ... AS WEIGHTS, ETC.

 

          See Tropfke 633.

          A special case of this is the use of such amounts to make regular unit payments, e.g. rent of one per day.  See:  Knobloch;  Fibonacci;  BR;  Widman;  Tartaglia;  Gori;  Les Amusemens.

 

Eberhard Knobloch.  Zur Überlieferungsgeschichte des Bachetschen Gewichtsproblems.  Sudhoffs Archiv 57 (1973) 142-151.  This describes the history of this topic and 7.L.3 from Fibonacci to Ozanam (1694).  He gives a table showing occurences of:  powers of two,  powers of three,  weight problem,  payment problem.  I am not entirely clear what he means in the first three cases _ I would have two kinds of weight problem corresponding to the first two cases and perhaps some of his references in the first case are listed under 7.L.2.  However, the last case clearly corresponds to the problem of making a payment of one unit per day as in Fibonacci.  He lists this as occurring in Fibonacci,  BR,  Widmann  and  Tartaglia  and notes that Sanford, H&S 91, only noticed Fibonacci.  Knobloch notes that Ball's citations are not very good and that Ahrens' note about them does not go much deeper.  I have a number of references listed below which were not available to Knobloch.

 

Fibonacci.  1202.  P. 298.  Uses  5  ciphi of value  1, 2, 4, 8, 15  to pay a man at rate of  1  per day for  30  days.

BR.  c1305.  No. 93, pp. 112‑113.  Use of  1, 2, 4  as payments at rate of one per year for  7  years.

Widman.  Op. cit. in 7.G.1.  1489.  Ff. 138v-139r.  ??NYS _ Knobloch says he uses values of  1, 2, 4, 8, 16  to pay for  31  days.

Tartaglia.  General Trattato, 1556, part 2, book 1, chap. 16, art. 32: Di una particolar proprieta della progression doppia geometrica, p. 17v.  Weights:  1, 2, 4, 8, ...  (See MUS I 89.).  Also does payments with  1, 2, 4, 8, 16, 29.  Knobloch also refers to art. 33-35 _ ??NYS _ and notes that the folios are misnumbered, but miscites 'doppia' as 'treppia' here.  This covers the powers of  3  also.

Buteo.  Logistica.  1559.  Prob. 91, pp. 309-312.  Use of  1, 2, 4, 8, 16, ...  as weights.  (Cited by Knobloch.)

Knobloch also cites  Ian Trenchant (1566),  Daniel Schwenter (1636),  Franz van Schooten (1657).

Gori.  Libro di arimetricha.  1571.  Ff. 71r‑71v (p. 76).  Use of cups weighing  1, 2, 4  to make all weights through  7,  to pay for days at one per day.

Bachet.  Problemes.  1612.  Addl. prob. V & V(bis), 1612: 143-146;  as one prob. V, 1624: 215-219;  1884: 154-156.  Mentions  1, 2, 4, 8, 16  and cites Tartaglia, art. 32 only.  This was omitted in the 1874 ed.  Knobloch cites 1612, pp. 127 & 143-146, but but p. 127 is Addl. prob. I, which is a Chinese Remainder problem?

van Etten/Henrion.  1630.  Notte to prob. 53, pp. 20‑21.  Refers to Bachet and compares with ternary weights.

Ozanam.  1694. 

Prob. 8, 1696: 33-35;  1708: 29-32.  Prob. 11, 1725: 68-75.  Section II, 1778: 68-74;  1803: 70-76;  1814: ??NYS;  1840: 34-36.  A discussion of geometric progression and a mention of  1, 2, 4, ...,  without any application to weighing.  1778 et seq. also mentions 1, 3, 9, ....

Prob. 12, vol. II, 1694: 18-19 (??NYS).  Prob. 12, 1696: 284 & fig. 131, plate 46, p. 275;  1708: 360 & fig. 26, plate 14, opp. p. 351.  Prob. 8, vol. II, 1725: 345‑348 & fig. 131, plate 46 (42).  Prob. 14, vol. I, 1778: 206‑207;  1803: 201-202.  Prob. 13, vol. II, 1814: 174-175;  1840: 90‑91.  Gives double and triple progressions.  Knobloch gives the 1694 citation.  The figure is just a picture of a balance and is not informative _ the same figure is also cited for other sets of weights.

Les Amusemens.  1749.  Prob. 8, p. 128.  Coins of value  1, 2, 4, 8, 15  to pay for a room at a rate of  1  per day for  30  days.

 

          7.L.3. 1 + 3 + 9 + ... AND OTHER SYSTEMS OF WEIGHTS

 

          See MUS I 88-98;  Tropfke 633.

 

Tabari.  Mift_h al-mu‘_mal_t.  c1075.  P. 125ff., no. 43.  ??NYS _ Hermelink, op. cit. in 3.A, and Tropfke 634-635 say this gives  1, 3, 9, ..., 19683  =  3⁹  to weigh up to 10,000.

Fibonacci.  1202. 

P. 297.  Weights  1, 3, 9, 27  and  1, 3, 9, 27, 81  'et sic eodem ordine possunt addi pesones in infinitum'.

Pp. 310‑311.  Finds  1 + 2 + 6 + 18 + ... + 2*3⁶²  =  3⁶³  by repeated squaring to get  3⁶⁴  and then divides by  3.

Gherardi.  Libro di ragioni.  1327.  P. 53.  Weights  1, 3, 9, 27, 80  to weigh up through  120.

Columbia Algorism.   c1370.  Prob. 71, pp. 92‑93.  Weights  1, 3, 9, 27.

AR.  c1450.  Prob. 127, pp. 67 & 182.  1, 3, 9, 27.

Chuquet.  1484.  Prob. 142. 

1, 2, 7  to weigh up to  10  (English in FHM 225);

1, 2, 4, 15  up to  22;

1, 3, 9  up to  13;

1, 3, 9, 27  up to  40;  (original of this and the next case reproduced on FHM 226)

1, 3, 9, 27, 81  up to  121.  Knobloch says Chuquet gives a general solution, but I don't see that Chuquet is general.

Pacioli.  Summa.  1494.  Ff. 97r-97v.  General discussion of  1, 3, 9, 27, 81, 243, ....

Cardan.  Practica Arithmetice.  1539.  Chap. 65, section 12, ff. BB.vii.r - BB.vii.v (p. 136).  Weights  1, 3, 9, 27, ....

Knobloch also cites:  Giel vanden Hoecke (1537);  Gemma Frisius (1540);  Michael Stifel (1553);  Simon Jacob (1565);  Ian Trenchant (1566);  Daniel Schwenter (1636);  Kaspar Ens (1628);  Claude Mydorge (1639);  Frans van Schooten (1657).

Tartaglia, 1556 _ see in 7.L.2.c.

Buteo.  Logistica.  1559.  Prob. 91, pp. 309-312.  Use of  1, 3, 9, 27, ...  as weights.  (Cited by Knobloch.)

John [Johann (or Hanss) Jacob] Wecker.  Eighteen Books of the Secrets of Art & Nature Being the Summe and Substance of Naturall Philosophy, Methodically Digested ....  (As:  De Secretis Libri XVII; P. Perna, Basel, 1582 _ ??NYS)  Now much Augmented and Inlarged by Dr. R. Read.  Simon Miller, London, 1660, 1661 [Toole Stott 1195, 1196];  reproduced by Robert Stockwell, London, nd [c1988].  Book XVI _ Of the Secrets of Sciences: chap. 19 _ Of Geometricall Secrets: To poyse all things by four Weights, p. 289.  1, 3, 9, 27;  1, 3, 9, 27, 81;  1, 3, 9, 27, 81, 243.  Cites Gemma Frisius.

Bachet.  Problemes.  1612.  Addl. prob. V & V(bis),  1612: 143-146;  as one prob. V, 1624: 215-219;  1884: 154‑156.  Weights:  1, 3, 9, 27, ...,  and the general case via the sum of a GP.  In the 1612 ed., Bachet only does the cases  40  and  121,  then does the general case.  Knobloch cites 1612, pp. 127 & 143-146, but p. 127 is Addl. prob. I, which is a Chinese Remainder problem.  He also says this is the first proof of the problem, excepting Chuquet, though I don't see such in Chuquet.

van Etten.  1624.  Prob. 53 (48), pp. 48‑49 (72).  1, 3, 9, 27;  1, 3, 9, 27, 81;  1, 3, 9, 27, 81, 243.  Henrion's Notte, pp. 20‑21, refers to Bachet and compares this with binary weights.

Ozanam.  1694. 

Prob. 8, 1696: 33-35;  1708: 29-32.  Prob. 11, 1725: 68-75.  Section II, 1778: 68-74;  1803: 70-76;  1814: ??NYS;  1840: 34-36.  A discussion of geometric progression and a mention of  1, 2, 4, ...,  without any application to weighing.  1778 et seq. also mentions 1, 3, 9, ....

Prob. 12, vol. II, 1694: 18-19 (??NYS).  Prob. 12, 1696: 284 & fig. 131, plate 46, p. 275;  1708: 360 & fig. 26, plate 14, opp. p. 351.  Prob. 8, vol. II, 1725: 345‑348 & fig. 131, plate 46 (42).  Prob. 14, vol. I, 1778: 206‑207;  1803: 201-202.  Prob. 13, vol. II, 1814: 174-175;  1840: 90‑91.  Gives double and triple progressions.  Knobloch gives the 1694 citation.  The figure is just a picture of a balance and is not informative _ the same figure is also cited for other sets of weights.

Les Amusemens.  1749.  Prob. 18, p. 140: Les Poids.  Weights  1, 3, 9, 27, 81, 243.

Vyse.  Tutor's Guide.  1771?  Prob. 2, p. 316 & Key pp. 356-357.  Weights  1, 3, 9, 27.

Bonnycastle.  Algebra.  1782.  P. 202, no. 13.  1, 3, 9, 27, 81, 243, 729, 2187  to weigh to  29 hundred weight _ an English hundred weight is  112  pounds.  c= 1815: p. 230, no. 33.  1, 3, 9, 27, 81  to weigh to a hundred weight.

Eadon.  Repository.  1794.  Pp. 297-298, no. 1.  1, 3, 9, ..., 3¹³.

Badcock.  Philosophical Recreations, or, Winter Amusements.  [1820].  P. 50, no. 77: To find the least Number of Weights that will weigh from One Pound to Forty.  1, 3, 9, 27.

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions.  No. 24, pp. 20 & 78‑79.  1, 2, 4, 8, 16, ...  and  1, 3, 9, 27, 81, ....

Endless Amusement II.  1826? 

P. 105: To find the least Number of Weights that will weigh from One Pound to Forty.  = Badcock.

Prob. 24, p. 201.  To name five weights, which, added together, make 121 pounds; by means of which may be weighed any intermediate weight, excluding fractions.  1, 3, 9, 27, 81.

Young Man's Book.  1839.  P. 242.  To name five weights, ....  Identical to Endless Amusement II, p. 201.

Boy's Treasury.  1844.  Amusements in arithmetic: To find the least number of weights which will weigh any intermediate weight, from one pound to forty, exclusive of fractions, p. 304.  Indicates that one can continue the progression.

Family Friend 1 (1849) 150 & 178.  Problems, arithmetical puzzles, &c. _ 5.  How can one divide  40  lb into four weights to weigh every value from  1  to  40?  Proposer says he can't do it.  Usual answer, but one solver gives weights  6, 10, 11, 13.  However, the latter weights will not weigh  22, 25, 26, 31, 32, 33, 35, 36, 37, 38, 39.

Boy's Own Book.  To weigh from one to forty pounds with four weights.  1855: 392;  1868: 430.  1, 3, 9, 27.  No generalizations.

Magician's Own Book.  1857.  The mathematical blacksmith, p. 230.  1, 3, 9, 27  and this can be continued.  = Boy's Own Conjuring Book, 1860, p. 200.

F. J. P. Riecke.  Op. cit. in 4.A.1, vol. 3, 1873.  Art. 3: Die Zauberkarten, p. 13.  Uses balanced ternary for divination.  See under 7.M.4

P. A. MacMahon.  Certain special partitions of numbers.  Quart. J. Math. 21 (1886) 367‑373.  Very technical.

P. A. MacMahon.  Weighing by a series of weights.  Nature 43 (No.1101) (4 Dec 1890) 113‑114.  Less technical description of the above work.

Lucas.  L'Arithmétique Amusante.  1895.  Pp. 166-168.  Notes that pharmacists, etc. use weights:  1, 1, 2, 5, 10, 10, 20, 50, 100, 100, 200, 500, 1000, 1000, 2000, 5000, ....  Discusses ternary.

Wehman.  New Book of 200 Puzzles.  1908.  P. 49.  1, 3, 9, 27, 81.

Ahrens.  MUS I.  1910.  Pp. 88-98 discusses this and some generalizations like MacMahon's.

 

          7.M.   BINARY SYSTEM AND BINARY RECREATIONS

 

          The binary system has several origins.

a)       Egyptian  &  Russian peasant multiplication.

b)       Weighing _ see 7.L.2.c.

c)       Binary divination _ see 7.M.4.

d)       The works below.

 

          See also:  5.E.2 for Memory Wheels;  5.F.4 for circuits on the  n‑cube;  5.AA for an application to card-shuffling;  7.AA.1 for Negabinary.

 

Anton Glaser.  History of Binary and Other Nondecimal Numeration.  Published by the author, 1971;  (2nd ed., Tomash, Los Angeles, 1981).  General survey, but has numerous omissions _ see the review by Knuth at Harriot, below, and MR 84f:01126.  He has no references to early Chinese material.

 

Shao Yung.  c1060.  Sung Yuan Hsüeh An, chap. 10.  Fu‑Hsi diagram of the 64 hexagrams of the I‑Ching, in binary order.  A version appears in Leibniz‑Briefe 105 (Bouvet) Bl. 27r/28r in the Niedersächsische Landesbibliothek, Hannover.  Needham, vol. 2, p. 341, notes that this had only been published in Japanese and Chinese by 1956.  See Zacher & Kinzô below for reproductions.  Also reproduced in:  E. J. Aiton; Essay Review [of Zacher, below]; Annals of Science 31 (1974) 575‑578.

Chu Hsi.  Chou I Pen I Thu Shuo.  12C.  Fu‑Hsi Liu‑shih‑ssu Kua Tzhu Hsü (Segregation Table of the symbols of the Book of Changes) _ reproduced in Hu Wei's I Thu Ming Pien.  An illustration is given in Needham, vol. 2, fig. 41 = plate XVI, opp. p. 276 _ he says it is based on the original chart of Shao Yung and that Tshai Chhen (c1210) gave a simplified version.  Also in Kinzô and in Aiton & Shimao, below.  Shows the alternation of 0 and 1 in each binary place. 

Thomas Harriot.  Unpublished MS.  c1604.  Described by  J. W. Shirley; Binary numeration before Leibniz; Amer. J. Physics 19 (1951) 452‑454;  and by  D. E. Knuth; Review of 'History of Binary and Other Nondecimal Numeration'; HM 10 (1983)) 236‑243.  This shows some binary calculation.  Shirley reproduces BM: Add MSS 6786, ff. 346v‑347r.  Knuth cites 6782, 1r, 247r;  6786, 243v, 305r, 346v, 347r, 516v;  6788, 244v.

Francis Bacon.  Of the Advancement of Learning.  1605.  ??NYS.  Describes his binary 5‑bit coding.

Francis Bacon.  De augmentis scientarum.  1623.  ??NYS.  Full description of his coding.  He does not have any arithmetic content, so he is not really part of the development of binary.

John Napier.  Rabdologiae.  Edinburgh, 1617.  ??NYS.  Describes binary as far as extracting square roots.  William F. Hawkins; The Mathematical Work of John Napier (1550‑1617); Ph.D. thesis, Univ. of Auckland, 1982, ??NYS, asserts this is THE invention of the binary system.

G. W. Leibniz.  De Progressione Dyadica.  3pp. Latin MS of Mar 1679.  Facsimile and translation into German included in:  Herrn von Leibniz' Rechnung mit Null und Eins; Siemens Aktiengesellschaft, (1966), 2nd corrected printing, 1969; facsimile between pp. 20 & 21 and translation on pp. 42-47, with an essay by Hermann J. Greve: Entdeckung der binären Welt on pp. 21-31.  His first, unpublished, MS on the binary system, showing all the arithmetic processes.

H. J. Zacher.  Die Hauptschriften zur Dyadik von G. W. Leibniz.  Klosterman, Frankfurt, 1973.  Gathers almost all the Leibniz material, notably omitting the above 1679 paper.  He does reproduce the Fu‑Hsi diagram sent by Bouvet (cf. Shao Yung above).  However Leibniz's letter of 2 Jan 1697 to Herzog Rudolf Augustus, in which he gives his drawing of his plan for a medallion commemorating the binary system, is now lost, but it was published in 1734.

G. W. Leibniz.  Two Latin letters on the binary system, 29 Mar 1698  &  17 May 1698, recipient not identified, apparently the author of a book in 1694 which occasioned Leibniz's correspondence with him.  Opera Omnia, vol. 3, 1768, pp. 183-190.  Facsimile and translation into German included in:  Herrn von Leibniz' Rechnung mit Null und Eins; Siemens Aktiengesellschaft, 2nd corrected printing, 1966; facsimile between pp. 40 & 41 and translation on pp. 53-60.

G. W. Leibniz.  Explication de l'arithmètique binaire.  Histoire de l'Academie Royale des Sciences 1703 (1705) 85-89.  Facsimile and translation into German included in:  Herrn von Leibniz' Rechnung mit Null und Eins; Siemens Aktiengesellschaft, 2nd corrected printing, 1966; facsimile between pp. 32 & 33 and translation on pp. 48-52.  Illustrates all the arithmetic operations and discusses the Chinese trigrams of 'Fohy' and his correspondence with Father Bouvet in China.

G. W. Leibniz.  Letter of 1716 to Bouvet.  ??NYS _ cited in Needham, vol. 2, p. 342.  Fourth section is: Des Caractères dont Fohi, Fondateur de l'Empire Chinois, s'est servi dans ses Ecrits, et de l'Arithmétique Binaire.

Gorai Kinzô.  Jukyô no Doitsu seiji shisô ni oyoboseru eikyô (Influence of Confucianism on German Political Thought [in Japanese]).  Waseda Univ. Press, Tokyo, 1929.  ??NYS.  First publication of Leibniz's correspondence with Bouvet which led to the identification of the Fu Hsi diagram with the binary numbers.  Gives a redrawn Fu‑Hsi diagram and a segregation table.

E. J. Aiton & Eikon Shimao.  Gorai Kinzô's study of Leibniz and the I Ching hexagrams.  Annals of Science 38 (1981) 71‑92.  Describes the above work.  Reproduces Kinzo's Fu‑Hsi diagram and segregation table.

 

Ahrens.  MUS I.  1910.  24-104 discusses numeration systems in general and numerous properties of binary and powers of 2.

Gardner.  SA (Aug 1972)  c= Knotted, chap. 2.  General survey of binary recreations.  The material in the book is much expanded from the SA column.

 

          7.M.1.          CHINESE RINGS

 

          See MUS I 61-72;  S&B 104‑107, 111 & 135.

          See also 4.A.4, 11.K.1.

 

S. N. Afriat.  The Ring of Linked Rings.  Duckworth, London, 1982.  This is devoted to the Chinese Rings and the Tower of Hanoi and gives much of the history.

 

Sun Tzu.  The Art of War.  c-4C.  With commentary by Tao Hanzhang.  Translated by Yuan Shibing.  (Sterling, 1990);  Wordsworth, London, 1993.  In chap. 5: Posture of Army, p. 109, the translator gives:  "It is like moving in a endless circle"  In the commentary, p. 84, it says:  "their interaction as endless as that of interlocked rings."  Though unlikely to refer to the puzzle,, this and the following indicate that interlocked rings was a common image of the time.

Needham, vol. II, pp. 189-197, describes the paradoxes of Hui Shih (‑4C).  P. 191 gives HS/8:  Linked rings can be sundered.  On p. 193, Needham gives several explanations of this statement and a reference to the Chinese Rings in vol. III, but he does not claim this statement refers to the puzzle.

Stewart Culin.  Korean Games.  Op. cit. in 4.B.5.  Section XX: Ryou‑Kaik‑Tjyo _ Delay Guest Instrument (Ring Puzzle), pp. 31‑32.  Story of Hung Ming (181‑234) inventing it.  States the Chinese name is  Lau Kák Ch'a  (Delay Guest Instrument) or  Kau Tsz' Lin Wain  (Nine Connected Rings).  Says there a great variety of ring puzzles in Japan, known as  Chie No Wa  (Rings of Ingenuity) and illustrates one, though it appears to be just 10 rings joined in a chain _ possibly a puzzle ring??  He says he has not found out whether the Chinese rings are known in Japan _ but see Gardner below.

Ch'ung‑En Yü.  Ingenious Ring Puzzle Book.  In Chinese:  Shanghai Culture Publishing Co., Shanghai, 1958.  English translation by Yenna Wu, published by Puzzles _ Jerry Slocum, Beverly Hills, Calif., 1981.  P. 6.  States it was well known in the Sung (960‑1279).

The Stratagem of Interlocking Rings.  A Chinese musical drama, first performed c1300.  Cited in:  Marguerite Fawdry; Chinese Childhood; Pollock's Toy Theatres, London, 1977, pp. 70-72.  Otherwise, Fawdrey repeats information from Culin and the story that it was used as a lock.

Needham.  P. 111 describes the puzzle as known in China at the beginning of the 20C, but says the origins are quite obscure and gives no early Chinese sources.  He also cites his vol. II, p. 191, for an early possible reference _ see above.

Gardner.  Knotted, chap. 2, says there are 17C Japanese haiku about it and it is used in Japanese heraldic emblems.

Cardan.  De subtilitate.  1550.  Liber XV.  Instrumentum ludicrum, pp. 294‑295.  = Basel, 1553, pp. 408‑409.  = French ed., 1556, et les raisons d'icelles; Book XV, para. 2, p. 291, ??NYS.  = Opera Omnia, vol. 3, p. 587.  Very cryptic description, with one diagram of a ring.

In England, the Chinese Rings were known as Tarriours or Tiring or Tyring or Tarrying Irons.  The OED entry at Tiring-irons gives 5 quotations from the 17C:  1601, 1627, 1661, 1675, 1690.

J. Wallis.  De Algebra Tractatus.  1685, ??NYS.  = Opera Math., Oxford, 1693, vol. II, chap. CXI, De Complicatus Annulis, 472‑478.  Detailed description with many diagrams.

Ozanam.  1725: vol. 4.  No text, but the puzzle with  7  rings is shown as an unnumbered figure on plate 14 (16).  Ball, MRE, 1st ed., 1892, p. 80, says the 1723 ed., vol. 4, p. 439 alludes to it.  The text there is actually dealing with Solomon's Seal (see 11.D) which is the adjacent figure on plate 14 (16).

Minguét.  Engaños.  1733.  Pp. 55-57 (1755: 27-28; 1822: 72-74): Juego del ñudo Gordiano, ò lazo de las sortijas enredadas.  7  ring version clearly drawn.

Alberti.  1747.  No text, but the puzzle is shown as an unnumbered figure on plate XIII, opp. p. 214 (111), copied from Ozanam, 1725.

Catel.  Kunst-Cabinet.  1790.  Der Nürnberger Tand, p. 15 & fig. 41 on plate II.  Figure shows  7  rings, text says you can have  7,  9,  11  or  13. 

Bestelmeier.  1801.  Item 298:  Der Nürnberger Tand.  Diagram shows  6  rings, but text refers to  13  rings.  Text is partly copied from Catel.

Endless Amusement II.  1826?  Prob. 29, pp. 204-207.  Cites Cardan as being very obscure.  Shows example with  5  rings and seems to imply it takes 63 moves.

The Boy's Own Book.  The puzzling rings.  1828: 419‑422;  1828‑2: 424‑427;  1829 (US): 216-218;  1855: 571‑573;  1868: 673-675.  Shows  10  ring version and says it takes 681 moves.  Cites Cardan.

Crambrook.  1843.  P. 5, no. 9: Puzzling Rings, or Tiring Irons.

Magician's Own Book.  1857.  Prob. 45: The puzzling rings, pp. 279-283.  Identical to Boy's Own Book, except 1st is spelled out first, etc.  = Book of 500 Puzzles, 1859, pp. 93-97.  = Boy's Own Conjuring Book, 1860, prob. 44, pp. 243‑246.

Magician's Own Book (UK version).  1871.  The tiring-irons, baguenaudier, or Cardan's rings, pp. 233-235.  Quite similar to Boy's Own Book, but somewhat simplified and gives a tabular solution.

L. A. Gros.  Théorie du Baguenodier.  Aimé Vingtrinier, Lyon, 1872.  (Copy in Radcliffe Science Library, Oxford _ cannot be located by them.)  ??NYS

Lucas.  Récréations scientifiques sur l'arithmétique et sur la géométrie de situation.  Troisième récréation, sur le jeu du Baguenaudier, ...   Revue Scientifique de la France et de l'étranger (2) 26 (1880) 36‑42.  c= La Jeu du Baguenaudier, RM1, 1882, pp. 164‑186 (and 146‑149).  c= Lucas; L'Arithmétique Amusante; 1895; pp. 170-179.  Exposition of history back to Cardan, Gros's work, use as a lock in Norway.  He says that Dr. O.-J. Broch, former Minister and President of the Royal Norwegian Commission at the Universal Exposition of 1878, recently told him that country people still used the rings to close their chest and sacks.  RM1 adds a letter from Gros.

The French term 'baguenaudier' has long mystified me.  A 'bague' is a ring.  My large Harrap's French‑English dictionary defines 'baguenaudier' as "trifler, loafer, retailer of idle talk;  ring‑puzzle, tiring irons;  bladder‑senna", but none of the related words indicates how 'baguenaudier' came to denote the puzzle.  However, Farmer & Henley's Dictionary of Slang gives 'baguenaude' as a French synonym for 'poke', so perhaps 'baguenaudier' means a 'poker' which has enough connection to the object to account for the name??  MUS I 62-63 discusses Gros's use of 'baguenodier' as unreasonable and quotes two French dictionaries of 1863 and 1884 for 'baguenaudier' which he identifies as an ornamental garden shrub, Colutea arborescens L.

Cassell's.  1881.  Pp. 91-92: The puzzling rings.  = Manson, 1911, pp. 144-145: Puzzling rings.  Shows  7  ring version and discusses  10  ring version, saying it takes 681 moves.  Discusses the Balls and Rings puzzle.

Peck & Snyder.  1886.  P. 299: The Chinese puzzling rings.  9  rings.  Mentions Cardan & Wallis.  Shown in Slocum's Compendium.

Ball.  MRE, 1st ed., 1892, pp. 80-85.  Cites Cardan, Wallis, Ozanam and Gros (via Lucas).  P. 85 says:  "It is said _ though a priori the fact would have seemed very improbable _ that Chinese rings are used in Norway to fasten the lids of boxes, ....  I have never seen them employed for such purposes in any part of the country in which I have travelled."  This whole comment is dropped in the 3rd ed.

Hoffmann.  1893.  Chap. X, no. 5: Cardan's rings, pp. 334‑335 & 364‑367.  Cites Encyclopédie Méthodique des Jeux, p. 424+.

H. F. Hobden.  Wire puzzles and how to make them.  The Boy's Own Paper 19 (No. 945) (13 Feb 1896) 332-333.  Magic rings (= Chinese rings) with  10  rings, requiring 681 moves.  (I think it should be 682.)

Dudeney.  Great puzzle crazes.  Op. cit. in 2.  1904.  "... it is said to be used by the Norwegians as a form of lock for boxes and bags ..."

Ahrens.  Mathematische Spiele.  Encyklopadie article, op. cit. in 3.B.  1904.  Note 60, p. 1091, reports that a Norwegian professor of Ethnography says the story of its use as a lock in Norway is erroneous.  He repeats this in MUS I 63.

Adams.  Indoor Games.  1912.  Pp. 337‑341 includes The magic rings.

Bartl.  c1920.  P. 309, no. 80: Armesünderspiel oder Zankeisen.  Seven ring version for sale.

Collins.  Book of Puzzles.  1927.  The great seven-ring puzzle, pp. 49-52.  Cites Cardan and Wallis.  Says it is known as Chinese rings, puzzling rings, Cardan's rings, tiring irons, etc.  Says  3 rings takes 5 moves,  5 rings takes 21  and  7 rings takes 85.

Rohrbough.  Puzzle Craft.  1932.  The Devil's Needle, p. 7 (= p. 9 of 1940s?).  Cites Boy's Own Book of 1863.

R. S. Scorer, P. M. Grundy & C. A. B. Smith.  Some binary games.  MG 28 (No. 280) (Jul 1944) 96‑103.  Studies the binary representations of the Chinese Rings and the Tower of Hanoi.  Gives a triangular coordinate system representation for the Tower of Hanoi.  Studies Tower of Hanoi when pegs are in a line and you cannot move between end pegs.  Defines an  n‑th order Chinese Rings and gives its solution.

E. H. Lockwood.  An old puzzle.  With Editorial Note by H. M. Cundy.  MG 53 (No. 386) (Dec 1969) 362‑364.  Derives number of moves by use of a second order non‑homogeneous recurrence.  Cundy mentions the connection with the Gray code and indicates how the Gray value at step  k,  G(k),  is derived from the binary representation of  k,  B(k).  [But he doesn't give the simplest expression:  G(k)  =  B(k)  EOR  B(ëk/2û).  I noted this a few years ago and am surprised that it does not appear to be old.  Gardner's 1972 article describes it but not so simply.]  This easily gives the number of steps.

Marvin H. Allison Jr.  The Brain.  This is a version of the Chinese Rings made by Mag-Nif since the 1970s.  [Gardner, Knotted.]

William Keister.  US Patent 3,637,215 _ Locking Disc Puzzle.  Filed 22 Dec 1970;  patented 25 Jan 1972.  Abstract + 3pp + 1p diagrams.  This is a version of the Chinese Rings, with discs on a sliding rod producing the interaction of one ring with the next.  Described on the package.  Keister worked on puzzles of this sort since the 1930s.  It was first produced by Binary Arts in 1986.

Ed Barbeau.  After Math.  Wall & Emerson, Toronto, 1995.  Protocol, pp. 163-166.  Gives seating and standing problems which lead to the same sequence of moves as for the Chinese rings, but one is in reverse order.

 

          7.M.2.          TOWER OF HANOI

 

          See MUS I 52-61, S&B 135.

          See also 5.F.4 for connection with Hamiltonian circuits on the n‑cube and 5.A.4 for the Panex Puzzle.

          All the following have three pegs unless specified otherwise.

 

The Conservatoire National des Arts et Métiers _ Musée National des Techniques, 292 Rue St. Martin, Paris, has two examples _ No. 11271 & 11272 _ presented by Edouard Lucas, professeur de mathématiques au Lycée Saint-Louis à Paris, in 1888.  The second is a 'grand modèle pour les cours publics' 1.05 m high!  Elisabeth Lefevre has kindly sent details and photocopies of the box, instruction sheet (one sheet printed on both sides) and an article.  Both versions have  8  discs.

                    The box is  157 x 180 mm  and has an elaborate picture with the following text:

                              La Tour d'Hanoï / Veritable casse‑Téte Annamite / Jeu / rapporté du Tonkin / par le Professeur N. Claus (de Siam) / Mandarin / du College / Li‑Sou‑Stian / Brevete / S. G. D. G.

          This cover is shown in Claus [Lucas] (1884) and Héraud (1903).  I will refer to this as the original cover.  (The Museum does not know of any patent _ they have looked in 1880-1890.  S. G. D. G.  stands for  Sans garantie du gouvernement.)  The bottom of the box has an ink inscription:  "Hommage del'auteur   Ed Lucas   Paris  1888"  ?? _ the date is not legible on the photocopy.  Inside the cover, apparently in the same hand (that is, in Lucas's writing), is an ink inscription:

                                        La tour d'Hanoï, --

                                        Jeu de combinaison pour

                              expliquer le systeme de la numération

                              binaire, inventé par M. Edouard Lucas,

                              (novembre 1883).  --  donné par l'auteur.

                    The Museum describes the puzzle as  15 cm long  by  14.5 cm wide by  10 cm high.  There is no photo available, but the examples are shown in catalogues of 1906 and 1943.

                    The instruction sheet reads as follows.  

                                                                La Tour d'Hanoï  

                                                       Véritable casse‑tête annamite  

                                                           Jeu rapporté du Tonkin  

                                                  par le Professeur N. Claus (de Siam)  

                                                   Mandarin du Collège Li‑Sou‑Stian! 

                                                                          ....

                                                        Paris, Pékin, Yédo et Saïgon

                                                                          ....

                                                                        1883

                    The sheet mentions the Temple of Bénarès where there are  64  discs.  A prize of a million (= a thousand thousand) francs is offered for a demonstration of the solution with  64  discs!  The second sheet of the instructions gives the rules and the number of moves for  2, 3, ..., 8  discs and the general rule.  It also refers to RM.

Edward Hordern has an example with the original instruction sheet, but in a simple box with just  'La Tour d'Hanoï'  in Chinese‑style lettering, not like the box described above.  Also, my recollection is that it is much smaller than the example above.  It has  8  discs.

G. de Longchamps.  Variétés.  Journal de Mathématiques Spéciales (2) 2 (1883) 286-287.  (The article is only signed G. L., but the author is further identified in the index on p. 290.  Copy provided by Hinz.)  Solves the recurrence relation  u_n = 2u_n-1 + 1,  u₀ = 0.  Says he was 'inspired by a letter which we have recently received from professor N. Claus.'  Describes the Tower of Hanoi briefly and says the above solution gives the number of moves when there are  n  discs.

Henri de Parville.  Column:  Revue des sciences.  Journal des Débats Politiques et Littéraires (27 Dec 1883) 1-2.  On p. 2, he reports receiving an example in the post with a cover like that of the original.  Gives the Benares story.  Wonders who the mandarin could be and notes the anagrams on  Lucas d'Amiens  and  Saint-Louis.

N. Claus (de Siam) [= Lucas (d'Amiens)].  La tour d'Hanoï.  Jeu de calcul.  Science et Nature 1:8 (19 Jan 1884) 127-128.  Says it takes  2ⁿ - 1  moves  "que M. de Longchamps l'a démontré (1)."  "(1) Journal de mathématiques spéciales.".  Observes that each of the discs always moves in the same cycle of pegs and hence gives the standard rule for doing the solution, which is attributed to the nephew of the inventor, M. Raoul Olive, student at the Lycée Charlemagne.  Asks for the minimum number of moves to restore an arbitrary distribution of discs to a Start position.  Says this is a complex problem in general, depending on binary and refers to RM 1 for this idea. 

                    (This paper is not in Harkin's bibliography (op. cit. in 1).  Hinz, 1989, cites it.)

Henri de Parville.  Récréations Mathématiques:  La Tour d'Hanoï et la question du Tonkin.   La Nature (Paris) 12 (No. 565, part 1) (29 Mar 1884) 285‑286.  Illustration by Poyet.  Asserts Lucas is the inventor. 

R. E. Allardice & A. Y. Fraser.  La Tour d'Hanoï.  Proc. Edin. Math. Soc. 2 (1883‑1884) 50‑53.  Includes de Parville from J. des Débats.  Then derives number of moves.

Anton Ohlert.  US Patent 303,946 _ Toy.  Applied 24 Jul 1884;  patented 19 Aug 1884.  1p + 1p diagrams.  8  discs.  Ohlert is a resident of Berlin.

Edward A. Filene, No. 4 Winter St, Boston Mass.  Eight Puzzle.  Copyrighted in 1887.  ??NYS _ described and illustrated in S&B, p. 135.  8  disc advertising version.

Tissandier.  Récréations Scientifiques.  5th ed., 1888, La tour d'Hanoï et la question du Tonkin, pp. 223-228.  Not in the 2nd ed. of 1881 nor the 3rd ed. of 1883.  Essentially de Parville's article with same illustration, but introduces it with coments saying it has had a great success and comes in a box labelled:  "la Tour d'Hanoï, véritable casse-tête annamite, rapporté du Tonkin par le professeur N. Claus (de Siam), mandarin du collège Li-Sou-Stian".  This would seem to be the original box.

                    = Popular Scientific Recreations; [c1890]; Supplement: The tower of Hanoï and the question of Tonquin, pp. 852‑856. 

Anon.   Jeux, Calculs et Divertissements.   Récréations Mathématiques.   La Tour d'Hanoï.  Liberté (9 Dec 1888) no page number on clipping.  Gives the story of N. Claus and says it was invented by Lucas and comes in a box decorated with annamite illustrations.  This would seem to be the original box.

Lucas.  Nouveaux jeux scientifiques ...,  op. cit. in 4.B.3.  1889.  Describes a series of games under the title of the next item, so the next may refer to the game or its instruction booklet.  A later version of the Tower of Hanoi is described on pp. 302‑303, having  5  pegs, and is shown on p. 301.  He says you can have  3, 4 or 5  pegs.

Lucas.  Jeux scientifiques pour servir à l'histoire, à l'enseignement et à la pratique du calcul et du dessin.  Première série: No. 3: La Tour d'Hanoï.  Brochure, Paris, 1889.  ??NYS.  (Not listed in BNC, but listed in Harkin, op. cit. in Section 1.  I wonder if these were booklets that accompanied the actual games??  Hinz says he has found some of these, but not the one on the Tower of Hanoi.  The booklet for La Pipopipette (= Dots and Boxes) is reproduced in his L'Arithmétique Amusante of 1895 _ see 4.B.3.)

Alfred Gartner & George Talcott.  UK Patent 20,672 _ Improvements in Games and Puzzles.  Applied 18 Dec 1890;  patented 21 Feb 1891.  2pp + 1p diagrams.  Shunting puzzle equivalent to Tower of Hanoi with 7 discs.

Ball.  MRE, 1st ed., 1892, pp. 78-79.  "... described by M. Tissandier as being common in France but which I have never seen on sale in England."  Gives the Benares story from De Parville in La Nature.

Lucas.  L'Arithmétique Amusante.  1895.  La tour d'Hanoï, pp. 179-183.  Description, including Benares story.  Says the nephew of the inventor, Raoul Olive, has noted that the smallest disc always moves in the same direction.  Says the whole idea of the mandarin and his story was invented a dozen years ago at  56 rue Monge,  which was built on the site where Pascal died.  [I visited this site recently _ it is a hotel and they knew nothing about Pascal.  Another source says Pascal died at  67 Rue Cardinal Lemoine,  which is several blocks away, but also is not an old building.]

Ball.  MRE, 3rd ed., 1896, pp. 99-101.  Omits the Tissandier reference and says:  "It was brought out in 1883 by M. Claus (Lucas)."

A. Héraud.  Jeux et Récréations Scientifiques _ Chimie, Histoire Naturelle, Mathématiques.  Not in the 1884 ed.  Baillière et Fils, Paris, 1903.  Pp. 300‑301 shows the original cover.

Burren Loughlin  &  L. L. Flood.  Bright-Wits  Prince of Mogador.  H. M. Caldwell Co., NY, 1909.  The five shields, pp. 20-24 & 59.  Five discs.

Tom Tit??.  In Knott, 1918, but I can't find it in Tom Tit.  No. 165: The tower of Hanoi, pp. 382‑383.  Describes it with cards  A ‑ 10  on piles labelled with  J, Q, K.

Robert Ripley.  Believe It Or Not!  Book 2.  (Simon & Schuster, 1931);  Pocket Books, NY, 1948, pp. 52‑53.  = Believe It or Not!  Two volumes in one; (Simon & Schuster, 1934); Garden City Books, 1946, pp. 222-223.  = Omnibus Believe It Or Not!; Stanley Paul, London, nd [c1935?], pp. 256‑257.  The Brahma Pyramid.  Outlines the Benares story, says he didn't locate the temple when he was in Benares, but it 'really exists', and that it will take  2⁶⁴  moves, but he then writes out  2⁶⁴ ‑ 1.  He says the Brahmins have been at it for  3000  years!!

Scorer, Grundy & Smith.  1944.  Op. cit. in 7.M.1.  They develop the graph of all positions of the Tower of Hanoi.

Donald W. Crowe.  The  n-dimensional cube and the tower of Hanoi.  AMM 63 (1956) 29-30.  Describes the connection with Hamiltonian circuits and binary ruler markings.

M. Gardner.  The Icosian Game and the Tower of Hanoi.  (SA (May 1957)) = 1st Book, chap. 6.  Describes Crowe's work.

A. J. McIntosh.  Binary and the Tower of Hanoi.  MTg 59 (1972) 15.  He sees the connection between binary and which disc is to be moved, but he wonders how to know which pegs are involved.  [This is a valid query _ though we know each disc moves cyclically, alternate ones in alternate directions, I don't know any easy way to translate a particular step number into the positions of all the discs _ Hinz (1989) gives a method which may be as simple as possible, but I have a feeling it ought to be easier.]

Andy Liu & Steve Newman, proposers and solvers.  Problem 1169 (ii) _ The two towers.  CM 12 (1986) 179  &  13 (1987) 328‑332.  Three pegs, two identical piles of size  n  on two of them.  The object is to interchange the bottom discs and reform the piles (though the smaller discs may or may not be interchanged).  They find it takes  7*2ⁿ⁺¹ ‑ 3n ‑ (10 or 11)/3  steps, depending on whether  n  is odd or even.

Andreas M. Hinz.  The Tower of Hanoi.  L'Enseignement Math. 35 (1989) 289-321.  Surveys history and current work.  50 references.  Finds many properties, particularly the average distance from having all discs on a given peg and the average distance between legal positions.  He also studies illegal positions.  Uses the Grundy, Scorer & Smith graph.  Gives general results, such as Schwenk's below.  He asserts that the mimimality of the classic solution was not proven until 1981, but I think the classic method clearly implies the proof of its minimality.

Hugh Noland, proposer;  Norman F. Lindquist, David G. Poole & Allen J. Schwenk, solvers.  Prob. 1350 _ Variation on the Tower of Hanoi.  MM 63:3 (1990) 189  &  64:3 (1991) 199-203.  Three pegs,  2n  discs, initially with the evens on one peg and the odds on another.  How many moves to get all onto the empty peg?  Answer is  ë(5/7) 4ⁿû.  Schwenk gives a solution for any starting position of  N  discs and shows the average number of moves to get to a single pile is  (2/3)(2^N-1).

Andreas M. Hinz.  Pascal's triangle and the Tower of Hanoi.  AMM 99 (1992) 538-544.  Shows the Grundy, Scorer & Smith graph is equivalent to the pattern of odd binomial coefficients in the first  2ⁿ  rows and hence to Sierpinski's fractal triangle.  Gives some life of Lucas.

David Poole.  The bottleneck Towers of Hanoi problem.  JRM 24:3 (1992) 203-207.  Studies the problem when big discs can go on smaller discs, but not too much smaller ones.  11 references to recent work on variations of the classical problem.

Ian Stewart.  Four encounters with Sierpi_ski's gasket.  Math. Intell. 17:1 (1995) 52-64.  This discusses the connections between the graph of the Tower of Hanoi, the pattern of odd binomial coefficients, Sierpi_ski's gasket and Barnsley's iterated fractal systems.  Lots of references, including 11 on the Tower of Hanoi.

Vladimir Dubrovsky.  Nesting Puzzles _ Part I: Moving oriental towers.  Quantum 6:3 (Jan/Feb 1996) 53-59 & 49-51.  Outlines the history and theory of the Tower of Hanoi.  Misha Fyodorov, a Russian high school student, obaserved that the peg not used in a move always moves in the same direction.  Discusses Kotani's modification which prevents placing some discs on a particular peg.  Also discusses the Panex Puzzle _ cf Section 5.A.4.

Jagannath V. Badami.  Musings on Arithmetical Numbers  Plus Delightful Magic Squares.  Published by the author, Bangalore, India, nd [Preface dated 9 Sep 1999].  Section 4.15: The Tower of Brahma, pp. 123-124.  "The author has lived in Banaras for a number of years and does not find any basis for this legend."

 

          7.M.2.a.       TOWER OF HANOI WITH MORE PEGS

 

Lucas.  Nouveaux jeux scientifiques ..., op. cit. in 4.B.3, 1889.  (See discussion in 7.M.2.)

Dudeney.  The Reve's Puzzle.  The Canterbury Puzzles.  London Mag. 8 (No. 46) (May 1902) 367‑368  &  8 (No. 47) (Jun 1902) 480.  = CP, prob. 1, pp. 24‑25 & 163‑164.  4  pegs,  8, 10 or 21  discs.

Dudeney.  Problem  447.  Weekly Dispatch (25 May, 15 Jun, 1902) both p. 13.  4  pegs,  36  discs.

Dudeney.  Problem 494.  Weekly Dispatch (15 Mar, 29 Mar, 5 Apr, 1903) all p. 13.  5  pegs,  35  discs.

B. M. Stewart, proposer;  J. S. Frame & B. M. Stewart, solvers.  Problem 3918.  AMM 48 (1941) 216‑219.  k  pegs,  n  discs.  General solution, but editorial note implies there is a gap in each solver's work.

Scorer, Grundy & Smith.  Op. cit. in 7.M.1.  1944.  They give some variations on the Tower of Hanoi with four pegs.

Doubleday - II.  1971.  Keep count, pp. 91-92.  15  discs,  6  pegs  _ solved in  49  moves.

Ted Roth.  The tower of Brahma revisited.  JRM 7 (1974) 116‑119.  Considers  4  pegs.

Brother Alfred Brousseau.  Tower of Hanoi with more pegs.  JRM 8 (1975/76) 169‑176.  Extension of Roth, with results for  4  and  5  pegs.

Joe Celko.  Puzzle Column:  Mutants of Hanoi.  Abacus 1:3 (1984) 54‑57.  Discusses variants:  where a disc can only move to an adjacent peg in a linear arrangement;  with two or three colours of discs;  with several piles of discs;  where a disc can only move forward in a circular arrangment.

Grame Williams.  In:  Joe Celko; Puzzle column replies; Abacus 5:2 (1988) 70‑72.  Table of minimum numbers of moves for  k  pegs,  k = 3, ..., 8  and  n  discs,  n = 1, ..., 10.

Andreas Hinz.  An iterative algorithm for the Tower of Hanoi with four pegs.  Computing 42 (1989) 133-140.  Studies the problem carefully.  17 references.

A. D. Forbes.  Problem 163.2 _ The Tower of Saigon.  M500 163 (Aug 1998) 18-19.  This is the Tower of Hanoi with four pegs.  Quotes an Internet posting by Bill Taylor giving an algorithm and its number of moves up to 12 discs.  Asks if this is optimal.

 

          7.M.3.          GRAY CODE

 

          See Gardner under 7.M.

 

L. A. Gros.  Op. cit. in 7.M.1, 1872.  ??NYS.  (Afriat.)

J. Emile Baudot.  c1878.  ??NYS.  Used Gray code in his printing telegraph. (Described by F. G. Heath; Origins of the binary code; SA (Aug 1972) 76‑83.)

Anon.  Télégraphe multiple imprimeur de M. Baudot.  Annales Télégraphiques (3) 6 (1879) 354‑389.  Says the device was presented at the 1878 Exposition and has been in use on the Paris‑Bordeaux line for several months.  See pp. 361‑362 for diagrams and p. 383 for discussion.

George R. Stibitz.  US Patent 2,307,868 _ Binary Counter.  Applied 26 Nov 1941;  granted 12 Jan 1943.  3pp + 1p diagrams.  Has an electromechanical binary counter using the Gray code with no comment or claims on it.

Frank Gray.  US Patent 2,632,058 _ Pulse Code Communication.  Applied 13 Nov 1947, patented 17 Mar 1953.  9pp + 4pp diagrams.  Systematic development of the idea and its uses.

A. J. Cole.  Cyclic progressive number systems.  MG 50 (No. 372) (May 1966) 122‑131.  These systems are Gray codes to arbitrary bases _ e.g. in base  4,  the sequence begins:  0, 1, 2, 3, 13, 12, 11, 10, 20, 21, 22, 23, 33, 32, 31, 30, 130, 131, ....  For odd bases, the sequence is harder.  He gives conversion rules and rules for arithmetic.

William Keister.  US Patent 3,637,216 _ Pattern-Matching Puzzle.  Filed 11 Dec 1970;  patented 25 Jan 1972.  Abstract + 4pp + 2pp diagrams.  This has a bar to remove from a frame _ one has to move various bits in the pattern of the Gray code (or similar codes) to extract the bar.  Made by Binary Arts since about 1986.

 

          7.M.4.          BINARY DIVINATION

 

          The classic cards for this process have the numbers  1 - 2ⁿ  on them, the  i-th card containing those numbers whose binary expression has a  1  in its  (i-1)-st place _ e.g. the first card contains all the odd numbers.  Then one adds up the smallest numbers, i.e.  2^i-1  on the  i‑th card, on the chosen cards to get the number thought of.  If one replaces the numbers by holes in the corresponding positions, one can overlay the cards to read off the answer.  This takes a little more work though _ one has to have each card containing its set of holes to be used if it is chosen and also containing the complementary holes to be used if it is not chosen.  This can be achieved if the holes are centrally located _ then turning the card around produces the complementary set of holes.

          A very similar principle is used as a kind of logical device.  See:  Martin Gardner; Logic Machines and Diagrams; McGraw‑Hill, NY, 1958, pp. 117-124;  slightly extended in the 2nd ed., Univ. of Chicago Press, 1982, and Harvester Press, Brighton, 1983, pp. 117-124;  for discussion of this idea and references to other articles.

          A simple form of binary division is used to divine a card among sixteen cards arranged in two columns, but it is surprisingly poorly described.  This is related to the 21 card trick which is listed in 7.M.4.b.  I have only recently added this topic and may not have noticed many versions.

 

Pacioli.  De Viribus.  c1500.  Prob. 69: A trovare una moneta fra 16 pensata.  Divides 16 coins in half 4 times, corresponding to the value of the binary digits.

Bachet.  Problemes.  1612.  Prob. XVI, 1612: 87-92.  Prob. 18; 1624: 143-151; 1884: 72-83.  15 card trick.   His Avertissement mentions that other versions are possible and describes divining from sixteen cards in two columns and in four columns, but with no diagrams!

Yoshida K_y_ (= Mitsuyoshi Yoshida) (1598-1672).  Jink_‑ki.  2nd ed., 1634 or 1641??.  Op. cit. in 5.D.1.  ??NYS  Shimodaira (see entry in 5.D.1) discusses this on pp. 2-12 since there are several Japanese versions of the idea.  The Japanese call these Metsukeji (Magic Cards).  The binary version is discussed on pp. 4-7 where it is said that they are known since the 14C or even earlier.  The Japanese magic card shown on p. 6 has  1, 2, 4, 8, 16  associated with branchings on a picture of a tree used to divine one of the  21  characters written on the flowers and leaves.  The other kinds of magic cards are more complex, not involving binary, but just memorisation.  The recent transcription of part of Yoshida into modern Japanese does not include this problem.

Ozanam.  1725.  Prob. 39, 1725: 231-233.  Prob. 15, 1778: 164-165;  1803: 165-166;  1814: ??NYS.  Prob. 14, 1840: 74-75.  Sixteen counters being disposed in two rows, to find that which a person has thought of.  Similar to Bachet, but with some diagrams.

Ozanam-Hutton.  1814: 124-126;  1840: 64.  (This is an addition which was not in the 1803 ed.)  Six cards to divine up through 63. 

Endless Amusement II.  1826?  Pp. 180-181.  Sixteen Cards being disposed in Two Rows, to tell the Card which a Person has thought of.  c= Ozanam, with 'counter' replaced by 'card'.

Young Man's Book.  1839.  Pp. 202-203.  Identical to Endless Amusement II.

Crambrook.  1843.  P. 7, no. 5: A pack of cards by which you may ascertain any person's age.  Not illustrated, but seems likely to be binary divination _ ??

Magician's Own Book.  1857.  The mathematical fortune teller, pp. 241‑242.  Six cards each having  30  numbers used to divine a number up through  60.  Some cards have duplicate numbers in the  30th position.  = Boy's Own Conjuring Book, 1860, pp. 211-212.   = Illustrated Boy's Own Treasury, 1860, prob. 38, pp. 402 & 442.

Book of 500 Puzzles.  1859.  The mathematical fortune teller, pp. 55-56.  Identical to Magician's Own Book.  However, my example of the book has  61  in the last cell of the last card, but a xerox sent by Sol Bobroff has this as a  41,  as do all other versions of this problem.

Leske.  Illustriertes Spielbuch für Mädchen.  1864?  Prob. 563-VI, pp. 249-250: Überraschungen mittels sieben Zauberkarten.  Seven cards used to divine up to 100.

Adams & Co., Boston.  Advertisement in The Holiday Journal of Parlor Plays and Pastimes, Fall 1868.  Details?? _ xerox sent by Slocum.  P. 5: Magic Divination Cards.  For telling any number thought of, or a person's age.  Amusing, curious, and sometimes "provoking."   Not illustrated, but seems likely to be binary divination _ ??

Magician's Own Book (UK version).  1871.  The numerical fortune teller, pp. 89-90.  Very similar to Magician's Own Book, pp. 241-242, with the same cards, but different text.

Bachet-Labosne.  Problemes.  3rd ed., 1874.  Supp. prob. X, 1884: 196-197.  Divine a number up to  100  with  7  cards.

F. J. P. Riecke.  Op. cit. in 4.A.1, vol. 3, 1873.  Art. 3: Die Zauberkarten, pp. 11‑13.  Describes  5  cards giving values up to  31,  with explanation.  Describes how to use balanced ternary to construct  7  cards giving values up to  22.  The limitation to  22  is due to the size of the cards _ the method works up to  40.

Mittenzwey.  1879?  Prob. 43, pp. 11‑13.  Seven cards.  Calls them "Boscos Zauberkarten".

Hoffmann.  1893.  Chap. IV, no. 68: The magic cards, pp. 160‑161 & 216‑217.  Photo in Hordern, p. 73, showing a German set 'Zauber‑Karten'.

Lucas.  L'Arithmétique Amusante.  1895.  L'éventail mystérieux, pp. 168‑170.  Shows five cards for divining  1  through  31  and notes it is based on binary.

Mr. X.  His Pages.  The Royal Magazine 10:2 (Jun 1903) 140-141.  To tell a lady's age.  Six cards to divine up through 63.

Ahrens.  MUS I.  1910.  Pp. 39-40 does  5  cards.  Pp. 43-48 develops a set of four see-through templates.

Adams.  Indoor Games.  1912.  How to divine ages, pp. 349-350.

William P. Keasby.  The Big Trick and Puzzle Book.  Whitman Publishing, Racine, Wisconsin, 1929.  Magic columns, pp. 176-177.  Applies the idea with letters, so you add the column heads to get the number of the letter.

Rohrbough.  Brain Resters and Testers.  c1935.  How to Mystify People, pp. 10-11.  = Keasby, whom he cites elsewhere.

John Fisher.  John Fisher's Magic Book.  Muller, London, 1968.  Think‑a‑Drink, pp. 27-29.  5  cards with cut out holes to divine  32  types of drink.  The last card seems to need one side reversed.

 

          7.M.4.a.       TERNARY DIVINATION

 

          This is much less common than binary divination and I have just added it.  Gardner, op. cit. at beginning of 7.M.4, describes one set of triangular cards.

 

Riecke.  1873.  See in 7.M.4.

Martin Hansen.  Mind probe.  MiS 21:1 (Jan 1992) cover & 2-6.  Adapts binary divination to locate a number among  1 - 80  with four cards _ but each card has its numbers half in black and half in white.  Describes how to make  4 x 4  and  8 x 8  binary cards with holes so the chosen number will appear in the hole.  Adapts to ternary to produce triangular cards with holes so that the chosen number appears in the hole.  Hansen has kindly given me a set of these:  'The Kingswood Mathemagic Club's Window Cards'.  He also describes 'logic cards' which display the truth values of three basic quantities which are consistent with various statements _ these were previously described by Gardner and Cundy & Rollett _ see the Gardner item at the beginning of 7.M.4 for details and references.

 

          7.M.4.b.       OTHER DIVINATIONS USING BINARY OR TERNARY

 

          New section.  Discussion with Bill Kalush has revealed that the classic 21-card trick is based on the ternary system.  Another trick involving twice taking  3/2  of a number, which goes back to at least Pseudo-Bede, makes a little use of binary.

          The  21-card trick can also be done with  15  or  27  cards and it is easiest to explain for  27  cards.  One deals  27  cards out, face up, into three columns and asks the spectator to mentally choose one card and tell you which column it is in.  You pick up the three columns, carefully placing the chosen column in the middle of the other two, then deal out the deck into three columns again.  Ask the spectator to tell you what column the card is now in and pick up the cards again with the chosen column in the middle and deal them out again.  Repeat the whole process a third time, then ask the spectator to turn over the middle card, which will be his chosen card.  The first process puts the chosen card among the positions whose initial ternary digit is a  1.  The second process puts it among the position whose ternary representation begins  11.  The third process puts it at the position whose ternary representation is  111.  For any number of cards which is a multiple of three, each process puts the chosen card in the 'middle third' of the previously determined portion of the deck and the trick works, though the patterns are less systematic than with  27  cards.  When the trick is done with, e.g.  24  cards, there is no middle card and you have to expose the chosen card yourself or develop some other way to do it.  One can also adapt the idea to other numbers of columns _ the two column case will usually be connected to binary and may be listed in 7.M.

          In the  3/2  method, you ask a person to think of a number,  x,  then take  3/2  of it (or itself plus half of itself).  If there is a half present, he is to round it up and let you know.  Do this again on the result.  Now ask how many nines are contained in his result.  You then tell the number thought of.  The process actually converts  x = a + 2b + 4c  to  3a + 5b + 9c,  where  a, b  =  0  or  1.  We have that   a = 1  iff  there is a rounding at the first stage   and   b = 1  iff  there is a rounding at the second stage   and the person then tells you  c.  Sometimes the second stage is omitted or the division by two in it is omitted, which makes things simpler.  The latter case takes  a + 2b  to  6a + 9b  and the person gives you  b  and there is a remainder if and only if  a = 1.

 

Pseudo-Bede.  De Arithmeticis propositionibus.  c8C, though the earliest MS is c9C.

                    IN: Venerabilis Bedae, Anglo‑Saxonis Presbyteri.  Opera Omnia: Pars Prima, Sectio II _ Dubia et Spuria: De Arithmeticis propositionibus.  Tomus 1, Joannes Herwagen (Hervagius), Basel, 1563, Columns 133-135, ??NYS.  Folkerts says Hervagius introduced the title De Arithmeticis propositionibus.

                    Revised and republished by J.‑P. Migne as:  Patrologiae Cursus Completus: Patrologiae Latinae, Tomus 90, Paris, 1904, columns 665‑668. 

                    Critical edition by Menso Folkerts.  Pseudo-Beda: De arithmeticis propositionibus.  Eine mathematische Schrift aus der Karolingerzeit.  Sudhoffs Archiv 56 (1972) 22-43.  A friend of Bill Kalush has made an English translation of the German text, 1998?, 11pp.

                    See also: Charles W. Jones; Bedae Pseudepigrapha: Scientific writings falsely attributed to Bede.  Cornell Univ. Press & Humphrey Milford, OUP, 1939, esp. pp. 50‑53.

                    This has three divination problems and Folkerts says these are the first known western examples.

                    1.       Triple and halve, then triple and tell quotient when divided by nine and whether there is a remainder in this.

                    2.       3/2  twice and tell if there is a rounding up at each stage.

                    3.       Divine a digit  a  from  a * 2 + 5 * 5 * 10,  where the operations are performed sequentially from left to right.  The result is  100a + 250.  This is not of the type considered in this section, but is the prototype of most later divination methods.

                    Folkerts mentions several later occurrences of these methods.

                    (The fourth and last part of the text is probably slightly later in the 9C and describes adding postive and negative numbers in a way not repeated in the west until the 15C.)

Fibonacci.  1202.  Pp. 303-304.  Take  3/2  twice.

Chuquet.  1484.  Prob. 155.  English in FHM 230-231.  Divination by doing  3/2  once, then tripling the result and telling the quotient when divided by  9.  Text only indicates what happens when  x  is even.  Marre notes that this appears in de la Roche, 1520: f. 218v ‑ 219r; 1538: f. 150v - 151r.  FHM say Chuquet gives an example, but it is not in Marre.

Pacioli.  De Viribus.  c1500. 

Prob. VII: Trovare un numero pensato intero.  Take  1 + 1/2  twice.  Agostini only indicates the case when  a = b = 0.

Prob. IX: A trovare un numero senza rotto.  Take  3/2  twice.  Same comments as for Prob. VII.

Prob. X: De trovare un numero senza rotto.  Take  3/2  and then triple and tell the quotient on division by  9.  Agostini only indicates the case when  x  is even.

Ghaligai.  Practica D'Arithmetica.  1521.  Prob. 34-35, f. 67r-67v.  Take  3/2  twice, but he rounds down each time and he is not at all clear how the roundings relate to finding the number, and it is not nearly as elegant as when one rounds up.

Apianus.  Kauffmanss Rechnung.  1527.  Ff. M.vii.r - M.viii.r.  Divination by  3/2  once followed by multiplying by  3  and then dividing by  9.

Tartaglia.  General Trattato.  1556.  Book 16, art. 197-198, f. 263v-264r.  Divination by  1 + 1/2  twice and by  3/2  twice.

Prévost.  Clever and Pleasant Inventions.  (1584), 1998. 

Pp. 166-168.  Divination by twice taking  3/2. 

Pp. 185-189.  Divination from  16  marked counters by a two column version of the  21‑card trick.  The method of rearranging the  16  counters is not entirely clear, but the principle is clearly explained.

Io. Baptiste Benedicti (= Giambattista Benedetti).  Diversarum Speculationum Mathematicarum, & Physicarum Liber.  Turin, (1580), Nicolai Bevilaquæ, Turin, 1585; (Venice, 1599).  [Rara 364.  Graves 141.f.16.]  Theorema CXVI, pp. 78-79.  Divination by twice taking  3/2.

Bachet.  Problemes.  1612. 

Prob. I.  1612: 14-17;  1624: 53-55;  1884: 15-16.  Divination by doing  3/2  once, then tripling the result and telling the quotient when divided by  9.  = Pacioli X, but with more explanation.

Prob. II.  1612: 17-27;  1624: 56-65;  1885: 17-22.  Divination by twice taking  3/2.  = Pacioli IX, but with more explanation.

Prob. III.  1624: 66-74;  1885: 23-26.  Take  3/2  twice and then tell all but one digit.  The Advertisesement in the 1624 ed. says this was invented by R. P. Jean Chastelier, S.J.  Not in the 1612,

Prob. XVI, 1612: 87-92.  Prob. 18,  1624: 143-151;  1884: 72-83.  15-card trick.   His Advertissement mentions that other versions are possible and describes a two column version.  Labosne adds several diagrams which make the process much clearer and discusses the general idea, illustrating it with 27 cards and then with 45 cards in 5 columns.

Ozanam.  1694.

Prob. 14 [part 9],  1696: 58;  1708: 52.  Prob. 17, part 10,  1725: 146.  Prob. 1, part 1,  1778: 139-140;  1803: 137;  1814: ??NYS;  1840: 62.  = Pacioli X.  In 1840, the algebraic proof is given.  In 1725: 170‑173,  he adds, as a remark to Prob. 17, the usual version taking  3/2  twice, with a detailed explanation.

Prob. 16,  1696: 63-64;  1708: 56-57.  Prob. 19,  1725: 154-156.  Prob. 1, part 3,  1778: 140-142; 1803: 138-139;  1814: ??NYS;  1840: omitted.  Take  1 + 1/2  twice but then subtract  2x  to get  r,  then divide  r  repeatedly by  2  until one gets down to  1.  Observing when he has to discard gives you the binary expansion of  r.  Subtracting  2x  yielded  r = a + b + c,  so  x =   4r ‑ (2b + 3c).

Prob. 31.  1696: 85;  1708: 76-77.  Prob. 35,  1725: 220-221.  21-card trick done with  36  cards.  He says the desired card will be in the middle of its row, i.e. in the 6th place! 

Prob. 15.  1778: 164-166;  1803: 165-166.  Prob. 14, 1840: 74-75.  Replaces the above with the  16  counter binary version as in Prevost and notes that one can use other powers of two.

Henry Dean.  The Whole Art of Legerdemain, or Hocus Pocus in Perfection.  11th ed., 1790?  ??NX _ seen at UCL Graves 124.b.36.  Pp. 89-90 (89 is misprinted as 87).  21 card trick.  "This trick may be done by an odd number of cards that may may be divided by three."

Badcock.  Philosophical Recreations, or, Winter Amusements.  [1820].  Pp. 81-83, no. 129: To tell the Number a Person has fixed upon, without asking him any question.  A variant of the  3/2  method.  Pick a number from  1  to  15,  add  1  to it.  (Triple the number; if it is odd, add  1;  then halve it) thrice.  Finally, if the result is odd, add one, and again halve.  From observing when the person has to add one, you determine the number.  If   n + 1  =  8a + 4b + 2c + d,   where the coefficients are zero or one, the process is straightforward up to the third tripling which yields  54a + 27b + 15c + 9d  which is odd if and only if  b + c + d  is odd.  The next steps are no longer simply expressible in terms of the coefficients.  One has to add one at the third stage if and only if  n + 1 º 1, 2, 4, 7 (mod 8)  and one has to add one at the fourth stage if and only if  n + 1 = 1, 6, 8, 10, 11, 12, 13, 15.  He gives tables to determine  n  from the observations.

Manuel des Sorciers.  1825.  ??NX 

Pp. 47-48, art. 20.  15 card trick.

Pp. 60-61, art. 33.  16 card trick using awkward binary.

Magician's Own Book.  1857.  Which counter has been thought of out of sixteen?, pp. 226‑227.  Like Prevost but clearer.  Uses  16  cards and repeated halving to bring the thought of card to the top in four steps.  Says it can be extended to  32  cards.  = Boy's Own Conjuring Book, 1860, pp. 196-197.

Indoor Tricks and Games.  Success Publishing, London, nd [1930s??].  The wonderful twenty-seven  &  Variants of the  27-card trick, pp. 47-51.  Uses ternary to locate and position a card among  27  _ but I find the description of how to perform the trick a bit cryptic.

 

          7.M.5.  LOONY LOOP  =  GORDIAN KNOT

 

          This is a configuration of wire loops attached to a rod with a loop of string to be removed.  The solution method uses the binary pattern of the Chinese Rings.

 

George E. Everett, via George Barker.  UK Patent 15,971 _ An Improved Puzzle.  Applied 21 Sep 1891;  accepted 24 Oct 1891.  1p + 1p diagrams.  Barker states that the invention was communicated by Everett, of Grand Island, Nebraska.  There is no indication of the binary pattern in the puzzle.

Adams.  Indoor Games.  1912.  Pp. 337‑341.  The double link (= Loony Loop or Satan's Rings). 

Ch'ung‑En Yü.  Ingenious Ring Puzzle Book.  1958.  Op. cit. in 7.M.1.  P. 21 shows a simple version.

 

          7.M.6.          BINARY BUTTON GAMES

 

          New section.

          The problems here are usually electronic puzzles with an array of lightable buttons which can take on two states _ lit and unlit.  Pressing a button toggles the lights on a certain pattern of buttons.  The earliest commercial example I know is the XL-25.  The box says Patent No. 122-8201 061.  However, there is an older similar electrical switching problem by Berlekamp _ see below.  There are also mechanical versions, such as Game Jugo from Japan (mid 1980s?), which has 15 petals such that turning one over turns over some others, and Orbik _ see below.  A number of further versions have appeared and I have seen some incorporated in game packages on computers.  Rubik's Clock is essentially the same kind of problem except that the states of the clocks and the possible turns are (mod 12) instead of (mod 2), though the main interest centres on whether a clock is correct or not.  Likewise Orbik is essentially the same, but with values (mod 4).

 

Elwyn Berlekamp.  Switching Game.  Devised and actually built at Bell Labs, c1970.  This is an  n x n  array of lights with  2n  switches which will invert the lights in a row or column.  Berlekamp's version is  n = 10.  See Fishburn & Sloane below.

Lazlo Meero.  The XL-25.   Exhibited at the London Toy Fair in 1983 and marketed by Vulcan Electronics of London.  The box says Patent No. 122-8201 061. 

David Singmaster.  The XL-25.  Cubic Circular 7/8 (Summer 1985) 39-42.  The XL-25 has a  5 x 5  array of buttons and two choices of toggle patterns _ a simple cross of  5  cells and the pattern of a cell and the cells which are a knight's move away.  In both cases, cells off the edge are simply lost.  I show that such problems with  n  buttons can be specified by an  n x n  input/output or transition binary matrix  A = (a_ij),  with  a_ij = 1  if pressing the  j-th button toggles the  i-th button;  otherwise  a_ij = 0.  If  x  is a binary column vector showing which buttons are pushed, then  Ax  is a binary column vector showing which buttons have been toggled.  From a random start, the object is to light all buttons.  So the general solution of the problem is obtained by inverting  A (mod 2).  For the  5 x 5  array, the knight version is invertible and I sketch its inverse.  But the cross version gives a matrix of rank  23.  I note that when the matrix is singular, the Gauss-Jordan elimination method for the inverse yields the null space and range and a kind of pseudo‑inverse, allowing for complete analysis _ in theory.  (More recently, I have seen that this is a proper pseudo-inverse and all the pseudo-inverses of  A  are computible by this process.)  Again I sketch the solution process.  (I now incorporate this idea in my teaching of linear algebra as a handout  "On trying to invert a singular matrix".)  I corresponded with Meero who had obtained some similar results and showed that each feasible pattern for the cross version could be obtained in at most  15  moves.  He also showed that if the input/output matrix  A  is symmetric and reflexive (i.e.  a_ii = 1),  then one can turn on all the lights, starting with them all off.  He studied the cross version on  n x n  boards up to  n = 100.  A friend of Meero studied fixed point patterns such that  Ax = x  and anti-fixed points where  Ax  is the vector complementary to  x.  Meero asserts the knight version has  A  invertible if  n ¹ 6, 7, 8 (mod 9).  I wondered what happens for cylindrical or toroidal boards, but made no attempt to study them.

In Spring 1986, I applied this method to a Japanese puzzle called  Game Jugo  which Edward Hordern had.  This has  15  two-sided 'petals' around a centre which has four pointers.  When one pointer points to petal  1,  the others point to  4, 8, 11  and turning one of these over turns over the others in this set.  Since the sum of all rows of the transition matrix is the zero vector, it follows that the matrix is singular.  I found that it had rank  12,  so there are just  2¹² = 4096  achievable patterns, each of which has  2³ = 8  solutions.  It turns out that there are three groups of  5  petals, e.g.  {1, 4, 7, 10, 13},  such that the sum of the turns in a group must be  0  for an achievable pattern.  From this, I showed that any achievable pattern could be reached in at most  6  moves, determined how many patterns required each number of moves and showed that the average number of moves was  4.6875.  I am very keen to get an example of this and/or its instructions (perferably with an English translation).

Donald H. Pelletier.  Merlin's magic square.  AMM 94:2 (Feb 1987) 143‑150.  Merlin is a product of Parker Brothers and provides several games, including a  3 x 3  binary button game.  If we number the cells   1, 2, 3;  4, 5, 6;  7, 8, 9;  then:  pressing  1  toggles  1, 2, 4, 5;  pressing  2  toggles  1, 2, 3;  pressing  5  toggles  2, 4, 5, 6, 8  (same as the cross on the XL-25).  The object is to light all but button  5.  He develops the binary vectors as above, finds that the transition matrix is invertible, computes the inverse and answers a few simple questions.

T. E. Gantner.  The game of quatrainment.  MM 61:1 (Feb 1988) 29-34.  Considers a game on a  4 x 4  field where a corner move reverses the six cells closest to the corner;  an edge move reverse the the neighbouring cells and a centre move reverse the cell and its four neighbours (i.e. the  +  pattern).  Sets up the matrix approach and shows the transition matrix is invertible, finding inputs which reverse just one cell.  Modifies the moves and finds versions of the game with reanks  12  and  14.

Ern_ Rubik (or Studio?).  Rubik's Clock.  Produced by Matchbox, 1988.  This is essentially the same kind of problem except that the states of the clocks and the possible turns are  (mod 12)  instead of  (mod 2), though the main interest centres on whether a clock is correct or not.  18  clock dials _  9  on front and  9  on back, both in a  3 by 3  pattern.  Four drive wheels on the edges, connected to the corner clocks, but their effects are determined by the positioning of  4  buttons in the middle of the puzzle, giving  30  types of move.  The four corner front clocks are connected to the four corner rear clocks, so there are  14  independent motions to make and the input/output matrix is  30 by 14.

Daniel L. Stock.  Merlin's magic square revisited.  AMM 96:7 (Aug/Sep 1989) 608-610.  He gives an easy algorithm for solving the problem by doing edges, then corners, then middle.

P. C. Fishburn & N. J. A. Sloane.  The solution to Berlekamp's switching game.  Discrete Mathematics 74 (1989) 263-290.  They describe Berlekamp's game, with photo, as a coding theory problem.  The transition matrix  A  here is  2n x n².  For any given initial state  x₀,  consider all the states that can be achieved from it, say  S(x₀)  =  { x₀ + Ax | x Î Z₂²ⁿ }.  We might expect  S(x₀)  to have  2²ⁿ  states, but reversing all rows is the same as reversing all columns _ and there is no other dependence _ so there are  2^2n‑1  states.  Among all these states, there is one with a minimal number,  f(x₀),  of lights turned on.  The covering radius  R  of the code formed by the  2n  rows and columns, considered as words in  Z₂^{n^2},  is the maximum of these minimal numbers, i.e.  min { f(x₀) | x₀ Î Z₂^{n^2} }.  These codes are called 'light‑bulb' codes and have been investigated since c1970 since they have the smallest known covering radius.  From our game point of view, the problem corresponds to finding the most-unsolvable position and  R  is a measure of unsolvability.  The values for  R  were known for  n £ 5.  The authors use extensive hand computing to extend this up through  n = 9  and then a lot of computer time to get to  n = 10.  The values of  R  for  n  =  1, .., 10  are:  0, 1, 2, 4, 7, 11, 16, 22, 27, 34.  That is, for the  10 x 10  game, there is an array of  34  turned-on lights which cannot be reduced to a smaller number of turned-on lights by any inversion of rows and columns.

Orbik.  Orbik is a ring of  12  wheels, each having  4  colours but just one colour can be seen through the top cover of windows.  There are three marks.  When one mark is at  1,  the others are at  4  and  8.  When the top cover is turned ahead, the marked wheels move forward one colour.  A backward turn leaves everything fixed, but moves the position of the marks.  Made by James Dalgety.  I believe it was Edward Hordern's exchange gift for a puzzle party, c1993. 

Edward Hordern.  Orbik.  CFF 29 (Sep 1992) 26-27.  ??NYR.  Orbik is a ring of  12  wheels, each having  4  colours but just one colour can be seen through the top cover of windows.  There are three marks.  When one mark is at  1,  the others are at  4  and  8.  When the top cover is turned ahead, the marked wheels move forward one colour.  A backward turn leaves everything fixed, but moves the position of the marks.

Ralph Gasser.  Orbik.  CFF 32 (Aug 1993) 26-27.  He counts both forward and backward turns and finds there are  60  antipodal positions requiring  54  moves to solve.  The shortest processes for moving a single wheel by  1, 2, 3  colours take  29, 28, 25  moves.  If a sequence of turns in the same direction is counted as a single move, there are  4  antipodal positions requiring  23  moves to solve and the single wheel processes take  9, 7, 9  moves.

Dieter Gebhardt.  Cross pattern piling.  CFF 33 (Feb 1994) 14-17.  Notes that Dario Uri independently invented the XL-25 idea with the cross pattern _ he called it Matrix of Lights.  Gebhardt modifies the problem by making two ons remain on.  Thus his computation of  Ax  is an ordinary matrix product and he wants results with each entry the same.  If one thinks of the cross shapes as five cubes piled onto the board, the sought result is a uniformly stacked board of some height  h.  This also allows for some cell to be turned on several times.  Thus we are trying to solve  Ax = hJ,  where  J  is the vector of all  1s,  h  is a positive integer and  x  is a vector with non-negative integer entries.  Obviously the minimal value of  h  is wanted.  He determines solvability and all minimal solutions up to  8 x 8,  with  9 x 9  given as a contest.

Tiger Electronics, 980 Woodlands Parkway, Vernon Hills, Illinois, 60061, USA  &  Belvedere House, Victoria Avenue, Harrogate, UK.  Lights Out.  Model 7-574, 1995.  Essentially the same concept as the  XL-25  with its 'cross' pattern.  With lots of preprogrammed puzzles, random puzzles and option to input your own puzzles.  The longest solution is  15  moves, as found by Meero for the  XL-25.

Uwe Mèffert produced Orbix (or Light Ball) in 1995 for Milton Bradley.  I advised a bit on the design of the games.  This is a sphere with  12  light buttons in the pattern of a dodecahedron.  There are four different games.  The object is to turn all lights on, but in some games, one can also get all lights off.  However, only the first game is a linear transformation in the sense discussed above.  The later games have rules where the effect of a button depends on whether it is lit or not and even on whether the opposite button is lit or not.  Nonetheless all examples are solvable in  12  moves or less.

Edward Hordern.  What's up?  CFF 38 (1995) 38.  ??NYR.  Discusses Tiger Electronics'  Lights Out.

Dieter Gebhardt & Edward Hordern.  How to get the lights of "Lights Out" out.  CFF 39 (1996) 20-22.  ??NYR.  Sketches a solution.

Edward Hordern.  What's up?  CFF 41 (Oct 1996) 42.  Discusses Tiger Electronics'  Deluxe Lights Out  which has a  6 x 6  array with several options _ one can affect five lights in the form of a  +  or of a  x;  a button can have effect only if it is lit, or alternately lit/unlit.

 

          7.N.    MAGIC SQUARES

 

      4    9    2                    The  3 x 3  magic square is usually given in the form on the left.

      3    5    7          We denote each of the 8 possible forms by its top row.  I.e. this is the

      8    1    6          492  form.  All Chinese material seems to give only this form, called the

                              Lo Shu [Lo River Writing].

            7

            2                                    An unrelated diagram, shown on the left, is called the Ho Thu

8    3    5    4    9              diagram [River Plan].  See 7.N.5 for magic versions of this shape.

            1

            6

 

          Semi‑magic  denotes a square whose rows and columns add to the magic constant, but not the diagonals.

          Pandiagonal  means that the 'broken diagonals' also add to the magic constant.  Lucas called these  diabolic  and they are also called  Nasik,  as they were studied by Frost, who was then living in Nasik, India.

          Associated  or  complementary  means that two cells symmetric with respect to the centre add to  n² + 1.

          See 7.AC.3 for related pan-digital sums.

 

          The early history of magic squares remains rather obscure.  In particular, the first four Chinese sources below are extremely vague!  Cammann‑4 argues that magic squares had powerful mystic meanings to the Chinese, Indians and Arabs, hence were not explicitly described in writing.

          There are several surveys of some or all of the history of magic squares which I list first for later reference.  These provide many more references.

          17‑20C material has generally been omitted, but see Bouteloup.  Smith & Mikami discuss several workers in Japan, but I've omitted some of them.

 

                    SURVEYS

 

Wilhelm Ahrens - 1.  Studien über die "magischen Quadraten" der Araber.  Der Islam 7 (1917) 186‑250.

Wilhelm Ahrens - 2.  Die "magischen Quadrate" al-B_n_'s.  Der Islam 12 (1922) 157‑177. 

Schuyler Cammann ‑ 1.  The evolution of magic squares in China.  J. Amer. Oriental Soc. 80 (1960) 116‑124.

Schuyler Cammann ‑ 2.  The magic square of three in old Chinese philosophy and religion.  History of Religions 1 (1961) 37‑80.  ??NYS

Schuyler Cammann ‑ 3.  Old Chinese magic squares.  Sinologia 7 (1962) 14‑53.

Schuyler Cammann ‑ 4.  Islamic and Indian magic squares I  &  II.  History of Religions 8 (1968‑69) 181‑209  &  271‑299.

Heinrich Hermelink.  Die ältesten magischen Quadrate höher Ordnung und ihre Bildungsweise.  Sudhoffs Arch. 42 (1953) 199‑217.

Lam Lay Yong.  1977.  See under Yang Hui below.  Her commentary surveys the history.

Needham.  1958.  Pp. 55‑61.  See also:  vol. 2, 1956, pp. 393 & 442;  Vol. 5, Part IV, 1980, pp. 462-472.

J. Sesiano ‑ I  &  II.  Herstellungsverfahren magischer Quadrate aus islamischer Zeit (I)  &  (II).  Sudhoffs Arch. 64 (1980) 187‑196  &  65 (1981) 251‑265.

 

          A number of the entries in 5.A _ The Fifteen Puzzle _ refer also to magic squares.  See:  Loyd (1896);  Cremer (1880);  Tissandier (1880 & 1880?);  Cassell's (1881);  Hutchison (1891). 

          Some entries in 5.A and here give problems of sliding the Fifteen Puzzle into a magic square.  See:  Dudeney (1898);  Anon & Dudeney (1899);  Loyd (1914);  Dudeney (1917);  Gordon (1988) in 5.A  and  Ollerenshaw & Bondi below.

 

                    POSSIBLE EARLY REFERENCES

 

Anon.  Shu Ching.  c‑650.  Part V, book IV, The Great Plan _ commentary and book XXII, The Testamentary Charge.  IN:  J. Legge, trans.  The Texts of Confucianism, Translated: Part I.  The Shû King, the religious portions of the Shih King, the Hsiâo King.  OUP, 1879, pp. 138‑139 & 239.  P. 138 discusses the Lo Shu and says it does not occur.  On p. 139, we see:  "To [Yu], Heaven gave the Great Plan with its nine divisions ..."  Various commentators, from  Gan‑Kwo  on, have asserted that this was the Lo‑shu which appeared on the back of a tortoise in the river Lo.  Legge says there is no evidence to connect the Lo‑shu with the Great Plan and that the commentators are indulging in leger‑de‑plume.  P. 239 mentions the River Plan.

                    (See also:  J. Legge, trans.; The Chinese Classics, etc.; Vol. III _ Part II; Trübner, London, 1865; pp. 321‑325 & 554.  This gives the Chinese and the English, with extensive notes.)

                    At this time, the number 'nine' was used to describe the largest number and hence does not necessarily imply  3².

Anon.  Lun Yu (Confucian Analects).  c‑5C.  Book IX, Tsze Han; chap. VIII.   IN:  J. Legge, trans.  The Chinese Classics, etc. vol. 1, Confucian Analects, The Great Learning, and the Doctrine of the Mean.  Trübner, London, 1861, p. 83.  = The Life and Teachings of Confucius; Trübner, London, 1869, pp. 169‑170, ??NX.  Also in:  A. Waley; The Analects of Confucius; Allen & Unwin, London, 1949, p. 140.  "The river sends forth no map."

Chuang Tzu (= Kwang‑Sze).  The Writings of Kwang‑Sze.  c‑300.  Part II, sect. VII  = Book XIV, Thien Yu (The Revolution of Heaven).  IN:  J. Legge, trans.  The Texts of Tâoism.  OUP, 1891.  Vol. 1, p. 346.  Refers to "the nine divisions of the writing of Lo."

Anon.  Ta Chuan (= Hsi Tzhu Chuan) (The Great Commentary on the I Ching [= Yi Jing]).  c‑300?? (Needham, vol. 2, p. 307, says c‑100 and vol. 5:IV, pp. 462-463, says -2C)  IN:  J. Legge, trans.  The Texts of Confucianism, Part II: The Yî King.  OUP, 1882.  Appendix III, sect. 1, chap. 12, art. 73, p. 374 & note on p. 376.  [There is a 1963 Dover ed. of Legge's 1899 edition.]  Also as:  Part I, chapter IX _ On the Oracle.  IN:  The I Ching, translated by R. Wilhelm and rendered into English by C. F. Baynes, 3rd ed., 1968, Routledge and Kegan Paul, London, pp. 308‑310.  The text is: "The Ho gave forth the map, and the Lo the writing, of (both of) which the sages took advantage."  This occurs just after paragraphs on the origin of the hexagrams and legend says the Ho Thu inspired the creation of the 8 trigrams.  Legge says the original Ho Thu map was considered to be lost in the -11C and the earliest reconstruction of it was presented during the reign of Hai Zung in the Sung Dynasty (1101-1125).  The I Ching is often cited but only this later commentary mentions an association of numbers with concepts.  Later commentators interpret this association as referring to the Ho Thu and Lo Shu diagrams, though this is not obvious from the association _ the names were not associated with the diagrams until about the 10C _ see Xu Yiu below.  See Needham, vol. 2, pp. 393 & 442 for discussion of the interpolation of the diagrams into the I Ching.)

 

E. John Holmyard.  Alchemy.  Penguin, 1957, pp. 36-38, discusses magic squares in relation to Chinese alchemy and Taoism.  He says the Taoist emphasis on the number 5 is related to its being the central value of the order 3 magic square.  He says this relation has been studied by H. E. Stapleton, but gives no reference.  Stapleton says the square of order 3 was the ground plan of the Ming-Tang or Hall of Distinction.  This Hall was used for the proclamation of monthly ordinances and the Calendar (which was partly lunar and hence of variable length).  When in the Ming-Tang, the Emperor became the incarnation of the god and hence the ground plan became of major importance in Chinese alchemy.  Stapleton conjectures that the original numbering of the  3 x 3  array of rooms in the Hall may not have been magic, but would have had 5 in the middle and that the magic numbering may have occurred at some time and been recognised as having special properties.  Holmyard indicates the Ming-Tang arose about -1000.  All in all, these statements do not agree with most of the other material in this section and it would be good to locate the work of Stapleton (??NYS), which is presumably well-known to students of ancient chemistry/alchemy.

Needham, Vol. 5:IV, 1980, pp. 462-.  Cites Stapleton on p. 462 and indicates his work is a bit cranky, but I haven't got the details yet.  He goes on to discuss why 9 was so important to the Chinese.  He desribes the tour of the pole-star sky-god Thai I which went through the nine cells of the Lo Shu in the order: 5, 1, 2, 3, 4, 5, pause, 5, 6, 7, 8, 9, 5.  His fig. 1535 shows this from a Tang encyclopedia, though this has the  2 7 6  orientation of the square.  The Chinese could see Yin and Yang (= even and odd), the Four Seasons and the Five Elements, and the Nine Directions of space, all in the Lo Shu.  Consequently it was not revealed to the general public until the end of the Tang (618-907).  Needham then discusses the influence of the Lo Shu on Arabic alchemical thought.

Tai the Elder.  Ta Tai Li Chi  (Record of Rites).  c80.  Chap. 67,  Ming Thang.  ??NYS  (See Needham, p. 58.)  Chap. 8, p. 43 of  Szu‑pu ts'ung‑k'an  edition, Shanghai, 1919‑1922.  Describes the  492  form.  (See Cammann‑2.)  (Cammann‑1,  Lam  and  Hayashi  say this is the first clear reference.)

Anon.  I Wei Chhien Tso Tu.  c1C.  Chap. 2, p. 3a.  ??NYS  (Translated in Needham, p. 58.)

Nâgârjuna.  c1C.  Order 4 square called Nâgârjunîya described in a MS on magic called Kaksaputa, nd.  ??NYS.  (A. N. Singh; History of magic squares in India; Proc. ICM, 1936, 275‑276.)

Anon.  Lî Kî.  c2C.  Book VII _  Lî Yun,  sect. IV.  IN:  J. Legge, trans.  The Texts of Confucianism, Part III: The  Li Ki,  I‑X.  OUP, 1885.  Pp. 392‑393.  "The Ho sent forth the horse with the map (on his back) ¹.   ¹ The famous 'River Map' from which, it has been fabled,  Fû-hsî  fashioned his eight trigrams.  See vol. xvi, pp. 14-16."  This last reference is ??NYS.

Theon of Smyrna.  c130.  Part B:  Βιβλιov τα τησ εv Αριθμoσ Μoυσικησ θεωρηματα Περιεχov  (Biblion ta tes en Arithmos Mousikes Theoremata Periechon).  Art. 44.  IN:  J. Dupuis, trans.; Théon de Smyrne; Hachette, Paris, 1892; pp. 166‑169.  (Greek & French.)   Natural square _ often erroneously cited as magic and used to 'prove' the Greeks had the idea of magic squares.

Xu Yiu (= Hsu Yo  = Xu Yue).  Shu Shu Ji Yi  (= Shu Shu Chi I)  (Memoir on Some Traditions of Mathematical Art).  190(?).  ??NYS.  Ho Peng Yoke  [Ancient Chinese Mathematics;  IN:  History of Mathematics, Proc. First Australian Conf., Monash Univ., 1980; Dept. of Math., Monash Univ., 1981, pp. 91‑102], p. 94, says that this is the earliest Chinese text to give the order 3 square.

              The date and authorship of this is contentious.  Current belief is that this was written by  Zhen Luan  (= Shuzun)  in c570, using the name of  Xu Yue.  Li & Du, pp. 96‑97, say that this work first introduces the diagram.  The diagram was called the "nine houses computation".  The diagram was connected with the  Yi Jing  commentary in the 10C and then renamed  Lo Shu.  After the 13C, magic squares were called  zong heng tu  (row and column diagrams).

                    Needham, vol. 5:IV, p. 464, considers this as being c190, referring the Chen Luan as a commentator on it.  He calls the diagram "Nine Hall computing method".

Varahamihira (= Var_hamihira (II)).  Brhatsamhit_.  c550.  Hayashi, below, cites a Sanskrit edition (NYS) and the following.

                    M. Ramakrishna Bhat.  Var_hamihira's  Brhat Samhit_  with English Translation, Exhaustive Notes and Literary Comments.  2 vols, Motilal Banarsidass, Delhi, 1981‑1982.  Vol. II: Chapter LXXVII _ Preparation of perfumes, pp. 704-718.  On pp. 714-715 is the description of a  4 x 4  array:  

                    2, 3, 5, 8;   5, 8, 2, 3;   4, 1, 7, 6;   7, 3, 6, 1,   with some cryptic observations that any mixture totalling  18  is permitted, e.g. "by combining the four corners, or four things in each corner, or the central four columns, or the four central ones on the four sides."  As given, many of the groups indicated do not add up to  18,  nor do the columns.  However, the following article notes that the bottom row should read  7, 6, 4, 1!!

                    This material is described and analysed in:  Takao Hayashi; Var_hamihira's pandiagonal magic square of the order four; HM 14 (1987) 159-166.  He gives the book's name as  Brhatsamhit_  and says the material is in Chapter 76 (Combinations of perfumes).  He gives the correct form of the array.  He notes that the array is a pandiagonal magic square with constant  18,  except the entries are  1, ..., 8  repeated twice.  Hayashi believes that Var_hamihira must have known one of the actual magic squares which yield this square when the numbers are taken  (mod 8).  He shows there are only  4  such magic squares, two of which are pandiagonal.  One of the pandiagonal squares is a rotation of: 

           8, 11, 14,  1;    13,  2,  7, 12;     3, 16,  9,  6;    10,  5,  4, 15,  which he describes as the most famous Islamic square of order  4  described in Ahrens-2.  Hayashi feels that order  4  squares must originate in India, contrary to Cammann's thesis.  (See also Singh, op. cit. at  Nâgârjuna,  above, and  Ikhw_n al-Saf_’,  below.)

‘Ali ibn Sahl Rabb_n al-Tabar_ (d. 860).  Paradise of Wisdom.  This is a gynaecological text discovered by Siggel.  ??NYS _ described in Needham, vol. 5:IV, p. 463.  Example of a magic square used as a charm in cases of difficult labour.  Needham thinks this is the earliest Arabic magic square.

Jabir ibn Hayyan  (= Jâbir ibn Hayyân  = Geber) (attrib.).  Kitâb al‑Mawâzin  (Book of the Balances).  c800.  ??NYS _ discussed in Ahrens-1.  The Arabic and a French translation are in:  M. Berthelot; La Chimie au Moyen Age: Vol. 3 _ "L'Alchimie Arabe"; Imprimerie Nationale, Paris, 1893.  The text is discussed on pp. 19-20, where he refers to the magic square of Apollonius, with a footnote saying 'De Tyane'.  The text is given on p. 118 (Arabic section) & 150 (French section).  Gives  3 x 3  square in form  492. 

                    "Here is a figure divided into three compartments, along the length and along the width.  Each line of cells gives the number  15  in all directions.  Apollonius affirms this is a magic tableau formed of nine cells.  If you draw this figure on two pieces of linen [or rags], which have never been touched by water, and which you place under the feet of a woman, who is experiencing dfficulty in childbirth, the delivery will occur immediately." 

                    The French is also in Ahrens, who notes that the square does not appear in the few extant writings of Apollonius of Tyana (c100).

                    Hermelink mentions this as the earliest Arabic square, but gives no details.  Needham, vol. 5:IV, p. 463, says Cammann gave this, and dates this as c900.

                    Suter, pp. 3‑4, doesn't mention magic squares for ibn Hayyan, but this appearance is simply in a list of questions on properties of animals, vegetables and minerals, so hardly counts as mathematics.

                    Holmyard [op. cit. above, pp. 74-75] discusses the work of Kraus and Stapleton on Jabir.  Jabir considers the numbers  1, 3, 5, 8  as of great importance _ these are the entries in the lower left  2 x 2  part of his magic square.  These add to  17  and everything in the world is governed by this number!  He also attaches importance to  28  which is the sum of the other entries.  Holmyard asserts this magic square was known to the New-Platonists of 3C _ an assertion which I have not seen elsewhere.  Jabir uses ratios  1/3  and  5/8  extensively in his alchemical theories.

Tâbit ibn Qorra  (= Thâbit ibn Qurra).  c875.  This is the first reference to magic squares in Suter, on pp. 34‑38, but he seems to say that the work has not survived and Ahrens-1 confirms this.  Needham, vol. 5:IV, p. 463, says Cammann wonders if this ever existed.

Ikhw_n al‑Saf_’.  Ras_’il  (Encyclopedia) (??*).  c983.  Cairo edition, 1928, p. 69.  ??NYS.  Paris MS Arabe 2304 (formerly 1005) of this is the work translated by F. Dieterici as:  Die Propaedeutik der Araber im zehnten Jahrhundert; Mittler & Sohn, Berlin, 1865;  reprinted as vol. 3 of F. Dieterici; Die Philosophie bei den Araben im X.Jahrhundert n. Chr.; Olms, Hildesheim, 1969.  Pp. 42‑44 (of the 1969 ed.) shows squares of orders  3, 4, 5 and 6.  The order  3  square is in the form  276.  The text refers to orders  7, 8 and 9  and gives their constants.  On p. 44, the translator notes that the Arabic text has some further incomplete diagrams which are not understandable.  Hermelink and Cammann‑4 say that the Cairo ed. is the only version to give these diagrams.  Ahrens-1 says it continues with a cryptic description of the use of a  9 x 9  square on two sherds, which have not been sprinkled with water, for easing childbirth.  The prescription has several more details than ibn Hayyan's.

              The Arabic text and rough translation are given in:  van der Linde; Geschichte und Literatur des Schachspiels; op. cit. in 5.F.1, vol. 1, p. 203.  This is a description of the  3 x 3  magic square, form  492  or  294,  in terms of chess moves.  Ahrens-1 says that Ruska tells him that much, if not all, of the magic square and adjacent material in Dieterici was added later to the Encyclopedia.  Ruska says there are many errors in the translation and Ahrens cites several further errors in nearby material.

                    Hermelink describes the methods and reconstructs the squares of orders  7, 8, 9  from the 1928 Cairo ed. and van der Linde.  Cammann‑4 says he obtained the same squares independently, but he doesn't agree on all the interpretations.  He feels there are Chinese influences, possibly via India, and gives his interpretations.

                    The square of order  4  is given by Hayashi, op. cit. above at Var_hamihira, as: 

           4, 14, 15,  1;     9,  7,  6, 12;     5, 11, 10,  8;    16,  2,  3, 13.  This is not pandiagonal.  The square of order  7  is doubly bordered _ the first such.

(Abû ‘Alî el‑Hasan ibn el Hasan) ibn el‑Haitam.  c1000.  ??NYS.  Suter, p. 93, says he wrote:  Über die Zahlen des magischen Quadrates.  He cites Woepcke, ??NYS, for MS details.  Ahrens-1 indicates that the work does not exist.

J. H. Rivett‑Carnac.  Magic squares in India.  Notes and Queries (Aug 1917) 383.  Quoted in:  Bull. Amer. Math. Soc. 24 (1917) 106,  which is cited by:  F. Cajori; History of Mathematics; op. cit. in 7.L.1, pp. 92‑93.  The square is in the ruins of a Hindu temple at Dudhai, Jhansi, attributed to the 11C.  It is  4 x 4,  and each  2 x 2  subsquare also adds to  34, but the full square is not given.

                    Cammann‑4, p. 273, says this is the same as the Jaina square at Khajuraho described below and cites the archaeological report, ??NYS.  He is dubious about the date.

(Muhammed ibn Muhammed ibn Muhammed, Abû Hâmid,) el‑_azzâlî  (= al‑Ghazzali).  Mundiqh.  c1100.  ??NYS _ described by Ahrens-1.  Ahrens cites two differing French editions which give  3 x 3  forms  492  and  294.  He says the latter is a transcription error.  Al‑Ghazzali's  text is very similar to  ibn Hayyan's,  though one translator says the cloths are moistened.  Ahrens discusses this point.  He says that amulets with this magic square, called 'seal of Ghazzali' are still available in the Middle East.

                    Lam, p. 318, cites this as an early Arabic magic square, but doesn't give details.  Suter, p. 112, doesn't mention magic squares.

Abraham ibn Ezra.  Sepher Ha‑Schem  (Book of Names), 12C,  and  Jesod Mora,  1158.   ??NYS _ both are described in:  M. Steinschneider; op. cit. in 7.B  and excerpted in the next item.  The material is art. 13, pp. 95++.  The  672  form is shown on p. 98.  Steinschneider, p. 98, also gives the  492  form and says it appears in  Jesod Mora,  described on pp. 99‑101.  Ahrens-1 only mentions that  Sepher Ha‑Schem  gives an order  3  square.

Abraham ibn Ezra.  Sêfer ha‑Echad.  c1150.  Translated and annotated by Ernest Müller as:  Buch der Einheit; Welt-Verlag, Berlin, 1921, with excerpts from:  Jessod Mora,  Sefer ha‑Schem,  Sefer ha-Mispar  and his Bible commentary. 

Sefer ha-Echad,  p. 25, has a reference to areas of squares which Müller thinks may refer to magic squares.

Sefer ha-Schem,  pforte VI, p. 49, discusses the order  3  square.  A note says to see Fig. 6, which appears on p. 80 and is the  492  form.  Müller's notes, p. 64, observe that the magic square of order  3  is essentially unique and makes some mystic comments about this.

Abraham ibn Ezra.  Ta'hbula.  c1150.  ??NYS.  Some source says this has magic squares, but Lévi's comments in 7.B indicate that this book is only concerned with the Josephus problem.  Steinschneider's description of  Tachbula,  pp. 123‑124 of the above cited article, makes no mention of a magic square.

Anon.  Arabic MS, Fatih 3439.  c1150.  ??NYS.  Described in Sesiano‑I.  Construction of squares of almost all orders.  Describes:  a method of  ibn al‑Haytham  (c1000) for odd orders;  a method of  al‑Isfarâ’inî  (c1100) for evenly even orders;  a method of  ibn al‑Haytham  for oddly even squares which only works for order  º 2 (mod 8).  Suter, p. 93, mentions  ibn al‑Haitam  _ see above, c1000.

Tshai Yuan‑Ting.  Lo‑Shu diagram, c1160.  ??NYS _ Biggs cites this as being in Needham, but I can't find it there.

(Ahmed ibn ‘Alî ibn Jûsuf) el‑Bûni, (Abû'l‑‘Abbâs, el‑Qoresî)  = Abu‑l‘Abbas al‑Buni  (??= Muhyi'l‑Dîn Abû’l-‘Abbâs al‑Bûnî  _ can't relocate my source of this form.)  Kitâb et‑chawâss (= Kitab al Khawass  or  Sharkh ismellah el‑a‘zam??)  (The Book of Magic Properties).  c1200.  Suter, p. 136, mentions magic squares.  ??NYS _ described in:  Carra de Vaux; Une solution arabe du problème des carrés magiques; Revue Hist. Sci. 1 (1948) 206‑212.  Construction of squares of all orders by bordering.  Hermelink refers to two other books of  al‑Buni,  ??NYS.

al‑Buni.  Sams al‑ma‘ârif  = Shams al‑ma‘ârif al‑kubrâ  = Šams al-ma‘_rif.  c1200.  ??NYS.  Ahrens-1 describes this briefly and incorrectly.  He expands and corrects this work in Ahrens-2, which mainly deals with  3 x 3  and  4 x 4,  the various sources and the accumulated errors in most of the squares.  He notes that a  4 x 4  can be based on the pattern of two orthogonal Latin squares of order  4,  and  Al-Buni's  work indicates knowledge of such a pattern, exemplified by the square (discussed by Hayashi under Var_hamihira, c550)

           8, 11, 14,  1;    13,  2,  7, 12;     3, 16,  9,  6;    10,  5,  4, 15   considered  (mod 4).  Al‑Buni  gives several  4 x 4's,  including that of  Ikhw_n al‑Saf_’,  c983, which does not have the above pattern.  He also has Latin squares of order  4  using letters from a name of God.  He goes on to show  7  Latin squares of order  7,  using the same  7  letters each time _ though four are corrupted.  (Throughout, the Latin squares also have 'Latin' diagonals.)  These are arranged so each has a different letter in the first place.  It is conjectured that these are associated with the days of the week or the planets.  In Ahrens-1, Ahrens reported that he had recently been told that  Al-Buni  had an association of magic squares of orders  3  through  9  with the planets, but he had not been able to investigate this.  In Ahrens-2, he is clear that  al-Buni  has no such association _ indeed, there is no square of order  9  anywhere in the standard edition of the works of  al-Buni.  See 14C & 15C entries below.

                    Cammann‑4, p. 184, says this text is "deliberately esoteric ... to confuse people" and the larger squares are so garbled as to be incomprehensible.  On pp. 200‑201, he says this has the knight's move method for odd orders.  Later it was noted that any number could be in the centre and  1  was popular, giving the 'unit centred' square of symbolic importance.  These squares are also pandiagonal.  Al‑Buni  gives many variant  4 x 4  squares with the top row spelling some magical word _ e.g. one of the  99  names of God.  He mentions a "method of the Indians", possibly the lozenge method described in Narayana, 1356.

BM Persian MS Add. 7713.  1211?  Described in Cammann‑4, pp. 196ff.  On p. 201, Cammann says p. 23 gives unit centred squares of orders  5  &  9,  pp. 112‑114 gives a rule for singly even order and p. 164 has an order  20  square.  This also has odd order lozenge squares _ see Narayana, 1356.  It also has some examples of a form of the system of broken reversions.

Persian MS.  1212.  Garrett Collection, No. 1057, Princeton Univ.  See  Cammann‑1  &  Cammann‑4, p. 196.  ??NYS

                    Cammann‑4, pp. 196ff, says the above two MSS show new developments and describes them.  Diagonal rules for odd orders first appear here and give an associated square with centre  (n² + 1)/2  which acquired mystic significance as a symbol of Allah.

(Jahjâ ibn Muhammed ibn ‘Abdân ibn ‘Abdelwâhid, Abû Zakarîjâ Ne_m ed‑dîn,)  known as  Ibn el‑Lubûdî  (= Najm al‑Din  (or  Abu Zakariya)  al‑Lubudi.  c1250.  Essay on magic squares dedicated to al‑Mansur.  ??NYS.  Mentioned in Suter, p. 146.

Yang Hui.  Hsü Ku Chai Ch'i Suan Fa  (= Xugu Zhaiqi Suanfa)  (Continuation of Ancient Mathematical Methods for Elucidating the Strange [Properties of Numbers]) (Needham, vol. 5:IV, p. 464, gives: Choice Mathematical Remains collected to preserve the Achievements of Old).  1275.  IN:  Lam Lay Yong;  A Critical Study of the  Yang Hui Suan Fa;  Singapore Univ. Press, 1977.  Book III, chap. 1, Magic Squares, pp. 145‑151 and commentary, pp. 293‑322.  This is the only source for older higher order squares in China.  (See Cammann‑1 and Cammann‑3 for details of constructions.)  Bordered squares of order  5  and  7.  Magic squares of orders  3  through  10, the last being only semimagic.  Methods are given for orders  3  and  4  only.  Gives some magic circles and other forms.  (Lam's commentary, p. 313, corrects the first figure on p. 150.  Lam also discusses the constructions.)

                    Li & Du, pp. 166‑167, say that  6 x 6  and  7 x 7  'central' (= bordered) squares arrived from central Asia at about this time.  The  6 x 6  example on their p. 172 and the  7 x 7  example on their p. 167 are bordered.

              Cammann‑4 says one of the order  8  squares is based on a Hindu construction.

Jaina square.  Inscription at Khajuraho, India.  12‑13C.  Ahrens-1 (p. 218) says it first appears as:  Fr. Schilling; Communication to the Math. Gesellschaft in Göttingen [Mitteilung zur Math. Ges. in Göttingen], 31 May 1904;  reported in:  Jahresber. Deutschen Math.‑Verein. 13 (1904) 383‑384.  (Schilling is reporting a communication from F. Kielhorn.)

           7, 12,  1, 14;    2, 13,  8, 11;   16,  3, 10,  5;    9,  6, 15,  4.

          It is pandiagonal and associated.  Cammann‑4, p. 273, says this is the same as the square reported by Rivett-Carnac and there claimed to be 11C?  Cammann feels it may derive from an Islamic source.  See also:  Smith History II 594  and  Singh, op. cit. at Nâgârjuna above.

Five cast iron plates with  6 x 6  magic squares, late 13C(??), were found at Xian in 1956.  The numerals are similar to East Arabic numerals so these reflect the Arabic influence on the Mongol dynasty.  Li & Du, p. 172, reproduces one.  This is on display in the new Provincial History Museum in Xian.  Jerry Slocum has given me a facsimile, in reduced size, of the same one.  Can any one supply more information about the others??

Μαvoυηλ Μoσχoπoυλoυ  (Manuel Moschopoulos _ variously spelled in Greek and variously transliterated).  c1315.  MS 2428, Bibliothèque Nationale, Paris.

Greek version with discussion in:  Siegmund Günther; Vermischte Untersuchungen zur Geschichte der mathematischen Wissenschaften; Teubner, Leipzig, 1876;  reprinted by Sändig, Wiesbaden, 1968.  Chap. IV: Historische Studien über die magischen Quadrate, pp. 188-276.  Section 5, pp. 195-203 is the Greek.

Greek and French in:  Paul Tannery; Le traité de Manual Moschopoulos sur les carrés magiques; Annuaire de l'Assoc. pour l'Encouragement des Études Grecques en France 20 (1886) 88‑118.  (= Mémoires (??*) Scientifiques, Paris, 1916‑1946, vol. 4, pp. 27?‑61, ??NYS.)  English translation by:  J. C. McCoy; Manuel Moschopolous's treatise on magic squares; SM 8 (1941) 15‑26. 

Gives diagonal rules for odd order and two rules for evenly even order.  These rules are sometimes attributed to Moschopoulos, but see the MSS at 1211? & 1212.  Various examples up through order  9.

‘Abdelwahhâb ibn Ibrâhîm, ‘Izz eddîn el‑Haramî el Zen_ânî  = ‘Abd al‑Wahhâb ibn Ibrâhîm al‑Zinjânî.  Arabic MS,  Feyzullah Ef. 1362.  c1340.  ??NYS _ Described in Sesiano‑II.  Suter, p. 144, doesn't mention magic squares.  Construction of bordered squares of all orders.

N_r_yana Pandita.  Ganita‑kaumud_.  1356.  Edited by P. Dvivedi, Indian Press, Benares, 1942.  Part II: Introduction _ magic squares, pp. xv‑xvi (in English); Chap. 14: Bhadra ganita, esp. pp. 384‑392 (in Sanskrit).  Shows orders  6, 10, 14.  Shows the  8  forms of order  3.  Obviously an extensive section _ is there an English translation of this material??.  (Editor refers to earlier sources:  Bhairava  and  _iva T_ndava  Tantras,  ??NYS.  Cammann‑4 cites other MS sources.  Singh, op. cit. under Nâgârjuna, c1C, above, says this is the first mathematical treatment.  He says it  classifies into odd, evenly even and oddly even;  gives the superposition method of de la Hire;  gives knight's move method for  4n  and filling parallel to diagonal for odd, attributing both to previous authors.

                    Cammann‑4, pp. 274‑290 discusses this in more detail.  He gives another diagonal rule, sometimes beginning and ending at the middle of a side.  He then gives a quite different rule based on use of  x + y  with  x  =  0, n, 2n, ..., (n‑1)n,   y  =  1, 2, ..., n  with both sets of values cycling in the row, then reversing the  xs.  E.g., for  n = 5,  his first row of  y  values is:   4  5  1  2  3   and the second is:   5  1  2  3  4.   His first two rows of  x  values are:   15  20  0  5  10   and   20  0  5  10  15.   Reversing the  xs  and adding gives rows:   14  10  1  22  18   and   20  11  7  3  24.   This process gives a central lozenge (or diamond) pattern of the odds and has an extended knight's move pattern.  He extends this to doubly even squares.  He also gives the 'method of broken reversions' for singly‑even squares in three forms _ cf. C. Planck; The Theory of Reversions, IN:  W. S. Andrews, op. cit. in 4.B.1.a, pp. 295‑320.

Ahrens-1 gives references to further Arabic mentions of magic squares, usually as amulets, notably to  ibn Khaldun  (c1370).  He also gives many 14C and later examples of  3 x 3  and  4 x 4  squares, with rearrangement and/or constants added, used for magical purposes.

Nadrûnî.  Qabs al‑Anwâr.  pre-1384.  ??NYS _ described in Ahrens-1, but not mentioned in Ahrens-2.  Ahrens only knows of this from a modern article in Arabic.  This gives the association of planets with magic squares later given by Pacioli (c1500) and Agrippa (1533).  Cardan later gives the reverse association.

Arabic MS, 1446, ??NYS.  Discussed in Ahrens-1 and Ahrens-2, citing:  W. Ahlwardt; Verzeichniss der arab. Handschr. der Königl. Bibliothek zu Berlin; Berlin, 1891; Vol. III, pp. 505-506 (No. 4115).  This gives the association of planets with magic squares, later given by Cardan in 1539, with the unique addition of a  10 x 10  square for the zodiac coming after Saturn.  Pacioli (c1500) and Agrippa (1533) later give the reverse of this association.

Jagiellonian MS 753.  15C Latin MS in Cracow.  Described in Cammann‑4, pp. 291‑297.  Earliest European set of magic squares of orders  3  through  9  associated with the planets.  The order  4  square is Durer's.  These squares later appear in Paracelsus.

Sûfî Kemal al‑Tustarî.  Ghayat al‑Murâd.  1448.  MS at Columbia.  ??NYS _ cited by Cammann‑4, p. 192.  On p. 196 Cammann says this represents a Persian Sufi tradition which was lost in sectarian warfare and the Mongol invasion.  On p. 201 he says this has a unit‑centred square of order  7.  On pp. 205‑206 are squares of orders  20, 29, 30.  He describes two bordering methods beyond  al‑Buni's.

Hindu square in a temple at Gwalior Fort, 1483.  Cited by Cammann‑4, p. 275, where the original source is cited _ ??NYS.

Pacioli.  De Viribus.  c1500.  Prob. 72: Numeri in quadrato disposti secondo astronomi ch' per ogni verso fanno tanto cioe per lati et per Diametro: figure de pianeti et a molti giuochi acomodabili et pero gli metto.  Gives magic squares of orders  3  through  9  associated with planets in the sequence usually attributed to Agrippa (1533), i.e.  3 - Saturn;  4 ‑ Jupiter;  5 - Mars;  6 - Sun;  7 - Venus;  8 - Mercury;  9 - Moon.

Albrecht Dürer.  Melencholia.  1514.  Two impressions are in the British Museum.  4 x 4  square with  15, 14  in the bottom centre cells.  Surprisingly, this is the same as the  4 x 4  appearing in  Ikhw_n al‑Saf_’  (c983), with the two central columns interchanged and the whole square reflected around a horizontal midline.

Riese.  Rechnung.  1522.  1544 ed. _ pp. 106‑107;  1574 ed. _ pp. 71v‑72v.  Gives  3 x 3  square in  672  form and how to construct other  3 x 3  forms.  Also gives a  4 x 4  square, like Durer's but with inner columns interchanged.

Riese.  Rechenung nach der lenge ....  Op. cit. under Riese, Die Coss.  1525.  ??NYR.  Cammann‑4, p. 294, says pp. 103r‑105v gives a diagonal rule for odd orders.  A quick look shows the material starts on p. 102v.

Cornelius Agrippa von Nettesheim.  De Occulta Philosophia.  Cologne, 1531.  Chap. 22??.  ??NYS.  Gives association of planets with squares _ in the reverse of Cardan's order of 1539 _ as previously done by Pacioli, c1500.  See Cammann‑4, p. 293-294.  The squares do not appear in a 1510 draft of this book.  Bill Kalush has kindly sent Chap. 22 from a 1913 English version, but it doesn't have any squares.  His  4 x 4  is the same as that of  Ikhw_n al‑Saf_’,  c983.

Cardan.  Practica Arithmetice.  1539.

Chap. 42, section 39, ff. H.v.r ‑ H.vi.r (p. 55).  Gives association of planets with squares of orders  3 ‑ 9:  3 ‑ Moon;  4 ‑ Mercury;  5 ‑ Venus;  6 ‑ Sun;  7 ‑ Mars;  8 ‑ Jupiter;  9 ‑ Saturn.  See comments under al‑Buni.

Chap. 66, section 72, ff. FF.v.r - FF.v.v (p. 157).  Shows how to construct a  5 x 5  magic square from the natural  5 x 5  array.

Michael Stifel.  Arithmetica Integra.  Nuremberg, 1544.  ??NYS _ discussed in Cammann‑4, p. 194.  Pp. 25‑26a shows some some bordered squares.  Consequently he is sometimes credited with inventing the concept, but see  Ikhw_n al‑Saf_’  (c983),  al‑Buni  (c1200),  Yang Hui  (1275),  ‘Abdelwahhâb  (c1340),  Sûfî Kemal al‑Tustarî  (1448)  above.

M. Mersenne.  Novarum observationum physico‑mathematicarum.  Paris, 1647.  Vol. 3, chap. 24, p. 211.  ??NYS.  States Frenicle's result. (MUS II #29.)

Isomura Kittoku.  Ketsugi-sh_.  1660, revised in 1684.  ??NYS _ described in Smith & Mikami, pp. 65-77.  He gives magic squares of orders up to order 10.  The order 9 square contains the order 3 square, in the  618  form, in the top middle section.  He gives magic circles with  n  rings of  2n  about a central value of  1,  for n = 2 - 6.  The values are symmetrically arranged, so corresponding pairs add to  2n² + 3  and each ring adds up to  n (2n² + 3),  while each diameter adds to one more than this.  In the 1684 edition, he gives some magic wheels, but these are simply a way of depicting magic squares, though it is not clear where the diagonals are.

Muramatsu Kuday_ Mosei.  Mantoku Jink_‑ri.  1665.  ??NYS _ described in Smith & Mikami, pp. 79-80.  Gives a magic square of order 19.  Gives a magic circle of Isomura's type for  n = 8.  Smith & Mikami, p. 79, gives Muramatsu's diagram with a transcription on p. 80.  The central  1  is omitted and the corresponding pairs no longer add to  131,  but the pairs adding to  131  lie on the same radius.

Bernard Frénicle de Bessy.  Des Quarrez ou Tables Magiques,  including:  Table generale des quarrez de quatre.  Mem. de l'Acad. Roy. des Sc. 5 (1666‑1699) (1729) 209‑354.  (Frénicle died in 1675.  Ollerenshaw & Bondi cite a 1731 edition from The Hague??)  (= Divers Ouvrages de Mathématique et de Physique par Messieurs de l'Académie des Sciences; ed. P. de la Hire; Paris, 1693, pp. 423‑507, ??NYS.  (Rara, 632).  = Recueil de divers Ouvrages de Mathematique de Mr. Frenicle; Arkstèe & Merkus, Amsterdam & Leipzig, 1756, pp. 207-374, ??NX.)

                    Shows there are  880  magic squares of order  4  and lists them all.  Cammann‑4, p. 202, asserts that they can all be derived from one square!!

                    The list of squares has been reprinted in the following. 

M. Gerardin.  Sphinx‑Oedipe _ supplement 4 (Sep‑Oct 1909) 129‑154.  ??NX

K. H. de Haas.  Frenicle's  880  basic Magic Squares of  4 x 4  cells, normalized, indexed, and inventoried (and recounted as  1232).  D. van Sijn & Zonen, Rotterdam, 1935, 23pp.

Seki K_wa.  H_jin Yensan.  MS revised in 1683.  Known also as his  Seven Books.  ??NYS _ described in Smith & Mikami, pp. 116-122.  Describes how to border squares of all sizes.  Gives an easy method for writing down a magic circle of Isomura's type.

Thomas Hyde.  Mandragorias seu Historia Shahiludii, ....  (= Vol. 1 of De Ludis Orientalibus, see  4.B.5 for vol. 2.)  From the Sheldonian Theatre (i.e. OUP), Oxford, 1694.  Prolegomena curiosa.  The initial material and the Prolegomena are unpaged but the folios of the Prolegomena are marked (a), (a 1), ....  The material is on  (d 4).v - (e 1).v, which are pages 32-34 if one starts counting from the beginning of the Prolegomena. 

                    Seems to believe magic squares come from Egypt and gives association of orders of squares with planets as in Pacioli and Agrippa, but he only gives one example of a magic square _ an  8 x 8  which is associated.

Leybourn.  Pleasure with Profit.  1694.  Prob. 10, pp. 4-5.  Gives the  294  form and then says that each line can be rearranged four ways, e.g.  294, 492, 924, 942.  He writes these out for all eith lines, but I can't see any pattern in the way he chooses his four of the six possible permutations.

Ozanam.  1694.  1696: Prob. 9: Des quarrez magiques, 36-41.  Prob. 9: Of magical squares, 1708: 33-36.  Prob. 12: Des quarrez magiques, 1725: 82-102.  Chap. 12: Des quarrés magiques, 1778: 217-244.  Chap. 12: Of magic squares:  1803: 211-240;  1814: 183-207 & 366-367;  1840: 94-105.  Extended discussion, but contains little new _ except some comments on Franklin's squares _ see Ozanam-Hutton (1803).  Associates squares with planets, as done by Pacioli.

Philippe de la Hire.  Sur les quarrés magiques.  Mémoires de l'Académie Royale des Sciences (1705 (1706)) 377‑378.  Gives a method for singly‑even squares, but it uses so many transpositions that it is hard to see if it works in general.  ??NYS _ described in Cammann‑4, p. 286.

Muhammed ibn Muhammmed.  A Treatise on the Magical Use of the Letters of the Alphabet.  Arabic MS of 1732,  described and partly reproduced in:  Claudia Zaslavsky; Africa Counts; Prindle, Weber & Schmidt, Boston, 1973; chap. 12, pp. 137‑151.  Several of his magic squares are deliberately defective, presumably because of the Islamic belief that only God can create something perfect.  I do not recall any other mention of this feature.

Minguét.  Engaños.  1733.  Pp. 169-172 (1755: 122-123;  not noticed in 1822, but probably about p. 180.)  Magic squares of order three with various sums, made by laying out cards.

Dilworth.  Schoolmaster's Assistant.  1743.  Part IV: Questions: A short Collection of pleasant and diverting Questions, p. 168.  Problem 4.  Asks for a  3 x 3  magic square.

Benjamin Franklin.  c1750.  Discovery of some large magic squares and circles.  He described these in letters to Peter Collinson whose originals do not survive, so their dates are uncertain.  They were first published in the following.

                    James Ferguson.  Tables and Tracts, Relative to Several Arts and Sciences.  A. Millar & T. Cadell, London, 1767.  Pp. 309-317.  ??NYS.  Ferguson may be indicating that he is the first person to whom Franklin showed them. 

                    B. Franklin.  Experiments and Observations on Electricity.  4th ed., London, 1769.  Two letters to Peter Collinson, pp. 350-355(??).  ??NYS, but reprinted in:  Albert Henry Smyth; The Writings of Benjamin Franklin; Vol. II, Macmillan, 1907, pp. 456-461 and Plates VII (opp. p. 458) and VIII (opp. p. 460).

                    The squares are of order  8  and  16,  but are only semi-magic (see Ozanam-Hutton (1803) and Patel (1991)), and the circle has  8  rings and  8  radii.  Franklin said he could make these squares as fast as he could write down the numbers!

Caietanus Gilardonus.  9 x 9  square on a marble plaque on the Villa Albani, near Rome, dated 1766.  The square and the accompanying inscription are given in:  E. V. R.; Arranged squares; Knowledge 1 (27 Jan 1882) 273, item 231.  These are also given by  Catalan; Mathesis 1, p. 151 (??NYS)  and  Lucas; L'Arithmétique Amusante; 1895; pp. 224-225.  Lucas says they were discovered in 1881, and that the villa is now owned by Prince Torlonia and is outside the Porta Salaria.

Catel.  Kunst-Cabinet.  1790.

Das grosse Zauberquadrat, p. 16 & fig. 33 on plate II.  49  numbered pieces to make into a magic square.

Das kleine Zauberquadrat, p. 16 & fig. 34 on plate II.  3 x 3.  Diagram is only semimagic.

Bestelmeier.  1801.

Item 441: Das arithmetische Zauber‑Quadrat.  9  numbered pieces to into a magic square.  Diagram is shown disarranged.

Item 961.a: Das Sonderbarste aller magischen Zahlen=Quadrate.  64  numbers to form into a magic square, with various groups of four to add up to half the magic constant.

Item 961.b: Gewöhnliche Zauberquadraten von  64, 48, 36, 25, 16, 9  Zahlen.

Ozanam-Hutton.  1803.  Chap. 12: Remarks, 1803: 237-240 & fig. 1, plate 4.  1814: 203-207 & corrections: 366-367 & fig. 1, plate 4 & additional plate 5.  1840: 104-105, with no figure.  The 1814 corrections note that Franklin's square is only semi-magic and gives another large example _ 1840 omits this.

Unger.  Arithmetische Unterhaltungen.  1838.  Pp. 238-243, no. 907-926.  This gives lots of straightforward exercises _ e.g. find a  5 x 5  magic square with sum  96,  which he does by adding  6 1/5  to each entry of a normal example.

Young Man's Book.  1839.  P. 232.  The Magical Square.  "The Chinese have discovered mystical letters on the back of the tortoise, which is the common magical square, making each way 15, viz."  Gives the  294  form, while all Chinese forms have  492.  Pp. 236-238 is a straightforward section on Magic Squares.

Indoor & Outdoor.  c1859.  Part II, prob. 13: Franklin's magic square, pp. 132-133.  Gives Franklin's  16 x 16  square and states some of its properties, very similar to Ozanam‑Hutton.

Magician's Own Book (UK version).  1871.  The magic square oraculum, pp. 94-98.  This shows a square of order 11 and says it is "a magic square of eleven, with one in the centre", but it is not at all magic.  Initially it appears to be bordered, but it is a kind of arithmetical square.  1  is in the middle.  Then  2 - 9  are wrapped around the central square, going clockwise with the  2  above the  1.  Then  10 - 25  are wrapped around the central  3 x 3  area, with the  10  above the  2,  etc.  The 'oracle' consists of thinking of a number and consulting a list of fortunes, so the 'magic square' is never used!

T. Squire Barrett.  The magic square of four.  Knowledge 14 (Mar 1891) 45-47  &  Letter (Apr 1891) 71  &  Letter (Aug 1891) 156.  Says he hasn't seen Frenicle's list.  Classifies the  4 x 4  squares into  12  types, and obtains  880  squares, but doesn't guarantee to have found all of them.  The first letter notes that he erred in counting one type, getting  48  too many, but a friend has found  16  more.  Second letter notes that the missing squares have been found by another correspondent who has compared them with Frenicle's list and then Barrett corrects some mistakes.  The other correspondent notes that Frenicle had proceeded by a trial and error method and probably had found all examples.

Ball.  MRE, 1st ed., 1892, pp. 108-121: Chap. V: Magic squares.  Later editions amplify this material, but the material is too detailed and too repetitive to appeal to me at the moment.

Hoffmann.  1893.  Chap. IV, no. 8: The "thirty‑four" puzzle, pp. 146 & 183‑186.  Asserts that Heywood of Manchester publish a booklet, 'The Curiosities of the Thirty‑four Puzzle' which has instructions for obtaining all the solutions (i.e. of the  4 x 4  magic square), ??NYS.  Hordern has a photo of a version by Perry.

Stewart Culin.  Chinese Games with Dice and Dominoes.  From the Report of the U. S. National Museum for 1893, pp. 489‑537.  On pp. 536-537, he discusses the Lok Shü, citing Legge.  He reports that the  618  version is popular as a charm with both Hindus and Moslems in India, while the Chinese  492  form is used in Tibet. 

Dudeney.  A batch of puzzles.  Royal Magazine 1:3 (Jan 1899)  &  1:4 (Feb 1899) 368-372.  The eight clowns puzzle.  = CP; 1907; prob. 81, pp. 128 & 126.  3 x 3  array with pieces  x 2 3;  4 5 6;  7 8 9  to be rearranged into a magic square, the blank being counted as a  0.  Answer is that this is impossible, but the clown marked  9  is juggling balls which make his number  .9,  i.e.  .9  recurring,  which is  1!

Dudeney.  The magic square of sixteen.  The Queen (15 Jan 1910) 125-126.  Good derivation of the  880  squares of order  4,  classified into  12  types.  A condensed version with some extra information is in AM, pp. 119-121.

Loyd.  Cyclopedia.  1914.  The 14‑15 puzzle in puzzleland, pp. 235 & 371.  = MPSL1, prob. 21, pp. 19‑20 & 128.  c= SLAHP: The "14‑15" magic square, pp. 17‑18 & 89.  Given the  15  Puzzle with the  14  and  15  interchanged, move to a magic square.  The blank counts zero, so the magic constant is  30.

Collins.  Book of Puzzles.  1927.  Magic squares and other figures, pp. 79-94.  Brief survey.  Gives a number of variant forms.

D. N. Lehmer.  A complete census of  4 x 4  magic squares.  Bull. Amer. Math. Soc. 39 (1933) 764-767.  Here he is dealing with semi-magic squares and then any permutation of the rows or columns or transposition of the whole array preserves the row and column sums.  Hence there are  2(n!)²  arrays in each equivalence class and he describes a normalized form for each class.  For the  3 x 3  case, there are  72  semimagic squares and one normalized form.  For the  4 x 4  case, there are  468  normalized forms and hence  468 x 2 x 24²  =  539,136   semimagic squares.

D. N. Lehmer.  A census of squares of order  4,  magic in rows, columns, and diagonals.  Bull. Amer. Math. Soc. 39 (1933) 981-982.  Here he discusses Frenicle's enumeration of  880  4 x 4  magic squares and points out that there are additional equivalences beyond the symmetries of the square so that Frenicle only needed to find and list  220  squares.  Using ideas like those in his previous article, he finds  220  such squares, confirming Frenicle's result once again.

J. Travers.  Rules for bordered magic squares.  MG 23 (No. 256) (Oct 1939) 349‑351.  Cites Rouse Ball, MRE (no details), as saying no such rules are known.  He believes these are the first published rules.

Anonymous.  A Book of Fun with Games and Puzzles.  One of a set of three 12pp booklets, no details, [1940s?].  P. 7: Here is the magic square.  Gives a  4 x 4  square using the numbers  3, ..., 17  and asks for it to be dissected along the lines into four pieces which can be rearranged into a magic square.

D. H. Hallowes.  On  4 x 4  pan‑magic squares.  MG 30 (No. 290) (Jul 1946) 153‑154.  Shows there are only  3  essentially different  4 x 4  pan‑magic squares.

Gardner.  SA (Jan 1976)  c= Time Travel, chap. 17.  Richard Schroeppel, of Information International, used a PDP‑10 to find  275,305,224  magic squares of order  5,  inequivalent under the  8  symmetries of the square.  If one also considers the 'eversion' symmetries, there are  32  symmetries and  68,826,306  inequivalent squares.  (Gardner says there is an Oct 1975 report on this work by Michael Beeler, ??NYS, and gives Schroeppel's address:  835 Ashland Ave., Santa Monica, Calif., 90405 _ I believe I wrote, but had no reply??)

K. Ollerenshaw & H. Bondi.  Magic squares of order four.  Phil. Trans. Roy. Soc. Lond. A306 (1982) 443‑532.  (Also available separately.)  Gives a new approach to Frénicle's results.  Relates to Magic Card Squares and the Fifteen Puzzle.

Lee C. F. Sallows.  Alphamagic squares:  I  &  II.  Abacus 4:1 (Fall 1986) 28‑45  &  4:2 (Winter 1987) 20-29 & 43.  Reprinted in: The Lighter Side of Mathematics; ed. by R. K. Guy & R. E. Woodrow; MAA, 1994, pp. 305-339.  Introduces notion of alphamagic square _ a magic square such that the numbers of letters in the words for the numbers also form a magic square.  Simplest example is:   5, 22, 18;   28, 15,  2;   12,  8, 25.  He asserts that this appears in runes in an 1888 book describing a 5C charm revealed to King Mi of North Britain.  (This seems a bit farfetched to me or mi?)  Asks if there can be an alphamagic square using the first  n²  numbers and shows that  n ³ 14.  Notes some interesting results on formulae for  4 x 4  squares, including one with minimum number of symbols.  There was a letter and response in Abacus 4:3 (Spring 1987) 67-69.

Lalbhai D. Patel.  The secret of Franklin's  8 x 8  'magic' square.  JRM 23:3 (1991) 175-182.  Develops a method to make Franklin's squares as fast as one can write down the numbers!

Jacques Bouteloup.  Carrés Magiques, Carrés Latins et Eulériens.  Éditions du Choix, Bréançon, 1991.  Nice systematic survey of this field, analysing many classic methods.

Lee Sallows.  Alphamagic squares.  CFF 35 (Dec 1994) 6-10.  "... a brief synopsis of [his above article] which handles the topic in very much greater detail."

Martin Gardner.  The magic of  3 x 3.  Quantum 6:3 (Jan-Feb, 1996) 24-26.  Says that modern scholars doubt if the pattern in China is older than 10C!  In 1987, he offered $100 for the first  3 x 3  magic square using consecutive primes.  Harry L. Nelson won, using a Cray at Lawrence Livermore Laboratories.  The smallest value is  1,480,028,129  and the values all have the same first seven digits and their last three digits are:  129,  141,  153,  159,  171,  183,  189,  201,  213.  Says Martin LaBar [CMJ (Jan 1984) 69] asked for a  3 x 3  magic square whose entries are all possitive squares.  Gardner reiterates this and offers $100 for the first example.  He gives examples of  3 x 3  squares with various properties and the lowest magic sum, e.g. using primes in arithmetic progression.

Lee Sallows.  The lost theorem.  Math. Intell. 19:4 (1997) 51-54.  Gives an almost solution to Gardener's problem, but one diagonal fails to add up correctly.  Gives an example of Michael Schweitzer which is magic but contains only six squares.  Using Lucas' pattern, where the central number,  c,  is one third of the magic sum and two adjacent corners are  c + a  and  c + b,  he observes that these can be vectors in the plane or complex numbers, which allows one to correspond classes of eight magic squares with parallelograms in the plane.  This leads to perhaps the most elegant magic square by taking  c = 0,  a = 1,  b = i.

 

          7.N.1  MAGIC CUBES

 

          Note.  Historically, a  k³  has been called magic when all the  3k²  lines parallel to the axes and the  4  space diagonals have the same sum.  But there are also  6k  2‑dimensional diagonals _ if these also have the same sum, we will say that the cube is perfectly magic.  Pandiagonal (= pan‑n‑agonal) refers just to the space diagonals.  Perfectly pandiagonal refers to all the diagonals.  In higher space, the simpler words refer to the  2^n-1  'space' diagonals and perfect will include all the diagonals in intermediate dimensions. 

          A  k-agonal is a line which varies in  k  coordinates, so a  1-agonal is a row or column, etc.,  the  2-agonals of a cube include the diagonals of the faces,  while the  3-agonals of a  3‑cube are the space diagonals.

          Associated  or  complementary  means that two cells symmetric with respect to the centre add to  kⁿ + 1.

 

Pierre de Fermat.  Letter to Mersenne (1 Apr 1640).  Oeuvres de Fermat.  Ed. by P. Tannery & C. Henry.  Vol. 2, Gauthier‑Villars, Paris, 1894, pp. 186‑194.  Gives a magic  4³.

                    A shorter, undated, version, occurs in Varia Opera Mathematica D. Petri de Fermat, Toulouse, 1679;  reprinted by Culture et Civilization, Brussels, 1969; pp. 173‑176.  The version in the Oeuvres has had its orthography modernized.

               On p. 174 of the Varia (= p. 190 of the Oeuvres), he says:  "j'ay trouvé une regle generale pour ranger tous les coubes à l'infiny, en telle façon que toutes les lignes de leurs quarrez tant diagonales, de largeur, de longeur, que de hauteur, fassent un méme nombre, & determiner outre cela en combien de façons differentes chaque cube doit étre rangé, ce qui est, ce me semble, une des plus belles choses de l'Arihmetique [sic]..."  He describes a assembly of four squares making a magic cube.  [The squares are missing in the Varia.]  He says that the magic sum occurs on  72  lines, but it fails to have the magic sum on  8  of the  2‑agonals and all  4  of the  3‑agonals.

Lucas.  Letter.  Mathesis 2 (1882) 243‑245.  First publication of the magic  4³  described by Fermat above.  Says it will appear in the Oeuvres.

E. Fourrey.  Op. cit. in 4.A.1.  1899.  Section 317, p. 257.  Notes that Fermat's magic cube has only  64  magic lines.

Lucas.  L'Arithmétique Amusante.  1895.  Note IV:  Section IV: Cube magique de Fermat, pp. 225‑229.  Reproduces the  4³  from his Mathesis letter and gives a generalization by V. Coccoz, for which the same diagonals fail to have the magic sum, though he implies they do have the sum on p. 229.

 

Pierre de Fermat.  Letter to Mersenne, nd [Jun? 1641].  Opp. cit. above:  Oeuvres, vol. 2, pp. 195‑199;  Varia Opera, pp. 176‑178.  On p. 177 of the Varia (= p. 197 of the Oeuvres), he says:  "Pource qui est des cubes, je n'en sçay pas plus que Monsieur Frenicle, mais pourtant je puis les ranger tous à la charge que les Diagonales seules de quarrez que nous pouvons supposer paralleles à l'Horizon, seront égales aux côtez des quarrez, ce qui n'est pas peu de chose.  En attendant qu'une plus longue meditation découvre le reste, je dresseray celuy de  8.  10.  ou  12.  à ces conditions si Monsieur de Frenicle me l'ordonne."

Joseph Sauveur.  Construction générale des quarrés magiques.  Mémoires de l'Académie Royale des Sciences (1710 (1711)) 92‑138.  ??NYS _ mentioned by Brooke (below), who says Sauvier [sic] presented the first magic cube but gives no reference.  Discussed by Cammann‑4, p. 297, who says Sauveur invented magic cubes and Latin squares.  This paper contains at least the latter and an improvement on de la Hire's method for magic squares, but Cammann doesn't indicate if this contains the magic cube.

Charles Babbage.  Notebooks _ unpublished collection of MSS in the BM as Add. MS 37205.  ??NX.  See 4.B.1 for more details.  F. 308: Essay towards forming a Magick Cube, c1840??  Very brief notes.

Gustavus Frankenstein.  [No title].  Commercial (a daily paper in Cincinnati, Ohio) (11 Mar 1875).  ??NYS.  Perfect  8³.  Described by Barnard, pp. 244‑248.

Hermann Scheffler.  Die magischen Figuren.  Teubner, Leipzig, 1882;  reprinted by Sändig, Wiesbaden, 1981.  Part III: Die magische Würfel, pp. 88‑101 & plates I & II, pp. 113 & 115.  He wants all  2‑  and  3‑agonals to add up to the magic constant, though he doesn't manage to construct any examples.  He gives a magic  5³  which has the magic sum on  14  of the  30  2‑agonals and many of the broken  2‑agonals.  He also gives a  4³  and a  5³,  but I haven't checked how successful they are.

F. A. P. Barnard.  Theory of magic squares and of magic cubes.  Memoirs of the National Academy of Science 4 (1888) 209‑270.  ??NYS. Excerpted, including the long footnote description of Frankenstein's  8³,  in Benson & Jacoby (below), pp. 32‑37, with diagrams of the result on pp. 37‑42.

C. Planck, on pp. 298 & 304 of Andrews, op. cit. in 4.B.1.a, says the first magic  6³  was found by W. Firth of Emmanuel College, Cambridge in 1889.

Pao Chi-shou.  Pi Nai Sahn Fang Chi  (Pi Nai Mountain Hut Records).  Late 19C.  ??NYS _ described by Lam (in op. cit. in 7.N under  Yang Hui),  pp. 321-322, who says it has magic cubes, spheres and tetrahedrons.  See also Needham, p. 60.

V. Schlegel.  ??  Bull. Soc. Math. France (1892) 97.  ??NYS.  First magic  3⁴.  Described by Brooke (below).

Berkeley & Rowland.  Card Tricks and Puzzles.  1892.  Magic cubes, pp. 99-100.  3³  magic cube with also the  6  2-dimensional diagonals through the centre having the same sum, so there are  37  lines with the magic sum.  4³  magic cube _ this has the  3 x 16 + 4 = 52  expected magic lines, but he asserts it has  68  magic lines, though I can only find  52.  The perfect case would have  76  magic lines.

C. Planck.  Theory of Path Nasiks.  Privately printed, Rugby, 1905.  ??NYS.  (Planck cites this on p. 363 of Andrews, op. cit. in 4.B.1.a, and says there are copies at BM, Bodleian and Cambridge.)  The smallest Nasik (= perfectly pandiagonal)  kⁿ  has  k = 2ⁿ.  If the cube is also associated, then  k = 2ⁿ + 1.  He quotes these results on p. 366 of Andrews and cites earlier erroneous results.  On p. 370 of Andrews, he says that a perfect  k⁴  has  k ³ 8.

Collins.  Book of Puzzles.  1927.  A magic cube, pp. 89-90.  3³,  different than that in Berkeley & Rowland, but with the same properties.

J. Barkley Rosser & Robert J. Walker.  MS deposited at Cornell Univ., late 1930s.  ??NYS.  (Cited by Gardner, loc. cit. below, and Ball, MRE, 11th ed., p. 220;  12th ed., p. 219.)  Finds a Nasik  8³  and shows that Nasik  k³  exist precisely for the multiples of  8  and for odd  k > 8.

G. L. Watson.  Note 2100:  To construct a symmetrical, pandiagonal magic cube of oddly even order  2n ³ 10.  MG 33 (No. 306) (Dec 1949) 299‑300.

Maxey Brooke.  How to make a magic tessarack.  RMM 5 (Oct 1961) 40‑44.  Cites Sauvier and Schlegel.  Believes this is the first English exposition of Schlegel.  The  3³  he develops is magic, but only the  6  2‑agonals through the centre have the magic sum.  The resulting  3⁴  is magic but not perfect.

Harry Langman.  Play Mathematics.  Hafner, 1962.  ??NYS _ cited by Gardner below.  Pp. 75‑76 gives the earliest known perfect  7³.

Birtwistle.  Math. Puzzles & Perplexities.  1971.  The magic cube, pp. 157 & 199.  3³  which is associated, but just the  6  2-agonals through the centre have the magic sum; the other  12  2-agonals do not.  The space diagonals also have the magic sum.

John Robert Hendricks.  The third‑order magic cube complete.  JRM 5:1 (1972) 43‑50.  Shows there are  4  magic  3³,  inequivalent under the  48  symmetries of the cube.  None of these is perfect.  (The author has published many articles on magic cubes in JRM but few seem appropriate to note here.)

Gardner.  1976.  Op. cit. in 7.N.  Gives Richard Lewis Myers Jr.'s proof that a perfect  3³  does not exist, and Richard Schroeppel's 1972 proof that a perfect  4³  does not exist.  (Gardner says Schroeppel published a memorandum on this, ??NYS.)  Says that perfect cubes of edge  5, 6, 7  are unknown and gives a perfect, associated  8³  found by Myers in 1970.  The Addendum in Time Travel cites Planck and Rosser & Walker for earlier  8³  and says that many readers found a perfect  7³  and refers to Langman.  Also  9³,  11³  and higher orders were found.

Johannes Lehmann.  Kurzweil durch Mathe.  Urania Verlag, Leipzig, 1980.  No. 8, pp. 61 & 160.  Arrange  0 - 15  on the vertices of a  2⁴  hypercube so that each  2-dimensional face has the sum  30.

John R. Hendricks.  The perfect magic cube of order 4.  JRM 13 (1980‑81) 204‑206.  Shows it does not exist.

William H. Benson & Oswald Jacoby.  Magic Cubes _ New Recreations.  Dover, 1981.  Summarises all past results on p. 5.  There are perfect  n³  for  n ¹ 6 (mod 12),  n ³ 7,  except  n = 10.  It is not clear if they have proofs for  n º 4 (mod 8)  or  n º 2 (mod 4).  They are unable to show the non‑existence for  n º 5  (cf. p. 29).  There are pandiagonal  n³  for  n ¹ 6 (mod 12),  n ³ 4,  though it is not clear if they have a proof for  n º 2 (mod 4)  (cf. p. 102).

Rudolf Ondrejka.  Letter:  The most perfect (8 x 8 x 8) magic cube?  JRM 20:3 (1988) 207‑209.  Says Benson & Jacoby sketch a perfectly pandiagonal  8³.  He gives it in detail and discusses it.

Joseph Arkin, David C. Arney & Bruce J. Porter.  A perfect  4‑dimensional hypercube of order 7 _ "The Cameron Cube".  JRM 21:2 (1989) 81‑88.  This is the smallest known order in four dimensions.

Allan William Johnson Jr.  Letter:  Normal magic cubes of order  4M+2.  JRM 21:2 (1989) 101‑103.  Refers to Firth, 1889, who is mentioned by Planck in Andrews.  Gives a program to compute  (4M+2)³  cubes and gives a  6³  in base 10  &  base 6.

John Robert Hendricks.  The magic tessaracts of order  3  complete.  JRM 22:1 (1990) 15‑26.  Says there are  58  of them and gives some history.

Jacques Bouteloup.  Carrés Magiques, Carrés Latins et Eulériens.  Éditions du Choix, Bréançon, 1991.  Nice systematic survey of this field, analysing many classic methods.  Includes some material on magic cubes.

 

          7.N.2. MAGIC TRIANGLES

 

          There are quite a number of possible types here and I have not been very systematic in recording them.

 

Frenicle de Bessy.  Letter to Mersenne, Mar 1640.  In:  Oeuvres de Fermat, op. cit. in 7.N.1, vol. 2, pp. 182‑185.  Discusses a magic triangle.

Scheffler.  Op. cit. in 7.N.1.  1882.  Part II: Das magische Polygon, pp. 47‑88 & Plate I, p. 113.  He considers nested  n‑gons with the number of numbers on each edge increasing  1, 3, 5, ...  or  2, 4, 6, ...,  such that each edge of length  k > 2  has the same sum and the diameters all have the same sum, though these sums are not all the same.  He develops various techniques and gives examples up to  26‑gons and  5‑level pentagons.

H. F. L. Meyer.  Magic Triangles.  In:  Adams; Indoor Games; 1912; pp. 357‑362.  He divides a triangle by lines so the triangle of order three has rows of  1, 3, 5  cells.  He gets some lines of  2  and of  3  to add to the same value, and then considers hexagons of six cells, but doesn't really get anywhere.

Peano.  Giochi.  1924.

Prob. 3, p. 2.  Gives triangle with sides  2, 5, 4;   4, 1, 6;   6, 3, 2.

Prob. 4, p. 2.  Gives triangle with sides  8, 1, 6, 5;   5, 4, 9, 2;   2, 3, 7, 8.  Gives a number of simple consequences of the magicness and also that the sum of the squares of the numbers on a side is  126.  [There are  18  magic triangles, but only this one has the sum of the squares constant.]

Collins.  Book of Puzzles.  1927.

Pp. 92-93: The magic triangle.  Consider a triangle with two points between the vertices.  Put the numbers  1 - 9  on the vertices and intermediate points so that the sum of the values on each edge is constant and the sum of the squares of the values is constant.  Gives one answer and is somewhat vague as to whether it is unique.  See Peano.

P. 93: A nest of magic triangles.  Says this occurs on a 1717 document of the Mathematical Society of Spitalfields.  Start with a triangle and join up its midpoints.  Repeat on the resulting triangle and continue to the fourth tie, getting five triangles with 18 points.  The values  1 - 18  are placed on these points to get various sums which are multiples of 19.

Collins.  Fun with Figures.  1928. 

The Siamese twin triangles, pp. 108-109.  Triangle with  4  cells along each side.  Place the digits  1  through  9  on the cells so each line adds to  20.  Two complementary solutions, with sums  19  &  21.  He gives a number of further properties about various sums of squares.

A magic hexagon within a circle, pp. 110-112.  This is really a pattern of six magic triangles like the above, with sums  20  and with sums all distinct, with a further property about sums of squares.

Perelman.  FFF.  1934.  1957: probs. 46-48, pp. 56-57 & 61-62;  1979: probs. 49-51, pp. 70‑71 & 77-78.  = MCBF, probs. 49-51, pp. 69-70 & 74-75.

49: A number triangle.  Triangle with  4  cells along each side.  Place the digits  1  through  9  on the cells so each line adds to  20 _ as in Collins, pp. 108-109.  One solution, but he notes that the two central cells in each line can be interchanged.

50: Another number triangle.  Same with total  17,  again one solution given.

51: A magic star.  Star of David with cells at the star points and the intersections.  Place numbers  1, .., 12  so each line of four and the six points all add to  26.  One solution given.

Jaime Poniachik, proposer;  Henry Ibstedt, solver.  Prob. 1776 _ Connected differences.  JRM 22:1 (1990) 67  &  23:1 (1991) 74-75.  Triangular lattice with edge  2.  Place numbers  1, ..., 15  on the  6  points and  9  edges so that each edge is the difference of its end points.  19  solutions found by computer.  [Not sure where to put this item??]

 

          7.N.3. ANTI‑MAGIC SQUARES AND TRIANGLES

 

          An antimagic  n x n  has its  2n+2  sums all distinct.  A consecutively antimagic  n x n  has its  2n+2  sums forming a set of  2n+2  consecutive integers.  Berloquin calls these heterogeneous and antimagic, respectively.  A heterosquare has all the  4n  sums along rows, columns and all broken diagonals being different.

 

Loyd Jr.  SLAHP.  1928.  Magic square reversed, pp. 44 & 100.  3  2  7;   8  5  9;   4  6  1  has all eight sums different.

Dewey Duncan.  ??  MM (Jan 1951) ??NYS _ cited by Gardner.  Defines a heterosquare as an arrangement of  1, 2, ..., n²  such that the  4n  sums along the rows, columns and all broken diagonals are all different.  (However, in the next item it appears that only the main diagonals are being considered??)  Asks for a proof that the  2 x 2  case is impossible and for a  3 x 3  example _ which turns out to be impossible.

John Lee Fults.  Magic Squares.  Open Court, La Salle, Illinois, 1974.  On p. 78, he asserts that Charles W. Trigg posed the problem of non‑existence of anti‑magic  2 x 2's  in 1951, that it was solved by Royal Heath and that Trigg gave the spiral construction for anti‑magics of odd order.  Unfortunately Fults gives no source, only noting that Trigg was editor at the time.  There is nothing in Heath's MatheMagic.  I now see that this is a corruption of the preceding item.  Madachy (see below at Lindon) refers to the problem in MM (1951) without further details, but restricted to just the main diagonals, and then says "An exchange of correspondence between Charles W. Trigg, then "Problems and Questions" editor for the magazine, and the late Royal V. Heath, ..., soon established some basic properties of potential heterosquares."

C. W. Trigg, proposer;  D. C. B. Marsh, solver.  Prob. E1116 _ Concerning pandiagonal heterosquares.  AMM 61 (1954) 343  &  62 (Jan 1955) 42.  The solution is also in:  Trigg; op. cit. in 5.Q; Quickie 160: Pandiagonal heterosquare, pp. 45 & 151.  There is no arrangement of  1, 2, ..., n²  such that the  4n  sums along the rows, columns and all broken diagonals are consecutive numbers.

Charles F. Pinska.  ??  MM (Sep/Oct 1965) 250-252.  ??NYS _ cited by Gardner.  Shows there are no  3 x 3  heterosquares, but gives two  4 x 4  examples.

Gardner.  SA (Jan 1961) c= Magic Numbers, chap. 2.  Notes that  1  2  3;   8  9  4;   7  6  5  is anti‑magic, i.e. all  8  sums are different, and it is also a rook's tour.  In Magic Numbers, Gardner says he had not known of anti-magic squares before seeing this one, but later discovered the Loyd example.  He summarises the knowledge up to 1971.

J. A. Lindon.  Anti‑magic squares.  RMM 7 (Feb 1962) 16‑19.  Summarised and extended in Madachy; Mathematics on Vacation, op. cit. in 6.D, 1966, pp. 101-110.  Author and editor believe this is the first such article.  Wants the  2n+2  sums for an  n x n  square to be all different and also to be a set of consecutive integers.  There are no such for  n = 1, 2, 3,  but they do exist for  n > 3.

M. Gardner.  Letter.  RMM 8 (Apr 1962) 45.  Points out his SA article and notes that  9  8  7,   2  1  6,   3  4  5  is even more anti‑magic in that the  8  lines and the  4  2 x 2  subsquares all have different sums.

Pierre Berloquin.  The Garden of the Sphinx, op. cit. in 5.N.  1981. 

Prob. 15: Heterogeneous squares, pp. 11 & 93.  Shows there is no antimagic square of order  2  and reports that a reader found  3120  inequivalent antimagic squares of order  3.  (The  8  symmetries of the square are the equivalences.)

Prob. 16: Antimagic, pp. 12 & 94.  Asserts there is no  3 x 3  consecutive antimagic square.  [Translation is unclear whether he is saying none is known or none exists.]  Gives examples of  4 x 4  and  5 x 5  consecutive antimagics and says there are  20  known examples for  4 x 4.

Prob. 148: Superior antimagic, pp. 81 & 184.  Gives a  6 x 6  consecutive antimagic square.

C. W. Trigg.  Special antimagic triangular arrays.  JRM 14 (1981‑82) 274‑278.  Says Gardner gives the following definition in a letter to Trigg on 22 Dec 1980.  Consider a triangular array of the numbers  1, 2, ..., n(n+1)/2.  We say this is anti‑magic if the sum of the three vertices and the  3(n‑1)  sums of rows of two or more, parallel to the sides, are all distinct.  Gardner then asks if these  3n‑2  sums can be consecutive.  Trigg asks when these sums can be in arithmetic progression and finds that this requires  n = 2, 3, 4, 10, 24  or  n > 99.  He resolves the existence problems for  n = 2, 3, 4.

M. Gardner.  Puzzles from Other Worlds.  Vintage (Random House), NY, 1984.  Problem 8: Antimagic at the number wall, pp. 19‑20, 96‑97 & 142‑143.  Notes his examples (which are complementary) are the only rook‑wise connected anti‑magic squares of order  3  and discusses anti‑magic triangles.

 

          7.N.4. MAGIC KNIGHT'S TOUR

 

          Note: for the  8 x 8,  the magic constant is  260.

 

G. P. Jellis.  Special Issue _ Magic Tours; Chessics 26 (Summer 1986) 113‑128  &  Notes on Chessics 26 (Magic Tours); Chessics 29 & 30 (1987) 163.  Says Kraitchik (L'Echiquier, 1926), ??NYS, showed  there is no magic tour unless both sides are even.  (Mentioned in his Math. des Jeux, op. cit. in 4.A.2, p. 388.)  Jelliss considers tours by other pieces including various generalized chess pieces.  He gives  8 x 8  magic king's and queen's tours.  Gives  97  inequivalent semi‑magic knight's paths of which  29  are tours.  These are derived from MSS of H. J. R. Murray at the Bodleian Library, Oxford.  He gives Beverley's square _ see below.  He says magic tours (paths?) exist on  8k x 8k  boards for  k ³ 2  and gives a  48 x 48  magic tour.  He attributes the method to Murray (Fairy Chess Review, Aug 1942).

                    The Notes report that one of the magic queen's tours was miscopied and that Tom Marlow has found that one of the  8 x 8  semi‑magic knight's tours is wrong.

 

William Beverley.  On the magic square of the knight's march.  (Letter of 5 Jun 1847.)  (The London, Dublin and Edinburgh) Philosophical Magazine & Journal (of Science) ?? (Aug 1848) 1-5.  Semimagic knight's path with diagonals of  210  and  282.  See:  G&PJ 1 (Sep 1987) 11  &  2 (Nov 1987) 17,  which describe what is known about the problem.  Tom Marlow reports that he has found that there are  101  examples, but he doesn't seem to consider the diagonals, so these are semimagic.  See also:  Murray, 1936, below.

C. F. de Jaenisch.  Op. cit. in 5.F.1.  1862.  Vol. 2, pp. 151-189.  ??NYS.  Semi‑magic knight's tour, with diagonal sums of  256  and  264.  [Given in Dickins, p. 27 _ which Dickins??.]

Berkeley & Rowland.  Card Tricks and Puzzles.  1892.  Magic squares in chess _ The knight's tour, pp. 101-102.  Semi-magic knight's tour with diagonals of  264  &  256,  taken from a BM MS: Bibl. Reg. 13, A. xviii., Plut. xx. c. _ ??NYS, but same as de Jaenisch's example.  Notes that symmetric cells differ by  32.  Gives a cryptic argument that this solution can be used to produce examples starting at  48  of the cells of the board.

Ball.  MRE, 5th ed., 1911, p. 133 gives a magic king's tour, which is hence a magic queen's tour.  The 11th ed., 1939, p. 185 refers to Ghersi, below.

Italo Ghersi.  Matematica Dilettevoli e Curiosa.  2nd ed., Hoepli, 1921.  Pp. 320‑321, fig. 261 & 265.  Magic king's tour _ different than Ball's.

H. J. R. Murray.  Beverley's magic S-tour and its plan _ probs 2106-2108.  Problemist Fairy Supplement (later known as Fairy Chess Review) 2:16 (Feb 1936) 166.  Discusses Beverley's 1848 semi-magic path and says the method leads to  28  solutions and many, including Beverley's example, have the property that each quarter of the board is also [semi-]magic.

H. J. R. Murray.  A new magic knight's tour _ Art. 68, prob. 5226.  Fairy Chess Review 5:1 (Aug 1942) 2-3.  A  16 x 16  semi-magic tour.  Cites Beverley and Roget.  Implies that Kraitchik asserted that no such  16 x 16  tours were possible.

J. S. Madachy.  Op. cit. in 6.D.  1966.  1979 reprint, pp. 87-89.  Order 16  magic knight's tour.

Stanley Rabinowitz.  A magic rook's tour.  JRM 18:3 (1985‑86) 203‑204.  Gives one.  Also gives Ball's magic king's tour.  Says the magic knight's tour is still unsolved.

David Marks.  Knight's Tours.  M500 137 (Apr 1994) 1.  Brief discussion of magic knight's tours, giving a semi-magic example due to Euler? and a magic example made up of  2  tours of  32  squares due to Roget.

 

          7.N.5. OTHER MAGIC SHAPES

 

          See also 7.Q and 7.Q.1.

 

Yang Hui.  Hsü Ku Chai Ch'i Suan Fa.  Op. cit. in 7.N.  1275.  Book 3, chap. 1, Magic squares, pp. 149-151, with Lam's commentary on pp. 311-322.  This includes  6  magic 'circles' which are diagrams of overlapping circles of values such that each circle adds to a constant value.  In two cases, the centres of the circles are used and in one case some lines also give the same total.  Needham, pp. 60-61, and Lam's commentary describe later work by:  Ch'êng Ta-Wei  (1593),  Fang Chung-Thung  (1661),  Chang Ch'ao  (c1680),  Ting I-tung  (Sung dynasty),  Wang Wên-su  (Ming Dynasty)  and  Pao Ch'i‑shou  (late 19C), who has magic cubes, spheres and tetrahedrons _ see 7.N.1.

Kanchusen.  Wakoku Chiekurabe.  1727.  Pp. 5 & 18-21 show two kinds of magic circles.  The first has two rings of  4  and one in the centre so that each ring adds to  22  and each diameter adds to  23.  This is achieved by putting  1  in the centre and then symmetrically placing the numbers in the pairs  2-9, 3-8, 4-7, 5-6.  The second example uses the same pairing principle to give three rings of six, with  1  in the centre, so each ring adds to  63  and each diameter to  64.

See Franklin, c1750, in 7.N for a magic circle.

Curiosities for the Ingenious selected from The most authentic Treasures of Nature, Science and Art, Biography, History, and General Literature.  2nd ed., Thomas Boys, London, 1822.  Magic circle of circles, pp. 56-57 & plate IV, opp. p. 56.  12 - 76  arranged in  8  rings of  8  sectors, with  12  in the centre, so that each ring and each radius, with the  12  in the centre, makes  360.

Manuel des Sorciers.  1825.  Pp. 82-83.  ??NX  4 x 8  semi-magic rectangle, associated.  The row sums are 132 while the column sums are 66.  I don't ever recall seeing a magic rectangle like this before.  For an  A x B  rectangle, we want  1 + ... + AB  =  AB(AB+1)/2  to be divisible by both  A  and  B  which holds if and only if  A º B (mod 2).  Since it doesn't make sense to talk about diagonal sums, this can only give semi-magic shapes, hence they should be easier to produce.  I find essentially one solution for the  2 x 4  case.

Hoffmann.  1893.  Chap IV, no. 10: The 'twenty‑six' puzzle, pp. 146‑147 & 191.  4 x 4  square with corners deleted.  Arrange  1 ‑ 12  so the  4  horizontal and vertical lines of  4  give the magic sum  26,  and so do the central  4  cells and hence the two sets of opposite outer cells.  Gives two solutions and says there may be more.

I. G. Ouseley.  Letters: Pentacle puzzle.  Knowledge 19 (Mar 1896) 63  &  (Apr 1896) 84.  Consider a pentagram, its five points and the five interior intersections.  Place the numbers  1  to  10  on these so that each line of five has the same sum and the five internal values shall add to the same sum, while the five outer values shall add to twice as much.  Second letter says it seems to be unsolvable and Editorial Note points out that the sum of all the numbers is  55,  which is not divisible by  3,  so the problem as stated is unsolvable.  [But what if we take the numbers  0  to  9 ??]

Pearson.  1907.  Part I, no. 35: A magic cross, p. 35.  Same pattern as Hoffmann, with numbers differently arranged.  Says there are  33  combinations that add to  26.

Williams.  Home Entertainments.  1914.  The cross puzzle, pp. 117-118.  Shape of the Ho Thu (the River Plan, see beginning of 7.N) to make have the same sum of  23  across and down.  [In fact one can have magic sums of  23, 24, ..., 27.]

Wood.  Oddities.  1927.  Prob. 8: A magic star, pp. 9-10.  Make an  8-pointed star by superimposing two concentric squares, one twisted by  45^o.   There are  8  vertices of the squares and  8  points of intersection, so there are  4  points along each square edge.  Arrange the numbers  1 - 16  on these points so each edge adds up to the same value.  This forces the sums of the 4 vertices to be the same (which he states as a given) and the magic constant to be  34.  He gives one solution and says there are  18  solutions, which I have confirmed by computer _ doing it by hand must have been tedious or I have overlooked some simplifications.

Collins.  Book of Puzzles.  1927.

Pp. 74-75: The card addition puzzle.  If one views cards as  1 x 2  rectangles, then a  6 x 6  frame can be formed by having three vertical cards on each side and two horizontal cards filling in the top and bottom.  Arrange the cards  1 - 10  so that the total along each side is the same.  Note that this adds four cards along the top and bottom but three along the sides.  Gives one solution and implies it is unique, but I have found ten solutions.

Pp. 82-84: Adding, subtracting, multiplying, dividing and fractional magic squares.  For a subtraction square, if  a, b, c  are the elements in a row, then  c ‑ (b ‑ a)  =  a ‑ b + c  is a constant.  An easy solution is obtained by replacing the odd or the even terms of a magic square by ten minus themselves.  He asserts this idea was invented by Dudeney.  He says the multiplying magic square dates from the last quarter of the 18C but was first published by Dudeney.  For a division square, we have  c/(b/a)  =  ac/b  is a constant.  One can find an easy solution from a multiplication square.  He says this was invented by Dudeney.  Fractional squares are ordinary additive squares with constant of one.

Pp. 90-92: A magic circle.  This is basically an  8 x 8  (semi?) magic square spread into a circular pattern.  He uses the numbers  12 - 75  to get a sum of  348 and then sticks some values of 12 in so as to yield 360.

Pp. 93-95: A magic pentagon.  Five  4 x 4  squares skewed to rhombi and fit together at a point to make a five-pointed star.  Each rhombus has constant  162  and various sums add up to  324.

Meyer.  Big Fun Book.  1940.  Mathematical fun, pp. 98 & 731.  Make the                                     ABC

          figure at the right magical.  I've called this a Magic Hourglass _ see below.                                D

          standard arguments show that the magic constant must be  12  and  D = 4.                            EFG

          The solution is essentially unique, with one horizontal line containing 

          7, 3, 2.

Birtwistle.  Math. Puzzles & Perplexities.  1971.                                                                               A  

          Pp. 11 & 13.  Consider the pattern at the right.  Place the digits                                               B  

          1, 2, ..., 8  so that the vertical and horizontal quadruples and the                                    E F G H

          inner and outer circles (i.e.  A, H, D, E  and  B, G, C, F)  all have the                                     C  

          same sum.  By exchanging  C, D  with  G, H,  we have a cube with four                                   D  

          magic faces, but the pattern has more automorphisms than the cube.  One

          sees that  A + D = F + G,  B + C = E + H,  so these pairs can be interchanged.  Also, one can interchange  B  and  C,  etc.  We can then assume  A = 1,  B < C,  B < E < H,  F < G  and then there are  6  solutions.  Each of these gives  16  solutions with  A = 1  and hence  128  solutions allowing any value of  A.  Hence there are  768  solutions in total.  He gives one.

Gareth Harries.  Going round in triangles.  M500 128 (Jul 1992) 11-12.  Consider the lattice triangle of side two.  This has  4  triangles,  6  vertices and  9  edges.  Place the numbers  1, .., 15  on the vertices and edges so that each edge number is the difference of the numbers at its ends.  He says his computer found  19  solutions.  For the side one problem, there are just two solutions.

David Singmaster.  Braintwister: Give the hour-glass some time.  The Weekend Telegraph (6 Feb 1993) xxxii  &  (13 Feb 1993) xxxvi.  The Magic Hourglass problem, as in Meyer, though I don't recall where I saw the problem _ possibly in one of Meyer's other books.  I recall that my source gave the value of  D  or  S.

Mirko Dobnik (Rošnja 5, 62205 Starše, Slovenia).  Letter and example of 31 Aug 1994.  Take a solid cube and make each face into a  2 x 2  array of squares.  He puts the numbers  1, 2, ..., 24  into these cells as shown below.  Each face adds up to  50  and numerous bands of  8  around the cube (at least  18)  add up to  100.  As he notes, the roundness of these values is notable.

                              10  13

                              11  16

                                1  22      8  17      3  24    20    5

                                4  23    19    6      2  21      7  18

                              14    9

                              15  12

Mirko Dobnik (Rošnja 5, 62205 Starše, Slovenia).  Letter and example of 24 Jan 1995.  Amends two faces of the above array to the following

                              10  13

                              11  16

                                1  22      8  19      3  24      6  17

                                4  23      5  18      2  21      7  20

                              14    9

                              15  12

 

          7.O.   MAGIC HEXAGON

 

          The unique solution is given at the right.  It is often presented                                     18  17    3       

with a point up.  I will describe versions by saying which point or                                  11    1     7   19   

edge value is up and whether it is a reflection or not.                                              9     6     5    2    16

E.g. This has 17 up.                                                                                                    14    8     4   12   

                                                                                                                                         15  13  10       

 

[Ernst] von Haselburg (of Stralsund).  MS of problem and solution in the City Archives of Stralsund, dated 5 May 1887, with note of 11 May 1887 saying it was sent to Illustrierten Zeitung, Leipzig.  Xerox kindly provided by Heinrich Hemme.

[Ernst] von Haselburg (of Stralsund), proposer and solver.  Prob. 795.  Zeitschrift für mathematischen und naturwissenschaftlichen Unterricht 19 (1888) 429  &  20 (1889) 263‑264.  Poses the problem for side  3  hexagon.  Solution deals with the three sums of six symmetric points to show the central number is at most  8  and then finds only  5  is feasible and it gives a unique solution.  He has  18  up.  Reported by Martin Gardner, 1988, without a diagram;  see also Hemme, Bauch.

William Radcliffe.  38 Puzzle.  UK & US patents, 1896.  ??NYS _ cited by Gardner, 1984.  I couldn't find the US patent in Marcel Gillen's compilation of US puzzle patents, but a version is reproduced in Tapson and in Hemme, which has  3  up.

The Pathfinder.  (A weekly magazine from Washington, DC.)  c1910.  ??NYS  Trigg says Clifford Adams saw the problem here, but with a row sum of  35!

Tom Vickers.  Magic Hexagon.  MG 42 (No. 342) (Dec 1958) 291.  Simply gives the solution, with  13  up, reflected.

M. Gardner.  SA (Aug 1963)  = 6th Book, chap. 3.  Describes Clifford Adams' discovery of it.  Shows  15  up, reflected.

C. W. Trigg.  A unique magic hexagon.  RMM 14 (Jan‑Feb 1964) 40‑43.  Shows the uniqueness in much the same way as von Haselburg, but in a bit more detail.  Shows  15  up, reflected.

Ross Honsberger.  Mathematical Gems.  MAA, 1973.  Chap. 6, section 2, pp. 69-76.  Describes Adams' work, as given in Gardner, and outlines Trigg's proof of the uniqueness.  A postscript adds that a Martin Kühl, of Hannover, found a solution c1940, but it was never published and cites Vickers note.

M. Gardner.  Puzzles from Other Worlds.  Vintage (Random House), NY, 1984, p. 141.  Describes Radcliffe's work _ see above.  Gardner says he was a school teacher at the Andres School on the Isle of Man and discovered the hexagon in 1895 and patented it in the US and UK.  He shows the pattern with 15 up, reflected.

Frank Tapson.  Note 71.25:  The magic hexagon: an historical note.  MG 71 (No. 457) (Oct 1987) 217‑220.  Says he has a book of correspondence of Dudeney's containing:  two letters from Radcliffe in 1902;  Dudeney's copy of Frénicle's letter to Fermat (or Mersenne?? _ see 7.O.1 & 7.N.2)  and a reproduction of Radcliffe's published solution, labelled  'Discovered 1895   Entered at Stationers Hall 1896'.  He shows  13  up, reflected, and a reproduction of Radcliffe's reflected form, which is  3  up.  Radcliffe's letters refer to the similar problem discussed by Dudeney in Harmsworth's Magazine (Jan 1902) and London Magazine (Feb 1902) _ see under Dudeney in 7.O.1.  [These are the same magazine _ it changed its name.]

Martin Gardner.  Letter:  The history of the magic hexagon.  MG 72 (No. 460) (Jun 1988) 133.  Describes von Haselburg, with no diagram, as communicated to him by Hemme.

Heinrich Hemme.  Das Kabinett:  Das magische Sechseck.  Bild der Wissenschaft (Oct 1988) 164-166.  Shows the solution is unique.  Describes Adams's discovery and Gardner's article.  Says R. A. Cooper then discovered Vickers' note, then Tapson discovered Radcliffe's version in 1973.  Hemme says Tapson said Radcliffe was a teacher at the Andreas-Schule on the Isle of Man and got a UK patent.  Then in the mid 1980s, Ivan Paasche of Stockdorf saw the German translation of Gardner and recalled Haselburg's work which he was able to locate.  Paasche found that there was a 'Stadtbaurat' with an interest in mathematics named von Haselberg in Stralsund at the time.  Stadtbaurat has several meanings _ it could be a member of the local planning board or a city architect, but it also was an honorific for a distinguished architect.

Hans F. Bauch.  Zum magischen Sechseck von Ernst v. Haselberg.  Wissenschaft und Fortschritt 40:9 (1990) 240-242 & a cover side.  This is the first source to give von Haselberg's given name and to give a picture of him.  He was born in 1827 and died in 1905.  The original MS of the problem and solution have been located in the City Archives of Stralsund (see above) and Bauch sketches the original method _ there is a fair amount of trial and error _ and reproduces some of the MS.  There are a number of sub‑configurations which must add up to the constant  38  and Bauch shows these, including a Star of David configuration _ see 7.O.1.  Von Haselberg submitted his MS to the Illustrierten Zeitung of Leipzig, but they didn't use it.  He was 'Stadtbaumeister' (= City Architect) in Stralsund and restored the facade of the City Hall.  He also published a five volume work on local architectural monuments. 

Ed Barbeau.  After Math.  Wall & Emerson, Toronto, 1995.  A magic hexagon, pp. 151-152.  When he gave this problem in his puzzle column, a solution came in only two weeks later and a second solution arrived six weeks after that.  Neither solver seems to have used a computer.  He cites only Honsberger.  The first solver obtained a number of identities which would simplify the solution.

 

          7.O.1. OTHER MAGIC HEXAGONS

 

Frenicle de Bessy.  Loc. cit. in 7.N.2.  1640.  Discusses a magic hexagon of White's form II (below).

I. G. Ouseley.  Letter:  The puzzle of "26".  Knowledge 18 (1895) 255-256.  Arrange  1 ‑ 12  on the points and crossings of a Star of David so that  the sum on each line  =  sum of vertices of each large triangle  =  26.  Says this has come from Mr. T. Ordish, who has arranged the numbers to sum to  26  in  30  different ways.  A number of these are consequences of the above conditions, e.g.  the sum on the inner hexagon  =  the sum of the vertices of a rhombus  which is 26,  as is the sum along certain angles,  but I can't understand what is meant by  '6  obtuse angles' or  '4  rhomboids'.  "I believe there are at least six ways ...."

                    In the next issue, p. 278, are three letters with a comment by Ouseley.  Un Vieil Étudiant sends four solutions with the sum of the vertices of the large triangles being  13,  and hence the sum on the inner hexagon being  52.  These are complementary to the solutions requested, though neither he nor Ouseley notes this.  These complementary solutions are rather easier to find though and I have indeed found six solutions, as does Ahrens below.  J. Willis sends one solution of the original form, possibly implying that it is unique.  T. sends one different solution.  Ouseley notes that in a solution, the value at a point is the sum of the values at the two opposite crossings.  ?? _ possibly more letters in the next issue.

                    In the next volume, pp. 35 & 84, are two notes saying that the publishers (T. Ordish & Co., London) and the proprietors (Joseph Wood Horsfield & Co., Dewsbury) of the puzzle have complained that the above notes are an infringement of their copyright in the "26" Puzzle and Knowledge apologizes for this.

Dudeney.  Puzzling times at Solvamhall Castle.  London Magazine 7 (No. 42) (Jan 1902) 580‑584  &  8 (No. 43) (Feb 1902) 53-56.  The archery butt.  = CP, prob. 35, pp. 60-61 & 187-188.  Hexagon of  19  numbers so that the  6  radii from the centre to the corners and the  6  sides each add to  22.  Problem is to rearrange them so each adds to  23.  Solution says one can get any number from  22  to  38,  except  30.

William F. White.  Op. cit. in 5.E.  1908.  Magic squares _ magic hexagons, pp. 187‑188.

I.        Arrange  1 ‑ 12  on the points and crossings of a Star of David so that  the sum on each line  =  sum of vertices of each large triangle  =  sum on inner hexagon  =  sum of vertices on each parallelogram  =  26  _ i.e. the problem of Ordish/Ouseley.  Gives one solution.  Quotes Escott:  "There are only six solutions."  "The first appeared in Knowledge, in 1895, and the second is due to Mr. S. Lloyd."  [Error for Loyd??]

II.       Arrange  1 ‑ 19  in a hexagon, consisting of six equilateral triangles "so that the sum on every side in the same".  This gives just  12  sums of three points, which is Dudeney's problem.  He gives solutions with the sum  =  22  and  23  and notes that subtracting from  20  gives sums  =  38  and  37.

Ahrens.  A&N.  1918.  Chap. XII: "Die wunderbare  26", pp. 133‑140.  Consider a hexagram (or six pointed star) formed from two triangles.  This has  12  vertices.  He finds  6  ways to place  1, 2, ..., 12  on these vertices so that each set of four along a triangle edge adds up to the same value (which must then be  26),  and the six corners of the inner hexagon also add to  26.  I.e. this is Ordish/Ouseley's problem, but with a slightly different statement of conditions.

                    On p. 134, a note says the puzzle "Wunderbare  26" is made and sold by Züllchower Anstalten, Züllchow bei Stettin, and they have registered designs 42,768 and 45,600 for it, though the 'Rustic  26' has already been on sale for many years.  S&B, p. 39, shows an English example with no identification.

Collins.  Fun with Figures.  1928. 

A nest of magic hexagons, pp. 109-110.  Central point surrounded by hexagons of  12, 24, 36  points.  The numbers  1, ..., 73  are placed on the points so the sides of each hexagon add up to  111, 185, 259  respectively  and  the diagonals and the midlines add up to  259.  In fact, opposite points in each hexagon add to  74  and the central value is  37.  Note that  111 = 3 x 37,  185 = 5 x 37,  259 = 7 x 37.

A magic hexagon within a circle, pp. 110-112.  This is really a pattern of magic triangles _ cf. 7.N.2.

Perelman.  1934.  See in 7.N.2 for the Star of David pattern with just one solution.

Birtwistle.  Math. Puzzles & Perplexities.  1971.

Digital sum 1, pp. 29, 163 & 188.  Take  12  points on the vertices and midpoints of a hexagon, so each edge contains three points, together with a point in the middle of the hexagon.  Place the numbers  1, ..., 13  so each straight line of three points has the same sum.  Standard arguments show the central point must be  7  and the magic sum is  21.  Factoring out symmetry, there are four solutions.  He gives one.

 Digital sum 2, pp. 29, 163 & 188.  Magic star of David, as in the  26  Puzzle, but with no condition on the sum of the inner hexagon.  This gives many more solutions _ see Singmaster, 1998.  He gives one solution, which has the sum of the inner hexagon being  26.

Bell & Cornelius.  Board Games Round the World.  Op. cit. in 4.B.1.  1988.  Marvellous  '26', p. 79.  Says this was sold for  6d  by T. Ordish "probably in about 1920".  [Above we see that it was known in 1895.]  They quote the instructions from the box:  "each of the six sides as well as the six spaces around the centre total up to  26  with perhaps the finding of several additional 'twenty sixes'."  This is Ordish/Ouseley's problem.

Bauch.  Op. cit. in 7.O, 1990.  In this, he gives the Star of David problem, which is a subproblem of the Magic Hexagon.  He asserts it has  96  'classical' solutions, but he gives no discussion or reference and it is not clear if this is for all possible magic sums _ he shows one solution with magic sum  26  and the central hexagon having total  26.

David Singmaster.  Magic Stars of David and the  26  Puzzle.  Draft written in May 1998.  There are  960  solutions for a magic Star of David.  There is clearly an equivalence given by the symmetries of the regular hexagon, so these solutions fall into  80  equivalence classes.  Six of these are solutions of the  26  Puzzle (i.e. the sum of the central hexagon is also  26) and six are solutions of the complemented problem (i.e. the sum of the outer points is  26).  One can use complementation to reduce the number of classes to  40.  However, there are more symmetries of the magic Star of David when one ignores the additional constraint of the  26  Puzzle _ indeed the pattern is isomorphic to labelling the edges of a cube so the sum of the edges around each face is  26.  Hence one can use the symmetries of the cube to produce  20  equivalence classes of solutions, but these symmetries do not interact simply with the additional sums used in the  26  Puzzle.

 

          7.P.    DIOPHANTINE RECREATIONS

 

          See also 7.E, 7.R.1, 7.R.2, 7.U.

 

          7.P.1. HUNDRED FOWLS AND OTHER LINEAR PROBLEMS

 

          See Tropfke 565, 569, 572 & 613.

          NOTATION:  n  for  p  at  a, b, c  means  n  items of three types, costing  a, b, c  were bought for a total of  p.  I.e. we want   x + y + z = n;   ax + by + cz = p,  with the conditions that  x, y, z  are positive (or non-negative) integers.

          (a, b)  solutions means  a  non‑negative solutions, including  b  positive solutions _ so  a £ b.  I have checked these with a computer program.  I also have a separate numerical index to these problems which enables me to tell whether problems are the same.

          When there are just two types of fowl, one gets two equations in two unknowns, but I have generally omitted such problems except when they seem to be part of an author's development or they represent a different context.  See:  MS Ambros. P114;  Tartaglia 17‑25 & 26;  Hutton;  Ozanam-Hutton 9;  Williams;  Collins.

          The medieval problems of alligation are related, but the solution need not be integral.  See:  Fibonacci;  Bartoli;  Lucca 1754;  Borghi;  Apianus;  Tropfke 569  for discussion and examples.  See:  Devi  for a modern version with integral solutions.  See  Williams  for a history of this aspect.

          Riese (1524);  Simpson (1745);  Euler (1770);  Ozanam-Montucla (1778);  Bonnycastle (1782 & 1815);  Hutton, 1798?;  Stewart (1802);  Bourdon (1834);  Unger (1838);  Colenso (1849);  Depew (1939);  McKay (1940);  Hedges (1950?);  Little Puzzle Book (1955);  Scott (1973)  are examples using the context of making change. 

          McKay (1940);  Little Puzzle Book (1955)  use the context of buying postage stamps.

          G.F. (1993)  uses the context of wheels of vehicles.

          Simpson (1745);  Dodson (1747?);  Euler (1770);  Ozanam-Montucla (1778);  Bonnycastle (1782 & 1815);  Ozanam-Hutton (1803);  Bourdon (1834);  Clark (1904);  McKay (At Home Tonight, 1940)  are the only examples here which consider problems with one equation in two unknowns.

          Problems involving heads and feet of a mixture of birds and beasts:  Clark;  Williams;  Collins.

          Impossible problems _ sometimes a problem is impossible only if positive solutions are required.  Abu Kamil;  Fibonacci 1202 & 1225;  Tartaglia;  Buteo;  Simpson;  Euler;  Ozanam‑Montucla;  Bonnycastle;  Perelman;  Depew;  Little Puzzle Book;  Scott;  Holt.

          I have recently realised that the relatively modern problem of asking how to hit target values to make a particular value is a problem of this general nature, especially if the number of shots is given.  E.g. a target has areas of value  16, 17, 23, 24, 39;  how does one achieve a total of  100?  These occur in Loyd, Dudeney, etc., but I haven't recorded them.  I may add some of them.

 

Zhang Qiujian.  Zhang Qiujian Suan Jing.  Op. cit. in 7.E.  468.  Chap. 3, last problem, pp. 37a ff (or 54 f ??).  ??NYS.   Hundred Fowls: 100 for 100 at  5, 3, _.  (Cocks, hens, chicks.)  This has (4, 3) solutions _ he gives (3, 3) of them, but only states the relation between the solutions _ no indication of how he found a solution.  (Translation in Needham 121-122 (problem only),  Libbrecht 277,  Mikami 43,  Li & Du 99.)

Ho Peng Yoke.  The lost problems of the  Chang Ch'iu‑chien Suan Ching,  a fifth‑century Chinese mathematical manual.  Oriens Extremus 12 (1965) 37‑53.  On pp. 46‑48, he identifies prob. 31 of  Yang Hui  (below at 1275) as being from Chang (= Zhang).  The solution involves choosing one value arbitrarily.

Zhen Luan (= Chen Luan).  Op. cit. in 7.N, also called  Commentary on  Hsü Yo's  Shu Shu Chi I.  c570.  ??NYS.  [See Li & Du, p. 100.]  100 for 100 at  5, 4, ¼.  (Cocks, hens, chicks.)  This has (2, 1) solutions _ he gives (1, 1).  Mikami says Chen's method would give one solution to Chang's problem.  Libbrecht, pp. 278‑279, gives the method, which is indeed nonsense, and states that other scholars noted that Chen's method is fortuitous.  He says Chen also gives 100 for 100 at  4, 3, _,  which has (3, 2) solutions, of which Chen gives (1, 1).

Liu Hsiao‑sun.  Chang ch'iu‑chien suan‑ching hsi‑ts'ao  (Detailed solutions of [the problems] in the  Chang ch'iu‑chien suan‑ching).  c600.  ??NYS.  Described in Libbrecht pp. 279‑280 as nonsense!!

Bakhshali MS.  c7C.  S_tra C7 (VII 11-12).  Hayashi 648-650 studies this.  General discussion and first example are too mutilated to restore.

Example 2 (VII 11).  20 for 20 at  ?, 4, ½.  From the working, the lost coefficient satisfies  0 £ ? £ 15/32.  One solution is given, but only partly readable:  ?, 3, ?.  Not mentioned by Kaye.

Example 3 (VII 11-12).  20 for 20 at  3, 3/2, ½.  (Earnings of men, women, children.)  (3, 1) solutions, (1, 1) given:  2, 5, 13.  (Also in Kaye I 42; III 191, f. 58v.  Bag; Mathematics in Ancient and Medieval India; 1979, p. 92 gives just the problem.  Datta, p. 50, says the answer is mutilated, but Hayashi does not comment on this _ Kaye III 191 displays the answer given as  2, 5, 1...,  so the mutilation is pretty minimal.)

Li Shun‑fêng.  Commentary on  Chang chiu‑chien.  7C.  ??NYS.  Libbrecht (p. 280) describes his comments as "unmitigated nonsense".

L. Vanhée.  Les cent volailles ou l'analyse indéterminée en Chine.  T'oung Pao 14 (1913) 203‑210  &  435‑450.  On pp. 204‑210, he gives 24 problems in Chinese and French, but doesn't identify the sources!!

Alcuin.  9C.

Prob. 5: Propositio de emptore in C denarius.  100 for 100 at  10, 5, ½.  (Boars, sows, piglets.)  (1, 1) solution, which he gives.

Prob. 32: Propositio de quodam patrefamilias distribuente annonam. 20 for 20 at  3, 2, ½.  (Dividing grain among men, women, children.)  This has (2, 1) solutions _ he gives (1, 1).

Prob. 33: Alia propositio.  30 for 30 at  3, 2, ½.  (Like Prop. 32.)  (3, 1) solutions, (1, 1) given.

Prob. 33a (in the Bede text): Item alia propositio.  90 for 90 at  3, 2, ½.  (Like Prob. 32.)  (7, 5) solutions, (1, 1) given.

Prob. 34: Item alia propositio.  100 for 100 at  3, 2, ½.  (Like Prop. 32.)  (7, 6) solutions, (1, 1) given.

Prob. 38: Propositio de quodam emptore in animalibus centum.  100 for 100 at  3, 1, 1/24.  (Horses, cows, sheep.)  (2, 1) solutions, (1, 1) given.

Prob. 39: Propositio de quodam emptore in oriente.  100 for 100 at  5, 1, 1/20.  (Camels, asses, sheep.)  (2, 1) solutions, (1, 1) given.

Prob. 47: Propositio de episcopo qui jussit XII panes in clero dividi.  12 for 12 at  2, ½, ¼.  (Dividing loaves among priests, deacons, readers.)  (2, 1) solutions, (1, 1) given.

Prob. 53: Propositio de homine patrefamilias monasteri XII monachorum.  This problem seems to be a major corruption of the following.  12 for 204 at  32, 8, 4.  (Dividing eggs among priests, deacons, readers.)  This problem has one solution: 5, 4, 3  and it seems like the problem was last and some scribe has used the result to reformulate the problem as giving  204/12 = 17  to each.

Mahavira.  850.  Chap. III, v. 133, pp. 67-68 is related to this general type.  Chap VI, v. 143‑153, pp. 130‑135.

  Chap. III.

133:  x(3/4)(4/5)(5/6) + y(1/2)(5/6)(4/5) + z(3/5)(3/4)(5/6)  =  1/2.  This is:  x/2 + y/3 + 3z/8  =  1/2  and he arbitrarily picks two of the values, getting:  1/3, 1/4, 2/3.

  Chap. VI.

143:  complex triple version with prices also to be found.

146:  general method.

147:  72 for 56 at  _, ¾, 4/5, 5/6.  (Peacocks, pigeons, swans, sârasa birds.)  (217, 169) solutions, (1, 1) is given.

150:  68 for 60 at  3/5, 4/11, 8.  (Ginger, long pepper, pepper.)  (1, 1) solutions, (1, 1) given.

151:  gives a complex method.

152:  100 for 100 at  3/5, 5/7, 7/9, 9/3.  (Pigeons, sârasa birds, swans, peacocks.)  (26, 16) solutions, (1, 1) given.  (See Sridhara & Bhaskara II below.)

Pseudo-Alcuin.  9C.  ??NYS _ cited by Hermelink, op. cit. in 3.A.

Sridhara.  c900.  V. 63‑64, ex. 78‑80, pp. 50‑52 & 95.

Ex. 78‑79.  Same as Mahavira's 152.  (Pigeons, cranes, swans, peacocks.)  (26, 16) solutions;  commentator gives (4, 4);  editor gives (16, 16).

Ex. 80.  100 for 80 at  (2, 3/5, ½).  (Pomegranates, mangoes, wood‑apples.)  Commentator gives (5, 5) solutions of the (6, 5).

Ab_ K_mil Shuj_‘ ibn Aslam ibn Muhammad ibn Shuj_‘, al-H_sib al-Misr_.  Kit_b al‑tar_’if fi’l his_b  (Book of Rare Things in the Art of Calculation).  c900.  Trans. by H. Suter as:  Das Buch der Seltenheiten der Rechenkunst von  Ab_ K_mil el‑Mi_r_;  Bibl. Math. (3) 11 (1910‑1911) 100‑120.  (I have a reference to an Italian translation by G. Sacerdote; IN:  Festschrift zum 80 Geburtstag M. Steinschneiders; Leipzig, 1896, pp. 169‑194, ??NYS.)  (Part is in English in Ore; Number Theory and Its History; 139‑140.)  Six problems of 100 for 100 at the following.

1.  5, 1, 1/20.  (Ducks, hens, sparrows.)  (2, 1) solutions, (1, 1) given.

2.  2, _, ½.  (Ducks, doves, hens.)  (7, 6) solutions, he gives (6, 6).

3.  4, 1/10, ½, 1.  (Ducks, sparrows, doves, hens.)  (122, 98) solutions, he gives (96, 96) and an early commentator pointed out the missing two positive solutions.

4.  2, ½, _, 1.  (Ducks, doves, larks, hens.)  (364, 304) solutions, he says (304, 304).

5.  3, _, 1/20.  (Ducks, hens, sparrows.)  (1, 0) solutions, he says there is no solution.

6.  2, ½, _, ¼, 1.  (Ducks, doves, ring‑doves, larks, sparrows.)  (3727, 2678)  solutions _ he says  2676  once and  2696  twice.  Suter notes that the Arabic words for  70  and  90  are easily confused.  Suter's comments say there are  2676  solutions.  Tom O'Beirne [Puzzles and Paradoxes; OUP, 1965; Chap. 12] discusses this and finds  2678  solutions _ he is apparently the first to find this number, but this probably appeared in his New Scientist column in 1960-1961, ??NYR.

The Appendix is a fragment of a commentary which the translator fills in to be Prob. A1:  400 for 400 at  1, 3, 2, 1/7.  (Doves, partridges, hens, sparrows.)  This has (1886, 1806) solutions. Suter estimates  c1700.  The commentator and Suter only check the case when the number of doves is divisible by  5.  This has (398, 342) solutions.  The commentator mentions  334  solutions and Suter says  341.

al‑Karkhi.  c1010.  Sect II, no. 10, p. 82.  Mix goods worth  5, 7, 9  to make one worth  8.

Tabari.  Mift_h al-mu‘_mal_t.  c1075.  P. 110f., part IV, no. 20.  ??NYS - cited by Tropfke 614, who says it has  (cows, sheep, hens).

Bhaskara II.  Bijaganita.  1150.  Chap VI, v. 158‑159.  In Colebrooke, pp. 233‑235.  Same as Mahavira's 152.  (Doves, cranes, geese, peacocks.)  (This also has  5x + 8y + 7z + 92  =  7x + 9y + 6z + 62.  ??)

Fibonacci.  1202.  Chap. 11: de consolamine monetarum, pp. 143‑166 deals with this problem in the more general form of combining metals.  (Other versions involve mixing spices, wines, etc.)  These lead to  a₁x₁ + ... + a_nx_n  =  b(x₁ + ... + x_n),  often with  x₁ + ... + x_n  specified.  x_i  is the weight and  a_i  is the purity or gold content, etc., of the  i-th metal.  We are mixing the metals to produce a total of weight  x₁ + ... + x_n  with purity  b.  Hence there is no need to consider only integral values but he usually gives one (or a few) integral solutions.  I describe a few examples.  A  denotes the vector of  a_i's.

Pp. 152‑154: De consolamine trium monetarum inter se.  A = (3, 4, 6),  b = 5.  Answers:  1, 1, 3  and  2, 5, 9.

Pp. 156‑158: De consolamine septum monetarum.  A = (1, 2, 3, 5, 6, 7, 8),  b = 4.  Answer has fractions.

Pp. 158‑159 deals with 240 metals!

P. 160: De homine qui emit libras  7  trium canium per denarios  7.  7 for 7 at  3, 2, ½.  (Pork, beef, hyrax [an animal somewhat like a rabbit].)  This has (0, 0) solutions!  He gives  1, _, 5_!

P. 165: De homine qui emit aves triginta trium generum pro denariis  30.  30 for 30 at  3, 2, ½.  (Partridges, doves, sparrows.)  (= Alcuin's 33.)  He gives the (1, 1) answer.

P. 165: De eodem.  12 for 12 at  2, ½, ¼.  (Same birds.)  (= Alcuin's 47.)  He gives the (1, 1) answer.

P. 165‑166: De eodem cum genera avium sint quattuor.  30 for 30 at  3, 2, ½, ¼.  (Partridges, doves, turtle doves, sparrows.)  (27, 19) solutions _ he gives (2, 2).

Pp. 322‑323 give some examples with two and four metals done by false position.

Fibonacci.  Epistola.  c1225.  In Picutti, pp. 332-336, numbers XI - XIII.  One of the problems is briefly mentioned in:  M. Cantor; Mathematische Beiträge zum Kulturleben des Völker; Halle, 1863; reprinted by Olms, Hildesheim, 1964; p. 345.  Surprisingly, none of these problems have appeared elsewhere!

P. 247: De avibus emendis secundum proportionem datam.  30 for 30 at  _, ½, 2.  (Sparrows, turtledoves, pigeons.)  This has (3, 1) solutions _ he gives (1, 1).

Pp. 247-248: De eodem.  29 for 29 at  _, ½, 2.  (Sparrows, turtledoves, pigeons.)  This has (2, 2) solutions _ he gives both.

P. 248: Item de avibus.  15 for 15 at  _, ½, 2.  (Sparrows, turtledoves, pigeons.)  This has (0, 2) solutions _ he says "hoc esse non posse sine fractione avium demonstrabo."  By eliminating the first variable, he gets  y + 10z = 60  and he notes that  0, 10, 5  solves the problem.  "Sed si volemus frangere aves", one can have  9/2, 5, 11/2.  However he fails to find  9, 0, 6.

P. 248: above continued.  15 for 16 at  _, ½, 2.  (Sparrows, turtledoves, pigeons.)  This has (1, 1) solutions which he gives.

P. 248: above continued.  30 for 30 at  _, 2, 3.  (Sparrows, pigeons, partridges.)  This has (2, 1) solutions _ he gives (1, 1).

P. 249: above continued.  24 for 24 at  1/5, _, 2, 3.  (Sparrows, turtledoves, pigeons, partridges.)  This has (6, 2) solutions _ he gives (2, 2).

Abbot Albert.  c1240.

Prob. 2, pp. 332‑333:  6 for 50 at  2, 9, 10.  (1, 1) solutions.  Actually he is using this for divination:  if  x + y + z = 6,  then the value of  2x + 9y + 10z  determines  x, y, z.  He gives a table of all the partitions of  6  into  3  non‑negative summands and computes  2z + 9y + 10z  for each.  This is more properly a problem for Section 7.AO.

Prob. 7, p. 334:  30 for 30 at  4, 2, ½.  (Geese??, ducks??, fig-peckers.)  (2, 1) solutions, he gives (1, 1).

Yang Hui.  Xu Gu Zhai Qi Suan Fa  (= Hsü Ku Chai Ch'i Suan Fa).  1275.  Loc. cit. in 7.N, pp. 165‑166, prob. 29‑31.

29.  100 for 100 at  5,3, _.  (Same as Chang's problem, but only gives (1, 1) of the (4, 3) solutions.)

30.  100 for 100 at  7, 3, _.  (Three types of tangerine.)  (6, 4) solutions, (1, 1) given.

31.  same as 30 in terms of wines and with different measures.

Gherardi.  Libro di ragioni.  1327.

Pp. 85‑86.  Chompera.  24 for 24 at  ¼, 2, 3.  (Sparrows, doves, geese.)  This has (1, 1) solutions, which he gives.

P. 86.  Chompera ucelli.  24 for 24 at  1/5, 1, 2, 3.  (Sparrows, thrushes, doves, geese.)  (1, 1) solution given of the (13, 6) solutions.

N_r_yana.  1356.  Op. cit. in 7.N, p. 1, lines 2‑5, p. 93.  (Same as Mahavira's 152.)  ??NYS _ see Bag, op. cit. under Bakhshali MS, p. 92.

Munich 14684.  14C.  Prob. XI, p. 79.

12 for 12 at  2, ¼, ½.  (2, 1) solutions, he gives (1, 1).  (= Alcuin's 47.)

20 for 20 at  2, ¼, ½.  (2, 2) solutions, he gives (1, 1).

Prob. XII, p. 79.  12 for 12 at  2, 1, ½, ¼.  (11, 4) solutions, he gives (1, 1).

Dell'Abbaco.  c1370.  Prob. 190, pp. 150‑151.  48 for 48 at  4, 2, ¼.  (1, 1) solution, given.

Lucca 1754.  c1390.

Ff. 9v‑10r, pp. 34‑35.  40 for 40 at  3, 2, ¼.  (Thrushes, larks, sparrows.)  (2, 2) solutions, both given.

Ff. 10r‑10v, pp. 35‑36.  100 for 100 at  3, 1, 1/20.  (Oxen, pigs, sheep.)  (2, 1) solutions, (1, 1) given.

F. 46v, pp. 96‑97.  60 for 600 at  7, 9, 11, 17.  (Mixing grain.) (314, 272) solutions, he gives one:  35, 5, 5, 15.

F. 46v, p. 97.  60 for 480 at  5, 9, 7.  (Mixing of metals.)  (16, 14) solutions, he gives one:  0, 30, 30.

F. 48v, p. 103.  775  for  162.75  at  .16,  .18, .20, .27, .31.  (Mixing of metals.)  This has  (3027289, 2966486)  solutions!  He gives one solution:  250, 150, 150, 100, 125.

(There are several similar problems here with solutions obtained by guessing.)

F. 56r, p. 125.  100 coins worth  2150  at  50, 33, 17, 25, 15.  (2160, 1536)  solutions, he gives one solution:  10, 10, 10, 10, 60.

Also the same with total value  3900.  (526, 388)  solutions, he gives one solution:  60, 10, 10, 10, 10.

F. 59v, pp. 135‑136.  24 for 24 at  ¼, 2, 3.  (Sparrows, doves, geese.)  (1, 1) solutions, which he gives.  = Gherardi, pp. 85‑86.

Bartoli.  Memoriale.  c1420.  Ff. 89v - 95r (= Sesiano, pp. 134-135).  Eleven problems of alligation, many identical to Lucca 1754.

Jamshid al-K_sh_  = _amš_d ibn Mas‘_d ibn Mahm_d, _ij_t ed-d_n al-K_š_  = Ghiy_th al-Din al-K_sh_.  Mift_h al-his_b  (The Calculator's Key).  1426.  Ed. by A. Demerdash & M. H. Hifna; Cairo, nd.  Facsimile and Russian translation by:  B. A. Rozenfeld', V. S. Segal' & A. P. Juškevi_ as:  Klju_ Arifmetiki _ Traktat ob Okru_nosti; Moscow, 1956.  ??NYS _ Hermelink, op. cit. in 3.A, says he gives a version with three fowls.

Provençale Arithmétique.  c1430.  Op. cit. in 7.E. 

F. 116r, p. 62.  12 for 12 at  2, 1, ½.  (Men, women, children.)  (5, 3) solutions, he gives (3, 3).

F. 117r, p. 63.  30 for 30 at  4, 2, ½.  (Costs of cloth.)  Same as Abbot Albert, p. 334.  (2, 1) solutions, he gives the positive one.  See Chuquet, 1484, prob. 83.

AR.  c1450.  Prob. 45, 81, 102, 119‑126, 309.  Pp. 40, 52, 60‑61, 66‑67, 137‑138, 175‑176, 221‑222.

45.  30 for 30 at  2, ½, 1/10.  (Pears, apples, nuts.)  (2, 1) solutions, (1, 1) given.  Vogel, p. 221, says this appears in Clm 8951, which is unpublished.

81.  Mix three kinds of wax worth  43, 29, 22  to make wax worth 32.

102.  Mix wines worth  5, 6, 8  to make wine worth 7.

119.  100 for 100 at  10, 5, ½.  (Oxen, cows, sheep.)  (1, 1) solution, which is given.  (= Alcuin 5.)

120.  20 for 20 at  3, 3/2, ½.  (Oxen, sheep, geese.)  (3, 1) solutions, (1, 1) given.  (= Bakhshali.)

121.  12 for 12 at  2, ½, ¼.  (Knights, boys, girls eating bread.)  (2, 1) solutions, (1, 1) given.  (= Alcuin 47.)

122.  12 for 12 at  2, 1, ½, ¼.  (Knights, men, boys, girls dividing a bill?)  (11, 4) solutions, (1, 1) given.  (= Munich 14684, XII.)

123.  30 for 30 at  3, 2, ½.  (Men women, children.)  (3, 1) solutions, (1, 1) given.  (= Alcuin 33.)

124.  100 for 100 at  3, 2, ½.  (Men, women, children.)  (7, 6) solutions, (1, 1) given.  (= Alcuin 34.)

125.  90 for 90 at  3, 2, ½.  (Men, women, children.)  (7, 5) solutions, (1, 1) given.  (= Alcuin 33A.)

126.  100 for 100 at  3, 1, 1/24.  (Horses, oxen, sheep.)  (2, 1) solutions, (1, 1) given.  (= Alcuin 38.)

309.  same as 81, done in three different ways.

Correspondence of Johannes Regiomontanus, 1463?-1465.  Edited by Maximilian Curtze as:  Der Briefwechsel Regiomontan's mit Giovanni Bianchini, Jacob von Speier und Christian Roder.  Part II of: Urkunden zur Geschichte der Mathematik im Mittelalter und der Renaissance.  AGM 12 (1902).

P. 262, letter to Bianchini, nd [presumably 1464].  240 for  16047  at  97, 56, 3,  specifically ruling out fractions, but not describing anything bought.

Pp. 293 & 296, letter to von Speier, nd [apparently early 1465].  On p. 293, Regiomontanus invites his friend to a dinner of pheasants, partridges and wine (see Hermelink, op. cit. in 3.A).  P. 296, same problem with (pheasants, partridges, doves).

P. 300, letter from von Speier, 6 Apr 1465.  Gives the only answer:  114, 87, 39.

Gottfried Wolack.  1468.  Dresden MS C80, ff. 301'-303.  This is a MS of lectures given at Erfurt, 1467 & 1468,  Transcribed and discussed in:  E. Wappler; Zur Geschichte der Mathematik im 15. Jahrhundert; Zeitschrift für Math. und Physik _ Hist.-Litt. Abteilung 45 (1900) 47-56.  P. 51 has:  100 for 100 at  2, 1, ½.  (Men, women, children.)  There is an obscure calculation leading to dividing the people in the proportion  4 : 2 : 1,  so there are  57 1/7  men!  This may be getting confused with 7.G.2 _ Posthumous twins.

P. M. Calandri.  c1480.  Pp. 151‑152.  Part of the text is lacking, but the problem must be 60 for 60 at  6, ½, _.  (Cows, calves, pigs.)  This has (2, 2) solutions, (1, 1) is given.

Luca Pacioli.  Aritmetica.  c1480.  ??NYS _ described in Sesiano.  F. 238r. 

10 for 10 at  2, 3.  Answer:  20, -10.

10 for 10 at  ½, _.  Answer:  40, -30.

Chuquet.  1484.

   Triparty, part 1.  Of apposition and remotion.  English in FHM 88-90.  Several examples treated as problems in numbers _ no mention of buying anything.  Gives a method of finding one positive solution.

12 for 12 at  2, 1, ½.  = Prov. Arith. no. 1, which has (5, 3) solutions.

12 for 60 at  8, 5, 3.  (3, 2) solutions.

12 for 36 at  4, 3, 2.  (7, 5) soltuions.

Two further examples are in the margin and FHM just gives the formulae and one solution.

12 for 35 at  6, 3, 1  {FHM has  6, 1, 1  but this doesn't agree with their answer}.  (2, 2) solutions.

20 for 20 at  6, 2, ½.  (1, 1) solutions

Chuquet says:  "This rule cannot be extended for the discovery of four or more numbers.  Also one should know that all such calculations have several answers, as many as you want, as appears [later in] this book.  Wherefore apposition and remotion is a science which has little to recommend it."

   Appendice.

Prob. 34.  15 for 160 at  (9, 13).  This is determinate and is only included as a lead-in to the next problem.

Prob. 35.  English in FHM 206-207.  15 for 160 at  (11, 13).  (Two kinds of cloth.)  Answer:  35/2,  -5/2.  Chuquet gives an interpretation of this.  See Sesiano, op. cit. in 7.R.  FHM says the "explanation fails to be entirely clear or convincing".

Prob. 40.  English in FHM 208.  Mixing of two kinds of waxes giving 100 for 1100 at  (9, 14).

Prob. 83.  English in FHM 213-214.  Reckoning which is done by the rule of apposition and remotion.  30 for 30 at  4, 2, ½.  Same as Abbot Albert, p. 334.  (2, 1) solutions.  He gives solutions  3,  3,  24  and  4,  _,  25_  and says one can have as many as one wants.  This holds because the products are cloth!

Prob. 84.  Mentioned in passing on FHM 214.  20 for  20 x 20  at  30, 25, 16, 18.  (10, 5) solutions _ he gives (1, 1).

Borghi.  Arithmetica.  1484.  Ff. 93v-101v.  Several problems of alligation, getting up to mixing five grains of values  44, 48, 52, 60, 66  per measure to mix to produce a product of value  50.

Johann Widman.  Op. cit. in 7.G.1.  1489.  ??NYS©  Glaisher, pp. 14 & 121, gives:  F. 109v: mix wines worth  20, 15, 10, 8  to make one worth 12.  Gives one solution:  6, 6, 11, 11.

Pacioli.  Summa.  1494.

F. 105r, prob. 14.  3 hens, 4 partridges and 5 geese cost  72;  while  2, 5, 7  cost  94_.  That is:  3x + 4y + 5z = 72,  2x + 5y + 7z = 94_.  He gives one solution:  (4, 16, 28)/3  with no indication that the problem is indeterminate.  _, 9, 7  is an easier solution.  There is no integral solution and the number of rational solutions is infinite!

Ff. 105r-105v, prob. 17.  100 for 100 at  ½, _, 1, 3.  (Sheep, goats, pigs, asses.)  This has (276, 226) solutions _ he gives (1, 1):  8, 51, 22, 19.

F. 105v, prob. 18.  20 for 20 at  4, ½, ¼.  (Men women, children) eating at a tavern.  This has (2, 1) solutions _ he gives (1, 1).  (See H&S 93.)

Riese.  Rechnung.  1522.  1544 ed. _ pp. 104‑106;  1574 ed. _ pp. 70r‑71v.  The 1544 ed. calls this section 'Regula cecis oder Virginum';  the 1574 ed. calls it 'Zech rechnen'.  There is first a simple problem with only two types, hence determinate.

20 for 20 at  3, 2, ½.  (Men, women, girls drinking.)  (2, 1) solutions, (1, 1) given.  (= Alcuin 32.)

100 for 100 at  4, 3/2, ½, ¼.  (Oxen, pigs, calves, goats.)  (265, 222) solutions, (1, 1) given:  12, 20, 20, 48.  The 1574 ed. has a nice woodcut illustration.

Tonstall.  De Arte Supputandi.  1522.  P. 240.  Repeats Pacioli's prob. 14, except there is a misprint _ in the second case, he has  3, 5, 7  costing  94_. 

Riese.  Die Coss.  1524.  No. 67, p. 49.  100 for 460 at  3, 5.  (Coins.)

H&S 93 says a tavern version is in Rudolff (1526?).

Apianus.  Kauffmanss Rechnung.  1527. 

Ff. H.viii.r - J.ii.v  is  Regula virginum.  He describes how to eliminate the least valuable variable and then gives a feeble attempt at describing how to find a solution from the result.

[No. 1.]  26 for 88 at  6, 4, 2.  (Men women, girls.)  (10,8) solutions.  He gives  (3,3)  solutions and says more (all?) can be found.

[No. 2.]  20 for 20 at  2, 1, 1/2, 1/4.  (Men women, girls, children.)  (25,14) solutions.  He gives  (3,3).

[No. 3.]  100 for 200 at  4, 3, 5/2, 1.  (Nutmeg, cinnamon, cloves, saffron & pepper.)  (289, 256)  solutions.  He gives (1,1).

[No. 4.]  300 for 2000 at  24, 12, 8, 4.  (Ranks of soldiers.)  (2081, 1921)  solutions.  He gives (1,1).

Ff. J.vvi.r - K.viii.v  is  Regula Alligationis

Ff. K.viii.v - L.ii.r  is  Munzschlagen.   These two sections deal with mixing of wine, spices, metals, etc., getting up to seven types.

Sesiano cites a 16C MS Ambros. P114 sup which gives  40 for 100 at  1/5, 1/10.  Answer:  960, -920.

Cardan.  Practica Arithmetice.  1539.

Chap. 47, f. L.iiii.v (p. 71).  End mentions Pacioli prob. 17.

Chap. 66, section 35, f. DD.v.r (pp. 145-146).  100 for 100 at  3, 2, ½.  (Pigs, asses, cows.)  (1, 1) solution given.  = Alcuin 34.

Chap. 66, section 67, ff. FF.ii.v - FF.iii.v (pp. 155-156).  (67 is not printed in the Opera Omnia.)  100 for 100 at  3, ½, _, 1/11.  (Turtledoves, thrushes, crested larks, sparrows.)  (18, 15) solutions, (1, 1) given.

Tartaglia.  General Trattato.  1556.  Book 16, art. 117‑129, pp. 254r‑255v  &  Book 17, art. 25, 26, 43, 44,  pp. 272v & 277r-277v.  18 versions.  The objects being bought are mostly not in the Italian dictionaries that I have consulted.  They are apparently 16C Italian, probably Venetian dialect.  Several Italian or Italian-speaking friends have helped to determine these _ my thanks to Jennifer Manco, Ann Maury, Ann Sassoon and especially Maria Grazia Enardu and a student of hers.

16-117.  60 for 60 at  4, 2, ½.  (Thrushes, larks, redstarts.)  (3, 2) solutions, he gives (1, 1).

16-118.  20 for 20 at  3, 2, ½.  (Partridges, pigeons, quails.)  (2, 1) solutions, he gives (1, 1).  (= Alcuin's 32.)  He discusses fractional solutions and gives one, but says it is not really acceptable.

16-119.  20 for 240 at  18, 10, 3.  (Sorghum, bran, grape seeds.)  (2, 0) solutions.  He finds  5, 15, 0,  then rejects it and finds a fractional solution but refers back to the previous discussion.

16-120.  40 for 480 at  36, 12, 1.  (Curlews or siskins, stock-doves, starlings.)  (2, 1) solutions, he gives (1, 1).

16-121.  40 for 40 at  3, 2, 1/5.  (Blackbirds, larks, sparrows.) (1, 1) solutions, which he gives.

16-122.  31 for 31 at  3, 2, _.  (Capons, ducks, thrushes.)  (1, 1) solutions, he gives (1, 1).

16-123.  100 for 100 at  3, 1, 1/20.  (Piglets, goats, weasels??)  (2, 1) solutions, he gives (1, 1).  (= Lucca 1754, p. 10r.)

16-124.  60 for 60 at  3, 1, 1/20.  (Same? animals.)  (2, 1) solutions, he gives (1, 1).

16-125.  100 for 100 at  3, 2, 1/20.  (Same? animals.)  (1, 1) solutions, which he gives.

          (Probs. 126‑129 involve men, women, children eating.) 

16-126.  12 for 12 at  2, ½, ¼.  (2, 1) solutions, he gives (1, 1).  (= Alcuin's 47.)

16-127.  15 for 15 at  4/3, 2/3, 1/3.  (I read the last  3  as a  2,  but the answer implies it is a  3.)  (3, 2) solutions, he gives (1, 1).

16-128.  18 for 18 at  2, 1, ½.  (7, 5) solutions, he gives (1, 1).

16-129.  20 for 20 at  4, ½, ¼.  (2, 1) solutions, he gives (1, 1).  = Pacioli 18.

17-25.  16 for 640 at  32, 50.  (Two types of cloth.)  This is determinate and he gives the solution.

17-26.  6 for 332 at  42, 66.  (Two types of cloth.)  Determinate _ he gives the solution.

17 _ 44-45.  See also Bachet, below. 

17-44.  100 for 100 at  3, 1, ½, _.  (Asses, pigs, sheep, goats.)  Quoted from Luca Pacioli, p. 105.  (276, 226) solutions _ he gives one but notes that the method of double false position does not totally resolve the problem, but that it can be solved by a combination of trying and of mathematics, but this is too long to describe here and he reserves it for another time.  = Pacioli 17

17-45.  200 for 200 at  12, 3, 1, ½, _.  (Mules, asses, pigs, goats, sheep.)  "Proposed to me in 1533 by a Genovese."  (8331, 6627)  solutions _ he gives one and indicates that more are possible, saying again that he will deal with this at another time.

Buteo.  Logistica.  1559. 

Prob. 66, p. 274.  50 for 160 at  (7, 2, 4, 1).  (Partridges, thrushes, quails, fig-peckers.)  Cites Pacioli for similar problems.  Gives (2, 2) of the (163, 144) solutions.

Prob. 67, pp. 274-275.   3x + 9y + 2z = 50;   7x + 3y + 6z = 70.   Cites Pacioli for a similar problem.  This has no non-negative integral solutions.  Gives three positive solutions, choosing different values to be integral.

Baker.  Well Spring of Sciences.  1562?  Prob. 11,  1580?: ff. 194r-195r;  1646: pp. 305‑306;  1670: pp. 346-347.  20 for 240 at  20, 15, 8.  (Payments to men, women, children.)  (1, 1) solutions, which he gives.

Bachet.  Problemes.  1612.  Addl. prob. X, 1612: 164-172;  1624: 237-247;  1884: 172‑179.

1612 cites  Tartaglia,  Pacioli,  de la Roche,  etc.

41 for 40 at  4, 3, _.  (1, 1) solution which he gives.

20 for 20 at  4, ½, ¼.  (2, 1) solutions _ he gives (1, 1).  (= Pacioli 18.)

He then describes the last two examples of Tartaglia.  For art. 44, he says there are 226 solutions and gives some.  For art. 45, my copy of Bachet has a defective type which made me think that the price of  3  was a  5,  but seeing Tartaglia has corrected this error.  Bachet says this has  6639  solutions and he gives the numbers for each given number of mules.  I find that for  7  mules, he has  571  instead of  570  and for  4  mules, he has  914  instead of  903,  which accounts for his extraneous numbers, but I cannot see why he might have miscounted.  Labosne adds a general argument for art. 44.

John Wallis.  A Treatise of Algebra, both Historical and Practical.  John Playford for Richard Davis, Oxford, 1685.  (Not = De Algebra Tractatus.)  Chap. LVIII, pp. 216-218.  ??NX

20 for 20 at  4, ½, ¼.  (Geese, quails, larks.)  = Pacioli 18.  He gives the positive solution of the (2, 1) solutions.  Cites Bachet and gives some general discussion.

100 for 100 at  3, 1, ½, 1/7.  Says the solutions are in Bachet, but they are not.  (121, 81) solutions.

Leybourn.  Pleasure with Profit.  1694.  Chap. XIII: Of Ceres and Virginum, pp. 51-55.

Quest. 1, p. 51.  8 for 20 at  4, 2.  (Geese, hens).  This is determinate.

Quest. 2, pp. 51-53.  21 for 26 at  2, 1, ½.  (Men, women, children.)  Some general discussion.  From  x + y + z  =  N  and  ax + by + cz  =  P,  assuming  a > b > c,  he gets  (a‑c)x + (b-c)y  =  P - cN  and deduces that  x  £  (P-cN)/(a-c).  He then assumes  x _ y  and hence that  x  cannot be much less than  (P-cN)/(a-c+b-c)  and he later drops the 'much' from this.  There are  (6, 5)  solutions; he finds all of them, but rejects the case  5, 16, 0  as the problem says there are 'some children'.

Quest 3, p. 53.  30 for 900 at  60, 40, 20.  (Ministers, lame soldiers, poor tradesmen.)  rom the above argument, he claims that  x ³ 3,  but then gives the solution  2, 11, 17  as though this showed that  x = 2  was impossible.  He finds  (6, 6)  of the  (8, 7)  solutions.

Quest 4, p. 54.  10 for 1000 at 50, 70, 130, 150.  (English, Dutch, French, Spanish) creditors.  Finds  (2, 2)  of the  (10, 4)  solutions and implies that all solutions are symmetric.

Quest 5, p. 55.  12 for 12 at 2, 1, ½, ¼.  (Different prices of loaves of bread.)  He finds  (2, 2)  of the  (11, 4)  solutions.

Ozanam.  1725.  [This is one of the few topics where the 1725, 1778 and 1803 editions vary widely _ see each of them.]

Prob. 24, question 8, p. 180.  41 for 40 at  4, 3, _.  (Men, women, children.)  (= Bachet's first.)  (1, 1) solutions which he gives.

Prob. 50, pp. 255‑256.  100 for 100 at  9, 1, ½, 3.  "Ce problême est capable d'un grand nombre de résolutions; ...."  It has  (73, 46)  solutions _ he gives  (3, 3).

Dilworth.  Schoolmaster's Assistant.  1743.  Part IV: Questions: A short Collection of pleasant and diverting Questions, p. 168.  Problem 3.  20 for 20 at  4, ½, ¼.  (Pigeons, larks, sparrows.)  This has (2, 1) solutions, none given.  (= Pacioli 18.)

Simpson.  Algebra.  1745.  Section XIII, quest. 2-8, 10-12, pp. 170-181  (1790: prob. II‑VII, IX, XI‑XV, pp. 183‑200).

2.  9x + 13y = 2000.  (17, 17) answers which he gives.

3.  Pay £100 (= 2000s) in guineas (= 21s) and pistoles (= 17s), i.e.  21x + 17y = 2000.  (6, 6) solutions which he gives.

4.  Pay £100 (= 2000s) in guineas (= 21s) and moidores (= 27s), i.e.  21x + 27y = 2000.  But  2000  is not a multiple of three, so there are no solutions, as he notes.

5.  Buy sheep at  17  and oxen at  140  to cost  2000.  (1, 1) solution which he gives.

6.  Men pay  42,  women  16,  to make  396.  (1, 1) solution which he gives.

7 (1745 only).                   20 for 16 at  2, 1, ¼.  (Loaves of bread of different values.)  Gives (1, 1) of the (2, 2) solutions.  He gets the equation  y = 14 - 2x + (2-x)/3  and neglects to allow  x = 2  in it.

8 (1745 only).                   20 for 20 at  3, 2, ½.  (Expenditure of men, women, children.)  Gives (1, 1) of the (2, 1) solutions.  = Alcuin 32.

10 (1790: XI).                   5x + 7y + 11z = 224.  This has (72, 59) solutions.  He gives (60, 60), but he has two erroneous solutions for  z = 14  instead of one.

11 (1790: XII).                  17x + 19y + 21z = 400.  (13, 10) solutions.  In 1745, there is an algebraic mistake and he gets (9, 9) solutions.  This is corrected to (10, 10) in 1790.

12 (1745 only).                 Pay £20 (= 400s) in pistoles (= 17s), guineas (= 21s) and moidores (= 27s).  Gives (9, 7) of the (9, 7) solutions, which is the only time here that he gives solutions with zeroes and is unusual for the time.  Cf. prob. XIV in the 1790 ed.

  The following are in the 1790 ed.

VII  12 for 12 at  2, 1, ¼.  (Loaves of bread of different values.)  (2, 1) solutions, he gives (1, 1).

IX.  87x + 256y = 15410  _  find the least solution.  In fact, there is only (1, 1) solution.

XIII.  7x + 9y + 23z = 9999.  This has  (34634, 34365)  solutions _ he describes the  34365  positive solutions.

XIV.  Pay £1000 (= 20000s) in crowns (= 5s), guineas (= 21s) and moidores (= 27s),  i.e.  5x + 21y + 27z = 20000.  This has  (70734,  70395)  solutions.  He describes all  70734.  Cf. Euler II.III.9;  Bonnycastle, 1782, no. 16.

XV.  12x + 15y + 20z = 100001.  There are  (1388611, 1388611)  solutions which he describes.

Dodson.  Math. Repository.  (1747?); 1775.  He also has several mixture problems and some simple problems which are mentioned at the entry for vol. II, below.

P. 16, Quest. XLI.  Pay  £50 (= 1000s) with  101  coins worth  21/2 s and  5s.

P. 139, Quest CCXXIII.  20 for 20 at  4, ½, ¼  (geese, quails, larks).  There are  (2, 1)  answers.  He gives the positive one.  = Pacioli 18.

P. 140, Quest CCXXIV.  Pay  £100 (= 2000s) with guineas (= 21s) and pistoles (= 17s).  Gives all (6, 6) solutions.  = Simpson 3.

P. 154, Quest. CCXXXVIII.  41 persons spent 40s; men spent 4s, women 3s, children _ s.  I.e.  41 for 40 at  4, 3, _.  He gives the (1,1) answer.  = Bachet's first.

Pp. 331-336, Quest CCI.  Divide 200 into five whole numbers  a, b, c, d, e,  so that  12a + 3b + c + e/2 + z/3 = 200.  I.e.  200 for 200 at  12, 3, 1, ½, _.  He finds 6627 answers.  There are  (8331, 6627)  answers.  = Tartaglia 45.

Les Amusemens.  1749.  Prob. 164, p. 310.  20 for 60 at  6, 4, 1.  (Men, women, valets.)  (3, 2) solutions _ he gives (1, 1).

James Dodson.  The Mathematical Repository.  Vol. II.  Containing Algebraical Solutions of A great Number of Problems, In several Branches of the Mathematics.  I.  Indetermined Questions, solved generally, by an elegant Method communicated by Mr. De Moivre.  II.  Many curious Questions relating to Chances and Lotteries.  III.  A great Number of Questions concerning annuities for lives, and their Reversions; wherein that Doctrine is illustrated in a Multitude of interesting Cases, with numerical Examples, and Rules in Words at length, for those who are unacquainted with the Elements of these Sciences, &c.  J. Nourse, London, 1753.  On pp. 1 - 63, he deals with 13 examples of determining the number of solutions of a linear equation in two (5 cases), three (6 cases) or four (2 cases) unknowns.  Because of the unusal extent of this, I will describe them all.  (I had to completely revise my programs for counting the number of solutions in order to deal with these.  The revision introduced double precision values for the counts and a way of computing the number of solutions for the two variable problem which speeded up the programs up by a factor of about 100, but one problem still took 2½ days!)  P. 1 has a subheading:  The Solution of indetermined Questions in Affirmative Integers, communicated by Mr. Abraham De Moivre, Fellow of the Royal Societies of London and Berlin.

Pp. 1-3, Quest. I.              35x + 43y = 4000.  (3, 3)  solutions which he gives.

P. 4, Quest II.                   71x + 17y = 1005.  (= p. 143, Quest CCXXVII of vol. I, 1775 version).  (1, 1)  solution which he gives.

Pp. 5-6, Quest III.   21x + 17y = 2000  _ pay  £100  with "guineas at  21,  and pistoles at  17  shillings each"  (= Simpson 3  = P. 140, Quest. CCXXXVIII of vol. I, 1775 version).  (6, 6)  solutions, which he gives.

Pp. 6-9, Quest IV.   5x + 8y = 1989.  (50, 50)  solutions which he gives.

P. 9, Quest V.                   3x + 5y = 173.  (12, 12)  solutions which he gives.

Pp. 10-14, Quest. VI.        3x + 5y + 8z = 10003.  (417584, 416250)  solutions of which he gives the positive ones.

Pp. 14-22, Quest. VII.       3x + 5y + 19z = 13051.  (299440, 298204)  solutions of which he gives the positive ones.  He describes De Moivre's analysis and then does a diffferent approach.

Pp. 23-34, Quest. VIII.      5x + 7y + 9z = 93256.  (13807365, 13801148)  solutions of which he gives the positive ones.  Again, he does this two different ways.

Pp. 34-35, Quest. IX.        3x + 5y + 9z = 1849.  (12710, 12546)  solutions.  He outlines the method, which leads to adding three arithmetic progressions, but he makes a simple error which leads to a considerably smaller number.

Pp. 36-37, Quest. X.         3x + 5y + 20z = 1849.  (5766, 5612)  solutions.  He outlines the method and it does yield the number of positive solutions.

Pp. 38-39, Quest. XI.        3x + 5y + 17z = 1849.  (6794, 6613)  solutions.  He outlines the method and it does yield the number of positive solutions.

Pp. 39-44, Quest. XII.       2x + 3y + 5z + 30w = 100003.  (185312986853, 185090752407)  solutions.  He sketches the method and says it all adds up to  160190378249  positive solutions.  I have not tried to locate his mistake, but one can estimate the number of solutions of  ax + by + cz + dw = e  as  e³/6abcd  which is  185201 x 10⁶  here, so it is clear that his answer is wrong.  Indeed, even his final addition is incorrect _ I get  160412356049.  He finds that the number of solutions for  w = 3333, 3332, ...  is a quadratic and sums this by Newton's interpolation formula.  I did some calculations and found that the number of positive solutions for w = 3333, 3332, ...  is  1, 24, 77, 160, 273, ...,  whose first differences are  23, 53, 83, ...  and second differences are constant at  30.  Hence the total number of positive solutions is obtained by adding  3333  terms, which is given by   0  +  1 * 3333  +  23 * 3333·3332/2  +  30 * 3333·3332·3331/6   and this gives the answer I found previously.  For reasons which I haven't tried to determine, Dodson gets   0  +  1 * 3333  +  20 * 3333·3332/2  +  26 * 3333·3332·3331/6.

Pp. 46-63, Quest. XIV.      3x + 57 + 19z + 143w = 91306.  (3121604438, 3104216955)  solutions.  He outlines the method but I haven't tried to see if it does yield the number of positive solutions.

Euler.  Algebra.  1770.

   II.I, pp. 302‑310.

Art. 8: Question 5.  Men pay  19,  women  13,  to total 1000.  He gives the general solution and all (4, 4) solutions.

Art. 9: Question 6.  Buy horses worth  31  and oxen worth  21  to cost  1770.  General solution and all (3, 3) solutions.

Art. 15: Question 9.  Men pay  25,  women pay  16.  All together the women pay  1  more than the men.  General solution and first few examples.

Art. 16: Question 10.  Horses cost  31  and oxen  20.  All together, the oxen cost  7  more than the horses.  General solution and first few examples.

Art. 17‑19.  General rule for  bp = aq + n.  Applies to the above problems and Chinese Remainder problems.

   II.II, pp. 311‑317.

Art. 25: Question 1.  30 for 50 at  3, 2, 1.  (Men, women, children.)  Gives all (11, 9) answers.

Art. 26: Question 2.  100 for 100 at  7/2, 4/3, ½.  This has (4, 3) solutions.  He gives (3, 3) answers and mentions  0, 60, 40  as another answer.

Art. 27.  Discusses when such problems are impossible, e.g.  100 for 51 at  7/2, 4/3, ½.

Art. 28: Question 3.  Combine silvers of quality  7, 11/2, 9/2  ounces per marc (= 8  ounces) to produce  30  marcs of quality  6.  I.e.  x + y + z = 30  and  7x + 11y/2 + 9z/2  =  6 x 30  or  30 for 180 at  7, 11/2, 9/2.  He gives all (5, 3) solutions.

Art. 29: Question 4.  100 for 100 at  10, 5, 2, ½.  (Oxen, cows, calves, sheep.)  Gives all (13, 10) answers.

Art. 30: Question 4.  3x + 5y + 7z = 560  and  9x + 25y + 49z = 2920.  Gives all (2, 2) answers.

   II.III: Questions for practice, p. 321.

No. 4.  Old guineas worth  21½  shillings and pistoles worth  17s  to make 2000s.  Gives all (3, 3) solutions.

No. 5.  20 for 20 at  4, ½, ¼.  (= Pacioli 18.)  Gives (1, 1) of the (2, 1) solutions.

No. 7.  Can one pay £100 (= 2000s) with guineas (= 21s) and moidores (= 27s)?  (= Simpson 4.)

No. 8.  How to pay  1s  to a friend when I only have guineas (= 21s) and he only has louis d'or (= 17s)?  I.e.  21x ‑ 17y = 1  with  x, y  positive.  He gives the least solution.  In fact there is only (1, 1) solution.

No. 9.  Same as Simpson XIV.

Ozanam-Montucla.  1778.  [This is one of the few topics where the 1725, 1778 and 1803 editions vary widely _ see each of them.]

Prob. 9, 1778: 195.  Pay  2000  with  450  coins worth  3  and  5,  i.e. 450 for 2000 at  3, 5.

Prob. 13, 1778: 204-205.  This is a complicated problem on the number of ways to make change _ see third section below.  He breaks it up into silver and copper coins as given in the first two sections.

                    Pay  60  with coins worth  60, 24, 12, 6.  There are (13, 0) solutions _ he says 13.

                    Pay  6k  with coins worth  2, 1½, 1, ½, ¼,  for  k = 1, 2, ..., 10.  There are (155, 2),  (1292, 194),  (5104, 1477),  (14147, 5615),  (31841, 15236),  (62470, 33832),  (111182, 65759),  (183989, 116237),  (287767, 191350),  (430256, 298046) solutions.  He gives the number of non-negative solutions in each case, but he gives four wrong values:  62400,  183999,  287777,  430264.

                    Pay  60  with coins worth  60, 24, 12, 6, 2, 1½, 1, ½, ¼.  There are  (1814151, 0)  solutions _ he says  1813899,  which corresponds to using all but the third of the above wrong values to do the final calculation.  Presumably the third wrong value is a misprint, rather than an error.

Prob. 23, 1778: 214-216.  A  must pay  B  31,  but  A  has only pieces worth  5  and  B  has only pieces worth  6,  i.e.  5x - 6y = 31.  Gives the general solution.

  The following are contained in the Supplement and have no solutions given.

Prob. 45, 1778: 433.  120 for 2400 at  12, 24, 60.  (Paying with coins.)  (21, 19) solutions.

Prob. 51, 1778: 434.  A,  B  and  C  have 100,  but nine times  A  plus  15  times  B  plus  20  times  C  equals  1500,  i.e. 100 for 1500 at  9, 15, 20.  Notes that this problem, and other similar problems, have several solutions and one should find them all.  This has (10, 9) solutions.

Prob. 52, 1778: 434-435.  120 for 20 at  3, 2, ½.  (Hares, pheasants, quails.)  [There must be a misprint here as this is clearly impossible.  Perhaps it should be 20 for 20,  which would  = Alcuin 32. ??]

Prob. 57, 1778: 435.  Pay 24 livres with demi-louis, pieces worth  6  livres and pieces worth  3  livres.  [I think a demi-louis is  10  livres,  so this would be  10a + 6b + 3c = 24.]  This has (5, 0) solutions.

Prob. 58, 1778: 435-436.  10a + 6b + 3c + 2d + e + ½f = 24.  [This is like the preceeding and I have assumed that the demi-louis is  10  livres.]  This has  (1178, 2)  solutions.

Bonnycastle.  Algebra.  1782.  Pp. 135-137 give a number of problems of finding some or all the integral solutions of one linear equation in two or three variables.

P. 135, no. 3.  Same as Simpson 2.  All answers given.

P. 136, no. 8.  Same as Simpson 10.  He says there are 60 solutions, but does not give them.

P. 136, no. 9.  Same as Simpson 4.

P. 136, no. 10.  Same as Simpson 3.

P. 136, no. 11.  Same as Bachet's first.

P. 136, no. 12 (1815: p. 158, no. 10).  Same as Euler II.III.8.  He gives the (1, 1) solution.

P. 136, no. 14.  Pay £351 (= 7020s) with guineas (21s) and moidores (27s).  Asks for the fewest numbers of pieces, and implies that there are  36  answers, but there are  (38, 37)  answers.  See p. 206, no. 49.

P. 137, no. 15.  Mix wines worth  18, 22, 24  per gallon to make  30  gallons worth  20,  i.e.  30 for 600 at  (18, 22, 24).  (6, 4) answers _ he gives the  4  positive ones.

P. 137, no. 16.  Pay £100 (=2000s) in crowns (5s), guineas (21s) and moidores (27s).  He says there are  70734  answers.  This is intended to be Simpson XIV, but that had £1000 (!).  This has only  (725, 691)  solutions.

P. 206, no. 49 (in 1805, no. 48 in 2nd ed., 1788.) 

                    "With guineas and moidores, the fewest, which way,

                    Three hundred and fifty-one pounds can I pay?

                    If paid every way 'twill admit of, what sum

                    Do the pieces amount to? _ my fortune's to come."

          This is the same problem as p. 136, no. 14, but here he says the answer is  9  guineas and  233  moidores (which was  257  in 2nd ed., 1788, but should be  253),  so he is ignoring the case with  0  guineas and  260  moidores _ or this is a misprint.  He says there are  37  solutions _ there are  (38, 37) solutions.

Hutton.  A Course of Mathematics.  1798?  Prob. 19,  1833: 221;  1857: 225.  Pay  £120 (= 2400s)  with 100 coins using guineas (= 21s) and moidores (= 27s), i.e.  100 for 2400 at  27, 21.

John Stewart.  School exercise book of 1801‑1802.  Described by:  W. More; Early nineteenth century mathematics; MG 46 (No. 355) (Feb 1962) 27‑29.  "Having nothing on me but guineas and having nothing on him but pistoles, I wish to pay him a shilling."  (= Euler II.III.8.)  (1, 1) solution given.

Ozanam-Hutton.  1803.  [This is one of the few topics where the 1725, 1778 and 1803 editions vary widely _ see each of them.]

Prob. 9, 1803: 192;  1814: 167;  1840: 86.  Pay  £2000  with  4700  half-guineas and crowns,  i.e.  4700 for 40000 at  10½, 5.  No solution.

Prob. 23, 1803: 209-210.  Prob. 22, 1814: 181-183;  1840: 94.  A  must pay  B  31,  but  A  has only pieces worth  7  and  B  has only pieces worth  5,  i.e.  7x - 5y = 31.  Gives the general solution.

  The following are contained in the Supplement and have no solutions given.

Prob. 45, 1803: 426;  1814: 361;  1840: 186.  Pay £100 (= 2000s) with guineas (= 21s) and pistoles (= 17s),  i.e.  21a + 17b = 2000.  (= Simpson 3.)

Prob. 51, 1803: 427;  1814: 362;  1840: 187.  Same as Ozanam-Montucla prob. 51.

Prob. 52, 1803: 427;  1814: 362;  1840: 187.  100 for 100 at  7/2, 4/3, ½.  (Calves, sheep, pigs.)  This has (4, 3) solutions.  (= Euler II.II.26.)

Prob. 57, 1803: 428;  1814: 362;  1840: 187.  Divide  24  into three parts  a, b, c  so that  36a + 24b + 8c = 516,  i.e. 24 for 516 at  36, 24, 8.  This has (3, 3) solutions.

Bonnycastle.  Algebra.  10th ed., 1815.

P. 158, no. 9.  Pay £20 (= 400s) with half-guineas (= 10½ s) and half-crowns (= 2½ s).  This gives  21x + 5y = 800.  (8, 7) solutions _ he says there are 7.

P. 158, no. 11.  Mix spirits worth 12, 15, 18 per gallon to make 1000 gallons worth 17.  This has no integral solutions _ he gives one fractional solution.

P. 228, no. 21.  Pay £100 (= 2000s) with 7s pieces and dollars (worth 4½ s).  I.e.  14x + 9y = 4000.  I find (31, 31) solutions, he says there are 21 _ a misprint?

P. 230, no. 34.  Spend 28s (= 336d) on geese worth 52d and ducks worth 30d.  (1, 1) solution, which he gives.

Augustus De Morgan.  Arithmetic and Algebra.  (1831 ?).  Reprinted as the second, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  Arts. 310-317, pp. 103-110, cover indeterminate problems in general. 

Art. 312, pp. 105-106.  Pay  2£ 16s  with crowns (= 5s) and three shilling pieces.  Describes the (4, 4) solutions.  Observes that negative solutions correspond to refunding.  Notes that paying in crowns (= 5s) and half-sovereigns (= 10s) is impossible.

Art. 314, pp. 107-108.  Discusses  4x + 9y + 10y  =  103  with no context.  Describes the (17, 13 solutions.

Art. 315, p. 108.  Discusses  14x + 11y + 9z  =  360;  x + y + z  =  30  without any context and describes how to solve such problems in general.  Gives (4, 3) of the (5, 3) solutions, omitting  18, 0, 12.

Louis Pierre Marie Bourdon (1779-1854).  Élémens d'Algèbre.  7th ed., Bachelier, Paris, 1834.  (The 6th ed., was 1831, but I haven't yet looked for the early editions _ ??) 

Art. 127, pp. 221-222.  Pay  78 fr.  with pieces worth  5 fr.  and  3 fr.  Sees that the number of  5 Fr.  pieces must be divisible by  3  and finds all (6, 5) solutions.

Art. 138, question 3, pp. 238.  = Euler II.II.28.

Art. 141, question 7, pp. 243-244.  = Euler II.II.29.

Unger.  Arithmetische Unterhaltungen.  1838.  Pp. 145-172 & 259-262, nos. 541-645.  He treats the topic at great and exhausting length, starting with solving  ax = by,  then  x + y = c,  then  ax + y = c,  then  ax = by + 1  and  ax = by + c,  also phrased as  ax º c (mod b).  He then does Chinese Remainder problems, but with just two moduli.  He continues with  ax + by = c,  initially giving just one solution, but then asking how many solutions there are.  He varies this in an uncommon way _ solve  x + y = e  with  x º a (mod b)  and  y º c (mod d).  He also varies the problem by letting  a, b  be rationals.  After 20 pages and 90 problems, he finally gets to one problem, no. 631, of the present type.  He then goes on to  ax + by = c (x + y),  but returns with 12 problems of our type, no. 634-645.  Of these, 640 and 645 have an answer with a zero, but he only gives the positive answers.  None of these problems are the same as any others that I have seen.

Pp. 153-154, no. 578.  Exchange fewer than  120  coins worth  1½  for coins worth  5_  with  a value of  14  left over.  I.e.  1½ x  =  5_ y  +  14.  Finds two answers.

P. 163 & 261, no. 616.  Change  50  into coins worth  3/8  and  2/9.  (9, 8) answers _ he gives the positive ones.  No. 617 also deals with coins.

Pp. 165-166, no. 631.  30 for 105 at  5, 3, 2 (unspecified goods).  (8, 7) answers _ he gives the positive ones.

PP. 167-168, no. 634.  50 for 395 at  5, 7, 12 (numbers).  (6, 5) answers _ he gives the positive ones.

Thomas Grainger Hall.  The Elements of Algebra: Chiefly Intended for Schools, and the Junior Classes in Colleges.  Second Edition: Altered and Enlarged.  John W. Parker, London, 1846.

P. 132, ex. 4.  55 crowns (= 5s) and shillings make £7 3s.

P. 133, ex. 18.  25 coins worth 1/30 and 1/15 to make 1.

Pp. 232-233, ex. 4.  How many ways can one pay £100 with sovereigns (= £1) and half-guineas (= 21/40 £)?  He gives all (5, 4) answers.

P. 233, ex. 5.  = Simpson 4.

John William Colenso (1814-1883).  The Elements of Algebra Designed for the Use of Schools.  Part I.  Longman, Brown, Green, Longmans & Roberts, London, (1849), 13th ed., 1858 [Advertisement dated 1849].  Exercises 63, p. 114 & Answers, p. 14.

No. 11.  Change a pound into  18  coins comprising half‑crowns (= 2½ s), shillings (= 1s) and six pences (= ½ s).  I.e.  20 for 18 at  5/2, 1, ½.  (4, 4) solutions, all given.

No. 16.  40 for 40 at  5, 1, ¼.  (Calves, pigs, geese.)  (3, 2) solutions, (2, 2) given.

Family Friend (Dec 1858) 357.  Arithmetical puzzles _ 3.  Same as Alcuin's 39 with (oxen, sheep, geese).  I haven't got the answer.

Leske.  Illustriertes Spielbuch für Mädchen.  1864?  Prob. 564-44, pp. 256 & 397: Arithmetisches Rätsel.  100 for 100 at  10, 3, ½.  (Geese, hares, partridges.)  (2, 1) answers, only (1, 1) given.

Editorial answer to F. Chapman, a correspondent.  Knowledge 2 (17 Nov 1882) 409.  100 for 100 at  10, 3, ½.  (Bullocks, sheep, pigs.)  (2, 1) answers, only (1, 1) given.  c= Leske.

Hoffmann.  1893.  Chap. IV, no. 35: Well laid out, pp. 152 & 200‑201.  21 for 24 at  2, 3, ½, 4.  (6, 2) solutions _ he gives (2, 2).

Clark.  Mental Nuts.  1904: no. 49;  1916: no. 63.  Turkeys and sheep.  "A drove of turkeys and sheep have 100 heads and feet.  How many are there of each?"  This leads to  3T + 5S = 100,  which has (7,6) answers.  He says there are  6  answers and gives one example.  His  1904: no. 20; 1916, no. 93   and  1904: no. 67  are ordinary versions, with only one positive answer in each case.

Pearson.  1907.  Part II, no. 155, pp. 144 & 222.  (= Alcuin's 39.)  Only the positive answer is given.

Loyd.  Cyclopedia.  1914.  Sam Loyd's candy puzzle, pp. 121 & 354.  = SLAHP: Assorted postcards, pp. 45 & 101.  = Pacioli 18.  Only the positive solution is given.

Williams.  Home Entertainments.  1914.  The menagerie, p. 127.  Menagerie of birds and beasts has  36  heads and  100  feet.  I.e.  36 for 100 at  2, 4.

Clark.  Mental Nuts. 

1916: no. 5.  How old is dad.  Dad, Ma, Bro & I.  D + M + B + I = 83;  6D = 7M;  M = 3I.  This yields  15I + 2B = 166,  This has  (6, 5)  solutions, but only one is possible.

1916: no. 13.  Change a quarter.  How many ways can you change a quarter (= 25¢) using  1¢, 5¢, 10¢  coins?  (12, 2) solutions; he says  12.

Collins.  Fun with Figures.  1928.  Four out of five have it, p. 183.  Hunter has shot birds and rabbits and has  36  heads and  100  feet.  I.e.  36 for 100 at  2, 4.

Loyd Jr.  SLAHP.  1928.  He gives a number of examples, usually with some extra feature.

Poultry profits, pp. 24 & 91.  100 for 100 at  .62, 1.02, 1.34.  If he makes profits of  .12, .22, .25  on each one, how does he maximize his profit?  (11, 10) solutions _ he says  9, 86, 5  is maximal, but  5, 95, 0  is better.

Easy come, easy go, pp. 25 & 91.  100 for 1000 at  100, 30, 5.  (2, 1) solutions _ he gives (1, 1).

Shooting mathematically, pp. 38 & 97.  Find least number  a + b + c  such that  6a + 12b + 30c  =  17 (a + b + c).

A puzzle in pants, pp. 42 & 99.  147 for 147 at  .49, .98, 2.45,  but he wants to maximize the minimum of the three numbers.  (37, 36) solutions in general.

An observing waiter, pp. 66 & 111.  This has men, women and couples, leading to  a + b + 2c  =  20,   .2a + .3b + 3c  =  20,    which has (1, 1) solutions which he gives.

Perelman.  FFF.  1934.  Hundred rubles for five.  1957: prob. 37, pp. 54 & 57-58;  1979: prob. 40, pp. 69‑70 & 72‑73.  = MCBF: prob. 40, pp. 67 & 70-71.  Magician asks for  20 for  500 or 300 or 200  at  50, 20, 5.  These are all impossible!

Adams.  Puzzle Book.  1939.  He has several straightforward problems, which I omit, and the following.  Prob. C.10, pp. 125 & 173.  Use  26d  in  florins (= 24d),  shillings (= 12d),  sixpence (= 6d),  pennies (= 1d)  and  half‑pences (= ½ d)  to measure  5_  inches.  The widths of the coins, in 16ths of an inch, are:  18,  15, 12, 19, 16,  respectively.  This leads to:   24a + 12b + 6c + d + e/2  =  26,   18a + 15b + 12c + 19d + 16e  =  90.  (1, 0) solution, which is given.

Depew.  Cokesbury Game Book.  1939. 

Change, p. 211.  Change a dollar into  50  coins.  I.e.  100 for 50 at  1, 5, 10, 25, 50.  (2, 0) answers, both given.

Six bills, p. 217.  Pay  $63  with six bills, no  $1  bills used.  I.e.  63 for 6 at  2, 5, 10, 20, 50.  (1, 0) answer, given.

McKay.  At Home Tonight.  1940. 

Prob. 3: A mixed bag, pp. 63 & 75-76.  50 for 50 at  5, 1, ½.  (Sheep, lambs, bundles of rabbits, where a bundle is one item.)  This has (6, 5) solutions, but it is added that he got the same number of two items and this has (2, 1) solutions of which he gives the positive one.

Prob. 5: A question of change, pp. 63 & 77.  I want to pay a friend  6/6  (= 78d),  but I only have  4s (= 48d)  pieces and he only has  half-crowns (= 30d).  I.e.  78 = 48x - 30y.  Gives the smallest solution.

Prob. 22: Brown at the market, pp. 66-67 & 81.  Spend 300 at  35, 25, 16.  (Cows, sheep, pigs.)  This has (5, 2) solutions, he gives the positive ones.

Prob. 32: A postage puzzle, pp. 69 & 82.  Spend  2s (= 24d) on  2d  and 4½d  stamps.  (2, 1) solutions, he gives the positive one.

McKay.  Party Night.  1940.  No. 24, pp. 181-182.  6 for 6 at  2, 1, ½.  (Men, women, children eating loaves of bread.)  Gives (1, 1) of the (3, 1) solutions.

Sid G. Hedges.  The Book of Stunts & Tricks.  Allenson & Co., London, nd [c1950?].  Shilling change, p. 45.  Change a shilling into  12  coins without using pennies, i.e.  12 for 12 at  6, 3, ½, ¼.  (2, 1) solutions _ he gives (1, 1).

The Little Puzzle Book.  Op. cit. in 5.D.5.  1955.  This has a number of examples which I include as illustrating mid 20C usage.

P. 6: Farm deal.  100 for 100 at  10, 5, ½.  (Cows, hogs, chickens.)  (1, 1) solutions, which he gives.  = Alcuin 5  = AR 119.

P. 19: Christmas savings.  100  coins make  500  cents, using  50, 10, 1.  (1, 1) solutions, which he gives.

Pp. 25-26: Arithmetic ability 1, 2, 3.   6  coins make  48, 52, 23  cents, using  25, 10, 5, 1.  There are  (1, 0),  (1, 1),  (1,0)  answers, which he gives.

P. 41: Buying stamps.  19 for 50 at  1, 2, 3,  with the condition that there be more  1s  than  2s.  (4, 3) answers, of which (1, 1) satisfies the condition and he gives it.

Joseph & Lenore Scott.  Master Mind Brain Teasers.  Op. cit. in 5.E.  1973.

Coin-counting, pp. 105-106.  100 for 500 at  1, 10, 25, 50.  (Coins.)  There are (9, 8) solutions _ only the one with a zero is given.

Change of a dollar, pp. 129-130.  50 for 100 at  1, 5, 10, 25, 50.  (Coins.)  (2, 0) solutions, both with no  50s.  Only the solution with a  25  is given.  = Depew, p. 211.

Shakuntala Devi.  Puzzles to Puzzle You.  Op. cit. in 5.D.1.  1976.  Prob. 143: The farmer and the animals, pp. 88 & 134.  Buy animals at  50, 40, 25, 10  (mules, sheep, goats, pigs)  to produce an average value of  30.  This is essentially an alligation problem as discussed under Fibonacci.  She gives one answer:  1, 1, 2, 1  and says "Other answers are possible" _ somewhat of an understatement since it has a three parameter set of solutions and two of the parameters can range to infinity!

Michael Holt.  Math Puzzles and Games.  Walker Publishing Co., NY, (1977), PB ed., 1983.  Hundred dollars for five, pp. 37-38 & 101-102.  10 for 500 at  10, 25, 50  _ making change, but requiring each type of coin to be used.  (1, 0) answers, hence impossible if each type of coin must be used.

Liz Allen.  Brain Sharpeners.  Op. cit. in 5.B.  1991.  Takeaway pay, pp. 83 & 133.  Nine employees of three types, earning  £5.00, £3.75 and £1.35  per hour earn  £333.60  in a shift of a whole number of hours.  How long was the shift?  This gives us   x + y + z  =  9,   500x + 375y + 135z  =  33360/n,   where  n  is the number of hours in a shift.  In the problem,  x  is given as  1,  but this information is not needed _ I have used this generalization for one of my problems.

G. F.  The crash.  Mathematical Pie, 129 (Summer 1993) 1022  &  Notes, p. 1.  Determinate version involving vehicles having  2, 4, 8  wheels  (motorcycles, cars, lorries),  giving   m + c + l  =  11,   2m + 4c + 8l  =  44,   m = 4.  Omitting the last equation would give us 11 for 44 at  (2, 4, 8),  which has (4, 3) solutions.

David Singmaster.  The hundred fowls, or how to count your chickens.  Accepted by Mathematics Review (Univ. of Warwick) for 1994, but it closed before the article was published.  General survey of the history.

J. Williams.  Mathematics and the alloying of coinage  1202-1700.  Annals of Science 52 (1995) 213-234 & 235-263.  ??NYS _ abstracted in BSHM Newsletter 29 (Summer 1995) 42 _ o/o.  Surveys the problem from Fibonacci through Kersey, etc. in the 17C.

Vladimir Dubrovsky.  Brainteaser B161 - Two-legged, three legged, and four legged.  Quantum 6:3 (Jan/Feb 1996) 15 & 48.  Room contains a number of people sitting on three-leeged stools and four-legged chairs.  There are no spare seats and the total number of legs is  39.  This gives  x = y + z,  2x + 3y + 4z = 39.

 

          7.P.2. CHINESE REMAINDER THEOREM

 

          See Tropfke 636.

          There are two forms of this.  The ancient Indians describe them as the residual pulveriser, where one is given the residues to several moduli, and the non-residual pulveriser, where one has  ax - c º 0 (mod b)  which is  ax - by = c.  The residual form is the classic Chinese Remainder Theorem, whose solution is generally found by reducing to the non-residual form, solved by the Euclidean algorithm.

          Notation.  Multiple congruences are written in an abbreviated notation.  E.g.  n  º  1 (mod 2)  and  n  º  2 (mod 3)  is written  n  º  1 (2),  2 (3)  or even more abbreviatedly as  n  º  -1 (2, 3).

          Standard problem types.

A‑k.     n  º  1 (2, 3, 4, 5, ..., k‑1),   0 (k).

          A-5.  See:  Tartaglia;  Baker;  Dilworth;  Jackson.

          A-7.  See:  Bhaskara I;  Ibn al-Haitam;  Fibonacci;  AR;  P. M. Calandri;  Chuquet; 

                    HB.XI.22;  Pacioli;  Tagliente;  Cardan;  Tartaglia;  Buteo;  Baker;  van Etten; 

                    Ozanam, 1725;  Vyse;  Dodson;  Badcock;  Magician's Own Book;  Wehman.

B-k.      n  º  0 (2, 3, 4, ..., k).

          B-9.  See:  Euler;  Bonnycastle 1782;  Jackson.

C-k.     n  º  -1 (2, 3, ..., k).

          C-5.  See:  Mahavira.

          C-6.  See:  Ladies' Diary, 1748;  Vyse;  Bonnycastle 1782.

          C-9.  See:  Tartaglia;  Leybourn;  Carlile.

          C-10. See:  Dell'Abbaco;  Muscarello.

          C-20. See:  Gentlemen's Diary, 1747;  Vyse.

C.        n  º  ‑1 (3, 4, 5, 6).  See: Brahmagupta;  Bhaskara I ??;  Bhaskara II.

D-k.     n  º  ‑1 (2, 3, 4, 5, ..., k‑1),   0 (k).

          D-5.  See: Baker;  Illustrated Boy's Own Treasury.

          D-7.  See: Fibonacci;  P. M. Calandri;  Pacioli;  Ghaligai;  Cardan;  Tartaglia;  Buteo; 

                    Baker;  van Etten;  Ozanam-Montucla;  Dodson.

E.         General result for  3, 5, 7.   See:  Sun Zi;  Fibonacci;  Yang Hui;  AR;  Pacioli; 

                    Tartaglia.

          I have a separate index of problems.

 

Sun Zi  (= Sun Tzu).  Sun Zi Suan Ching  (Master Sun's Arithmetical Manual).  4C.  [This is not the famous general and writer of The Art of War, c-4C.]  Chap. 3, prob. 26, p. 10b: There is an unknown number of things.  ??NYS.  n  º  2 (3),  3 (5),  2 (7)  & problem  E.  Only the least answer is given.  (See Needham, pp. 34 & 119.  English in Needham 119,  Mikami 32  and  Li & Du 93;  Chinese and English in Libbrecht 269.)

The problem has been transmitted as a folk rhyme in China and Japan.  It is known as  Sun Zi Ge  (The Song of Master Sun) or  Han Xin Dian Bin  (General Han Xin's Method of Counting Soldiers).  (Han Xin  was a general of c‑200.)  The rhyme is cryptic but gives the three multipliers  70, 21, 15  for problem  E.  An English version of a 1592 version of the rhyme and of  Sun Zi's  text is in:  Li Wenlin and Yuan Xiangdong; The Chinese Remainder Theorem, pp. 79‑110 of  Ancient China's Technology and Science, op. cit. in 6.AN.  Li & Yuan also note that such problems arose and were undoubtedly solved in calendrical calculations in the 3C.

Li & Du, pp. 93‑94, discuss several Chinese versions over the next centuries, including the rhyme of Li & Yuan.  Li & Du's p. 94 mentions the calculation of the Da Ming Calendar by Zu Chongzhi in 462 which probably used  11  simultaneous congruences, but the method has not survived.  (Li & Yuan say it was  10  congruences.)

Aryabhata.  499.  Chap. II, v. 32‑33, pp. 74-84.  (Clark edition: pp. 42‑50.)  Rule for the residual form, i.e. the Chinese Remainder Theorem.  The text is rather brief, but Shukla makes it clear that it is giving the Euclidean algorithm for the problem with two residues.  Shukla does an example with three residues and gives the general solution, though the text stops with one solution.  Shukla gives an alternative translation which would apply to the non-residual form of the problem and notes that later writers recognised the relation between the two forms.  (See Libbrecht 229,  Datta & Singh II 87‑99 & 131‑133  and  Bag, op. cit. under Bakhshali MS in 7.P.1, pp. 193‑204.)

Brahmagupta.  Brahma‑sphuta‑siddhanta.  628.  Chap. XVIII, sect. 1, art. 7.  In Colebrooke, pp. 326‑327.  After some astronomical data, he gives Problem  C  as an example.

                    The earlier part of Section 1 (pp. 325-326) deals with the general rule and the later part (pp. 327-338) gives astronomical examples.

Bhaskara I.  Mah_‑Bh_skar_ya.  c629.  Edited and translated by Kripa Shankar Shukla.  Lucknow Univ., 1960.  Chap 1, v. 41‑52, Sanskrit pages 7-9; English pages 29‑46.  These deal with the non-residual 'pulveriser' in its astronomical applications and it seems clearly illustrated.  Illustrative examples are provided by Shukla, from other works of Bhaskara, or from Chap 8 of stated, but not worked, examples to this work, e.g.   44789760000 x ‑ 101  =  210389 y  (chap. 8, no. 13).

Bhaskara I.  629.  Commentary to Aryabhata, chap. II, v. 32-33.  Sanskrit is on pp. 132-171; English version of the examples is on pp. 309-332.  Some further applications are discussed on pp. 191-196, 199-201, 332-334.  Bhaskara I gives 26 examples (and four further examples).  I will just give the simpler ones.  Shukla, pp. lxxxi-lxxxii, says Bhaskara I was the first to distinguish the residual and non-residual forms of the problem.  He also gives tables of solutions of  ax - 1 = by  for values of  a, b  of astronomical significance _ Shukla gives these in Appendix II, pp. 335-339.  Shukla asserts that the Indians were the first to find the general solution for the Chinese Remainder Problem and it was transmitted to China, c700.  However, see Li & Du above.

                    Ex. 1:  n  º  1 (5),  2 (7).

                    Ex. 2:  n  º  5 (12),  7 (31).

                    Ex. 3:  n  º  1 (7),  5 (8),  4 (9).

                    Ex. 4:  A-7.  First appearance; see beginning of section for list of later appearances.

                    Ex. 5:  8x + 6 º 0 (13).

                    Ex. 6:  11x - 3 º 0 (23).

                    Ex. 7:  576x - 86688 º 0 (210389). 

          All the later examples are similar, arising from astronomical calculations.

          (Bag, loc. cit, cites Examples 2 and 4 above and says  C  occurs, but it is not here.  Datta & Singh may imply that  C  occurs here.  Cf Brahmagupta.)  (See  Libbrecht 231‑232  and  Datta & Singh II 133‑135.)

Bhaskara I.  Laghu‑Bh_skar_ya.  c629.  Edited and translated by Kripa Shankar Shukla.  Lucknow Univ., 1963.  Chap 8, v. 17-18, Sanskrit pages 26-27; English pages 99‑102.  This is a simplified version of his earlier Mah_‑Bh_skar_ya, so it does not deal with the pulveriser, but these verses give some astronomical problems with number theoretic conditions which lead to uses of the pulveriser, e.g.  36641x - 24 º 0 (394479375).

Mahavira.  850.  Chap. VI, v. 121‑129, pp. 122‑123.  6 simple and 3 more complex examples, e.g. the following.

121.  n  º  7 (8),  3 (13).

122.  n  º  1 (2, 3, 4, 5).

123.  n  º  ‑1 (2, 3, 4, 5).

124.  n  º  1 (2),  0 (3),  3 (4),  4 (5).

127.  2n  º  3 (9),  3n  º  5 (11),  5n  º  2 (8).

Ibn al‑Haitam.  c1000.  ??NYS.  Problem  A‑7  (= Bhaskara I).  (English in Libbrecht, p. 234.)  (See also:  E. Wiedemann; Notiz über ein vom  Ibn al Haitam  gelöstes arithmetisches Problem; Sitzungsber. der phys. Soz. in Erlangen 24 (1892) 83.  = Aufsätze zur Arabischen Wissenschaftsgeschichte; Olms, Hildesheim, 1970, vol. 2, p. 756.)

Bhaskara II.  Bijaganita.  1150.  Chap. VI, v. 160 & 162.  In Colebrooke, pp. 235‑237 & 238‑239.

V. 160.  Problem  C  (= Brahmagupta).

V. 162.  n  º  1 (2),  2 (3),  3 (5).

Fibonacci.  1202.

Pp. 281‑282.  Problem  A‑7  (= Bhaskara I).  He says the answer is  301  but that one can add  420.  He doesn't mention this point in later problems.

Pp. 282‑283.  Problems  A‑11,  A‑23,  D-7,  D-10,  D-23.  These are all first appearances of these forms, and  A-23,  D-10,  D-23  never occur again, while  A‑11  only reappears in Tartaglia.

P. 304.  n  º  2 (3),  3 (5),  4 (7);  problem E  &  general result for  5, 7, 9.  (See Libbrecht, pp. 236‑238 for Latin and English.)

Chhin Chiu‑shao  (= Ch'in Chiu Shao  = Qin Jiushao).  Shu Shu Chiu Chang  (Mathematical Treatise in Nine Sections).  1247.  Complete analysis.  (See Libbrecht, passim.  See also Mikami 65‑69  and  Li & Yuan (op. cit under Sun Zhi).  (n  º  32 (83),  70 (110),  30 (135)  is given by Mikami 69.)

Yang Hui.  Hsü Ku Chai Ch'i Suan Fa.  1275.  Loc. cit. in 7.N, pp. 151‑153, problems 1‑5.

1.  n  º  2 (3),  3 (5),  2 (7)  & problem  E  (= Sun Zi).

2.  n  º  2 (3),  3 (5),  0 (7). 

3.  n  º  1 (7),  2 (8),  3 (9). 

4.  n  º  3 (11),  2 (12),  1 (13).

5.  n  º  1 (2),  2 (5),  3 (7),  4 (9).

Dell'Abbaco.  c1370. 

Prob. 114, p. 95.  n  º  ‑1 (2, 3, ..., 10).  Takes  10! ‑ 1.

Prob. 115, p. 96.  n  º  1 (2, 3, ..., 10).  Takes  7560 + 1  and says  75601  also works.

AR.  c1450.  Prob. 268, 311, 349.  Pp. 120‑121, 138‑139, 153, 181, 228‑229.

268.  Divinare.  Three moduli used for divination.  Gives the multipliers for triples;  3, 5, 7;  2, 3, 5;  3, 4, 5;  3, 4, 7;  2, 3, 7;  2, 7, 9;  5, 6, 7;  5, 8, 9;  9, 11, 13.

311.  Case  3, 5, 7  of prob. 268  = Problem  E.

349.  Problem  A‑7  (= Bhaskara I).

Correspondence of Johannes Regiomontanus, 1463?-1465.  Op. cit. in 7.P.1.

P. 219, letter to Bianchini, late 1463 or early 1464, question 8:  n  º  15 (17),  11 (13),  3 (10).

P. 237, letter from Bianchini, 5 Feb 1464.  Bianchini answers the above problem with  1103  and  3313  and says there are many more solutions, but he doesn't wish to spend the labour required to find more.  Curtze notes that Bianchini must not have understood the general solution.

P. 254, letter to Bianchini, nd [presumably 1464].  Notes that  1103  is the smallest solution and that the other solutions are obtained by adding the product of  17, 13 and 10,  namely  2210.  Curtze notes that Regiomontanus clearly understood the general solution.

P. 295, letter to von Speier, nd [apparently early 1465].  Prob. 6:  n  º  12 (23),  7 (17),  3 (10).

Muscarello.  1478.  Ff. 69r-69v, p. 180.  n  º  -1 (2, 3, 4, ..., 10)  (= dell'Abbaco).

P. M. Calandri.  c1480.  Pp. 68‑69.  Problems  A‑7 (= Bhaskara I),  D-7 (= Fibonacci).  He indicates the general answers.

Chuquet.  1484. 

Prob. 143.  English in FHM 227, with reproduction of original on p. 226.  Prob.  A‑7 (= Bhaskara I).  Gets  301  by trial and error and says there are other solutions, e.g.  721,  519841,  90601.  "Thus it appears that such questions may have several and divers responses."

Prob. 144.  n  º  2 (3, 4, 5, 6),  0 (7).  Mentioned on FHM 227, which erroneously implies  n  º  1 (3, 4, 5, 6).

HB.XI.22.  1488.  Pp. 52‑53 (Rath 247).  Prob. A‑7  (= Bhaskara I).

Pacioli.  De Viribus.  c1500. 

Prob. 22: A trovare un numero pensato non più de 105.  Problem  E. 

Prob. 23: same title with 105 replaced by 315.  General result for  5, 7, 9.

Prob. 24: Un numero ch' partito per  2, 3, 4, 5, 6  sempre avanzi  1  et partito per  7  avanzi nulla.  Problem  A‑7  (= Bhaskara I).

Prob. 25: A trovare un numero ch' partito in  2  avanza  1,  in  3,  2,  in  4,  3,  in  5,  4,  in  6,  5,  in  7  nulla.  Problem  D-7  (= Fibonacci).

Tagliente.  Libro de Abaco.  (1515).  1541.  Prob. 116, f. 57v.  Woman with basket of eggs _ problem  A‑7 (= Bhaskara I).

Ghaligai.  Practica D'Arithmetica.  1521.  Prob. 26, f. 66r.  Basket of eggs _ problem  D‑7 (= Fibonacci).

Cardan.  Practica Arithmetice.  1539.  Chap. 66, sections 63 & 64, ff. FF.i.v - FF.ii.r (pp. 154‑155).  Problems  A-7 (= Bhaskara I)  &  D-7 (= Fibonacci).

Tartaglia.  General Trattato, 1556, art. 146‑150, pp. 257v‑258v; art. 199, p. 264r. 

146.  Problem  A‑7  (= Bhaskara I).

147.  Problem  A‑5.  First appearance.

148.  Problem  A‑11  (= Fibonacci).

149.  n  º  ‑1 (2, 3, ..., 9).

150.  Problem  D-7  (= Fibonacci).

199.  Problem  E.

Buteo.  Logistica.  1559.  Prob. 70, pp. 279-280.  Problem  D-7 (= Fibonacci).  Discusses problem  A‑7 (= Bhaskara I)  and Cardan.

Baker.  Well Spring of Sciences.  1562?

Prob. 4,  1580?: ff. 190v-191r;  1646: pp. 300-301;  1670: pp. 342-343.  Probs.  A‑5 (= Tartaglia)  &  A‑7 (= Bhaskara I).

Prob. 5,  1580?: ff. 191v-192r;  1646: pp. 301-302;  1670: p. 342.  Problem  D‑5.  First appearance.

Prob. 6,  1580?: f. 192r;  1646: p. 302;  1670: pp. 343-344.  Problem  D-7 (= Fibonacci).

Bachet.  Problemes.  1612.  Prob. V: Faire encore le même d'une autre façon, 1612: 37-45.  Prob. VI, 1624: 84-93;  1884: 34‑37.  General solution for  3, 4, 5  used for divination.  Labosne adds case  2, 3, 5, 7  and a general approach.  1612 cites Forcadel, Gemma Frisius, Tartaglia, etc.

van Etten.  1624.  Prob. 51‑52 (46‑47), pp. 46‑48 (69‑71).  Problems  A‑7 (= Bhaskara I)  and  D-7 (= Fibonacci).  French ed. refers to Bachet for more detailed treatment.  Henrion's 1630 Notte to prob. 52, p. 18, says that Bachet has treated this problem. 

Seki K_wa.  Sh_i Shoyaku no H_.  1683.  ??NYS _ described in Smith & Mikami, pp. 123‑124.  Studies  ax - by = 1.   n  º  1 (5),  2 (7)  (= Bhaskara I).   Then extends to an number of congruences.  Then does the system   35 n  º  35 (42),  44 n  º  28 (32)  and   45 n  º  35 (50).

Leybourn.  Pleasure with Profit.  1694.  Prob. 17, pp. 40-41.  Prob.  C-9.  Constructs a table of  2·n! - 1,  n = 2, ..., 9,  and says  2·9! - 1  is the least solution.  But he then gives  2519 + 2520k,  k = 0, ..., 7  and says these are some of the infinite other solutions

Ozanam.  1694.  Prob. 23, 1696: 74-77;  1708: 65‑67.  Prob. 27, 1725: 188‑198.  Prob. 10, 1778: 195-198;  1803: 192-195;  1814: 167-169;  1840: 86-87.  1696 gives many examples, too numerous to detail, and some general discussion.  The following is common to all editions:   n  º  1 (2, 3, 5),  0 (7).  1725 has problem  A-7 (= Bhaskara I).  1778 et seq. has problem  D-7 (= Fibonacci)  and then notes that Ozanam would solve this as  119 (mod 5040)  rather than  119 (mod 420)  _ but the 1696 or 1725 ed. only have relatively prime moduli.

Dilworth.  Schoolmaster's Assistant.  1743.  Part IV: Questions: A short Collection of pleasant and diverting Questions, p. 168.  Problem 2.  Problem  A-5  (= Tartaglia).

Simpson.  Algebra.  1745.  Section XIII.

Quest. 1, p. 170 (1790: prob. I, p. 182).  n  º  7 (17),  13 (26).

Quest. 9, pp. 175-176.  (1790: prob. VIII, p. 187).  n  º  19 (28),  15 (19),  11 (15).

1790: Prob. X, pp. 189‑190.  General case for  28, 19, 15.  This is for computing the Julian year.  Cf Dodson.

Gentlemen's Diary, 1747 _ see Vyse, below.  C-20.

Ladies' Diary, 1748 _ see Vyse, below.  C-6.

Les Amusemens.  1749.  Prob. 171, p. 318.  n  º  1 (2, 3, 4, 6),  4 (5),  0 (7).

Euler.  Algebra.  1770.

   II.I.

Art. 13, pp. 305‑306.  n  º  2 (6),  3 (13).

Art. 14: Question 8, pp. 306‑307.  n  º  16 (39),  27 (56).

Art. 20: Question 11, p. 310.  n  º  3 (11),  5 (19).

Art. 21: Question 12, pp. 310‑311.  n  º  3 (11),  5 (19),  10 (29).

   II.III: Questions for practice.

No. 10, p. 321.  Problem  B-9.

Vyse.  Tutor's Guide.  1773?  The following are in a supplement in the Key only, but some are referred to earlier sources.

Prob. 1, Key p. 361.  C-20.  Attributed to the Gentlemen's Diary, 1747.

Prob. 3, Key p. 362.  A-7 in verse.

Prob. 4, Key pp. 362-363.  C-6.  Attributed to the Ladies' Diary, 1748.

Dodson.  Math. Repository.  1775.

P. 142, Quest. CCXXVI.  A-7  (= Bhaskara I).

Pp. 142-143, Quest CCXXVII.  D-7  (= Fibonacci).

Pp. 148-149, Quest. CCXXXVI.  x  º  n (28),  m (19).  Then gives an Example: "The cycle of the sun 17; and the cycle of the moon 13; being given; to find the year of the Dionysian period?"  This is the case  n = 17,  m = 13  of the general problem.

Pp. 150-151, Quest. CCXXXVII.  x  º  n (28),  m (19),  p (15).  Then applies to Julian period where the first is the cycle of the sun, the second is the cycle of the moon and the third is the Roman indiction.  Cf Simpson.

Pp. 151-153.  He continues the discussion to find the general solution for any number of moduli which are prime to each other.  Then does  x  º  1 (2),  2 (3),  3 (5),  4 (7),  5 (11).

Bonnycastle.  Algebra.  1782.  Pp. 137-140 (c= 1815: pp. 159-162) discuss the general method and give the following examples and problems.

No. 1.  n  º  7 (17),  13 (26).  (1815: no. 1,  = Simpson).

No. 2.  n  º  3 (11),  5 (19),  10 (29).  (1815: no. 2,  = Euler).

No. 3.  n  º  7 (19),  13 (28).

No. 4.  n  º  2 (3),  4 (5),  6 (7),  0 (2).

No. 5.  n  º  6 (16),  7 (17),  8 (18),  9 (19),  10 (20).

No. 6.  Problem  B-9.  (1815: no. 6,  = Euler).

No. 7.  n  º  1 (2),  2 (3),  3 (5),  4 (7),  5 (11)  (= Dodson).

P. 205, no. 35 (in 1805; 34 in 1788).  n  º  -1 (6, 5, 4, 3, 2).

Carlile.  Collection.  1793.  Prob. LXVIII, pp. 39-40.  n  º  ‑1 (2, 3, 4, 5, ..., 9)  (= Tartaglia).  He gives the answer:  2519.

Bonnycastle.  Algebra.  10th ed., 1815.  Pp. 159-162 is similar to the 1782 ed., but has the following different problems.

No. 3.  n  º  2 (6),  3 (13)  (= Euler).

No. 4.  n  º  5 (7),  2 (9).

No. 5.  n  º  16 (39),  27 (56)  (= Euler).

No. 6.  n  º  5 (7),  7 (8),  8 (9).

No. 8.  n  º  0 (2, 3, 4, 5, 6),   5 (7).

P. 230, no. 35.  n  º  -1 (6, 5, 4, 3, 2).

Badcock.  Philosophical Recreations, or, Winter Amusements.  [1820].  Pp. 49-50, no. 76: A quantity of eggs being broken, to find how many there were, without remembering the number.  Problem  A-7  (= Bhaskara I).

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions.

No. 15, pp. 17-18 & 74.  Problem  A-5  (= Tartaglia).

No. 20, pp. 19 & 76.  Problem  B-9  (= Euler).

Unger.  Arithmetische Unterhaltungen.  1838.  Pp. 145-172 & 259-262, nos. 541-645.  He treats the topic of 7.P.1 and 7.P.2 at great and exhausting length, starting with solving  ax = by,  then  x + y = c,  then  ax + y = c,  then  ax = by + 1  and  ax = by + c,  also phrased as  ax º c (mod b).  On pp. 156-158 & 260, nos. 588-597, are Chinese Remainder problems, but with just two moduli, so I will not describe them.

Magician's Own Book.  1857.  The basket of nuts, pp. 245-246.  Problem  A-7 (= Bhaskara I),  general solution.  = Book of 500 Puzzles, 1859, pp. 59-60.  = Boy's Own Conjuring Book, 1860, p. 217.

Illustrated Boy's Own Treasury.  1860.  Prob. 14, pp. 428 & 431.  Problem  D-5, one solution  (= Baker).

Wehman.  New Book of 200 Puzzles.  1908.  The basket of nuts, p. 55.  A7 (= Bhaskara I)  = Magician's Own Book, except that a line of the statement has been dropped by the typesetter, making the problem unintelligible.

 

          7.P.3. ARCHIMEDES' CATTLE PROBLEM

 

Archimedes?   Letter to Eratosthenes, c‑250. 

                    Greek text, with commentaries, first published by Gotthold Ephraim Lessing in Beiträge zur Geschichte und Literatur 1 (1773) 421‑???.

              Archimedes.  Opera Omnia.  Ed. by. J. L. Heiberg.  2nd ed., vol. II, Teubner, 1913.  Problema Bovinum, pp. 527‑534.  Heiberg gives the same classical references as Dijksterhuis (below), cites Lessing as the first editor of the problem (from Gud. Graec. 77, f. 415v), gives the later commentators and editors and says the problem also appears in Cod. Paris Gr. 2448, f. 57.  He then gives the Greek and a Latin translation.

              The first edition of Heiberg's edition is the basis of:

              T. L. Heath.  The Works of Archimedes (CUP, 1897)  +  The Method of Archimedes (CUP, 1912);  reprinted in one volume by Dover, 1953.  The Cattle‑Problem, pp. 319‑326, discusses the problem and the attempts at solving it.

              Dijksterhuis, p. 43, says Lessing's article occurs in the Zweiter Beitrag (1773), 2nd ed., Braunschweig, 1773 (??).  = Sämtliche Schriften; ed. by K. Lachmann, vol. IX, p. 285+; 3rd ed., much corrected by F. Muncker, Leipzig, 1897, vol. XIII (or 12??), pp. 99-115.

              English verse version in H. Dorrie; 100 Great Problems of Elementary Mathematics; Dover, 1965, pp. 5‑6.  Dorrie also cites the 19C historians on the question of authenticity.

                    Greek and English in SIHGM, vol. 2, pp. 202-207.  Thomas notes that the epigram is unlikely to have been actually written by Archimedes.  SIHGM, vol. 1, pp. 16-17, is a Scholium to Plato's Charmides 165 E which states "logistic ... treats on the one hand the problem called by Archimedes the cattle-problem" and Thomas gives some of the standard references in a note.

                    English translation in D. H. Fowler; Archimedes Cattle Problem and the Pocket Calculating Machine; Preprint, 1980, plus addenda.  (Based on SIHGM.)

Dijksterhuis.  Archimedes.  Op. cit. in 6.S.1.  P. 398 gives some classical references to the problem:  a scholium to Plato's Charmides;  Heron;  two references in Cicero _ ??all NYS.

T. L. Heath.  Diophantos of Alexandria.  Op. cit. as Diophantos.  1910.  Pp. 121‑124 discusses the problem.

 

J. F. Wurm.  Review of J. G. Hermann's pamphlet:  De archimedis problemate bovino; Leipzig, 1828.  In:  Jahn's Jahrbücher für Philologie und Pädagogik 14 (1830) 195-??.  ??NYS _ cited by Archibald.  Solves the easier interpretation, getting  5,916,837,175,686  cattle in all.

B. Krumbiegel.  Das problema bovinum des Archimedes.  Zeitschrift für Mathematik und Physik _ hist.-litterar. Abt. 25 (1880) 121-136.  ??NYS _ cited by Archibald.  Survey of earlier historical work.

A. Amthor.  ???.  Ibid, pp. 153-171.  ??NYS _ cited by Archibald.  Survey of the mathematics.

H. E. Licks.  Op. cit. in 5.A.  1917.  Art. 54, pp. 33‑39. discusses the work of Amthor & A. H. Bell (AMM (May 1895), ??NYS), who started the calculation of the answers.

[R. C. Archibald.]  Topics for club programs _ 14: The cattle problem of Archimedes.  AMM 25 (1918) 411-414.  He gives a detailed history, but some details vary from Dijksterhuis.  He cites Krumbiegel and Amthor as the basic works, Wurm as the first solver with the simpler interpretation and numerous other works.

H. C. Williams, R. A. German & C. R. Zarnke.  Solution of the cattle problem of Archimedes.  Math. Comp. 19 (1965) 671.  Plus comments by D. Shanks, ibid., 686‑687.  Describes full solution, but doesn't print it.

Harry L. Nelson.  A solution to Archimedes' cattle problem  &  Note.  JRM 13 (1980‑81) 162‑176  &  14 (1981‑82) 126.  First printed version of the solution _  206,545  digits.  The note clarifies the formulation of the problem.

 

          7.P.4. PRESENT OF GEMS

 

Bakhshali MS.  c7C.

Kaye I 40-42 interprets all of the following as examples of the form described for the third problem, which has some connection with 7.R.1.

Kaye I 40-42; III 170-171, f. 3v.  Hoernle, 1886, p. 129; 1888, pp. 33‑34, and Gupta describe it fully.  Three men have  7  horses,  9  ponies,  10  camels.  Each puts in three animals which are then divided equally, which is the same as each giving one to both others.  Then they are equally wealthy.  Solution gives  42, 28, 24  as values of the animals, though  21, 14, 12  is the smallest integral solution.  The original wealths ("capitals") of each merchant are  294, 252, 240  and the final wealth is  262.  (Kaye III 170 has  242  for  240.)

Kaye I 41; III 171, f. 3r says there is a fragment of another problem of this type with values  4, 5, 6.

Kaye I 40-42, III 168-169, ff. 1r-2r, sutra 11.  Gupta says the MS is poor and Kaye has misinterpreted it.  Gupta interprets it as leading to  x₁/2 + x₂ + x₃ + x₄ + x₅  =  h,  etc.,  where  h  is the price of a jewel.  The  i-th equation then reduces to  T  =  h + x_i (i‑1)/i,  where  T  is the total wealth.  Kaye considers it as  T ‑ x_i = h ‑ x_i/i.  Gupta converts this to a present of gems problem, but since the multipliers are non-integral, it takes a little more work. Answer:  120, 90, 80, 75, 72;  377.  I can't see that Kaye treats this any differently.

Kaye III 170, f. 2v is another example, with three values and diagonal coefficients  ‑7/12, ‑3/4, -5/6  and answer:  924, 836, 798;  1095.

Mahavira.  850.  Chap. VI, v. 162‑166, pp. 137‑138.  Gupta (op. cit. under Bakhshali MS) says the rule given is similar to the Bakhshali rule.

164.  6, 7, 8  giving one each. 

165.  16, 10, 8  giving two each.

Sridhara.  c900.  Gupta (op. cit. under Bakhshali MS) says that Sridhara gives the same rule as the Bakhshali MS, but allowing  n  people.  This rule is quoted in the Kriy_kramakari, a 1534 commentary on the Lilivati and Gupta quotes and translates it.  The Kriy_kramakari also quotes Mahavira, without attribution.  Gupta cannot locate this rule in Sridhara's extant works.

Bhaskara II.  Lilavati.  1150.  Chap. IV, sect. IV, v. 100.  In Colebrooke, p. 45.  Also in his Bijaganita, chap. IV, v. 111, pp. 195‑196.  8, 10, 100, 5  giving one each to others and all are equal.

 

          7.P.5. SELLING DIFFERENT AMOUNTS 'AT SAME PRICES' YIELDING

                                        THE SAME

 

          NOTATION:  (a, b, c, ...)  means the sellers initially have  a, b, c, ....   They all sell certain amounts at one price, then sell their remnants at a second price so that each receives the same amount.  Western versions give  a, b, c, ...  and sometimes the amount each receives.  The Indian versions give the proportion  a : b : c : ...  (by stating each person's capital, but not the cost price of the items; they invest their capital in the items and then sell them) and the larger price for selling the remnant of the items.  Further, the price for selling the first part of the items is the reciprocal of an integer.  (However the remnant price is sometimes a fraction.)  In both versions, the problem is indeterminate, with a  3  parameter solution set, but scaling or similarity or fixing the yield reduces this to  2.  There are also non‑negativity and integrality conditions.  The Indian version has infinitely many solutions, while the Western version gives a finite number of solutions.  I have recently found a relatively simple way to generate and count the solutions in the Western version, which is basically a generalization of Ozanam's example _ see my paper below.  The article by Glaisher discusses many of these problems.  As in 7.P.1,  (a, b)  solutions means  a  non-negative solutions of which  b  are positive solutions. 

          Versions where the earnings are different:  Ghaligai.

          See Tropfke 651.

 

                    Index of western versions.

 

(10, 20)                                      Abraham

(10, 30)                                      Fibonacci

(12, 32)                                      Fibonacci

(12, 33)                                      Fibonacci

(18, 40)                                      Labosne

(  7, 18, 29)                                McKay

(  8, 17, 26)                                Blasius

(10, 12, 15)                                Labosne

(10, 16, 22)                                Amusement

(10, 20, 30)                                Pacioli

(10, 25, 30)                                Ozanam

(10, 30, 50)                                Munich 14684, Dell'Abbaco, Provençale Arithmétique, Chuquet, HB.XI.22?, Widman, Demaundes Joyous, Tagliente, Ghaligai, Tartaglia, Jackson, Badcock, Boy's Treasury, Rowley, Hoffmann

(11, 33, 55)                                Tartaglia

(16, 48, 80)                                Tartaglia

(18, 40, 50)                                Labosne

(19, 25, 27)                                Williams & Savage

(20, 25, 32)                                Bachet

(20, 30, 40)                                Bachet, van Etten

(20, 40, 60)                                Tagliente

(27, 29, 33)                                Leske, Hoffmann, Pearson

(30, 56, 82)                                Widman

(31, 32, 37)                                Labosne

(60, 63, 66)                                Bath

(17, 68, 119, 170)                       Widman

(20, 30, 40, 50, 60)                     Dudeney

(305, 454, 603, 752, 901)            Widman

(20, 40, ..., 140)                         Glaisher, Gould

(10, 20, ..., 90)                           Tartaglia

 

Mahavira.  850.  Chap. VI, v. 102‑110, pp. 113‑116.  He gives a rule which gives one special solution of Sridhara's set of solutions.

V. 103.  Capitals:  2, 8, 36;  remnant price  6.

V. 104.  Capitals:  1½, ½, 2½;  remnant price  6.

V. 105.  Each receives  41;  remnant price  6.  What is the largest of the capitals?  (The other capitals are not determined.)

V. 106.  Each receives  35;  remnant price  4.  (Cf. v. 105.)

V. 108.  Capitals:  ½, _, ¼;  remnant price  6/5.

V. 110.  Capitals:  ½, _, ¾;  remnant price  5/4.

Sridhara.  c900.  V. 60‑62, ex. 76‑77, pp. 44‑49 & 94.  The verses are brief rules, which are expanded by editorial algebra, giving a one parameter family of solutions.

Ex. 76.  Capitals:  1, 3, 5  or  _, ¼, ½;  remnant price  3.

Ex. 77.  Capitals:  3/2, 2, 3, 5;  remnant price  ½.

Bhaskara II.  Bijaganita.  1150.  Chap. 6, v. 170.  In Colebrooke, pp. 242‑244.  Capitals  6, 8, 100;  remnant price  5.  Solution given is  3294, 4392, 54900,  which is one solution from Sridhara's set of solutions, but not by the same method as Mahavira.  The method is not clearly described.  Bhaskara says:  "Example instanced by ancient authors ....  This, which is instanced by ancient writers as an example of a solution resting on unconfirmed ground, has been by some means reduced to equation; and such a supposition introduced, as has brought out a result in an unrestricted case as in a restricted one.  In the like suppositions, when the operation, owing to restriction, disappoints; the answer must by the intelligent be elicited by the exercise of ingenuity."

Fibonacci.  1202.  Pp. 298‑302: De duobus hominibus, qui habuerunt poma.  He clearly states that there are two forums where the same prices are different. 

(10, 30)  _ he gives  5  solutions, there are  (55, 36).

(12, 32)  _ he gives  6  solutions, there are  (78, 55).

(12, 33)  _ he gives  1  solution, there are  (78, 55).

                    Glaisher, pp. 79-107, analyses this text in detail and finds that Fibonacci gives a reasonably general method which would give the  24  positive solutions with smaller price  1  in the first example.  He then considers the amounts received as being fixed.  He then permits the amounts received to differ, one receiving a multiple of what the other receives.  He also considers whether a solution exists for given amounts and prices and how to find solutions with one price given.

Abraham.  Liber Augmenti et Diminutionis.  Early 14C.  ??NYS _ cited by Tropfke 651.  (10, 20).  This has  (55, 36)  solutions.

Munich 14684.  14C.  Prob. XIII, pp. 79‑80.  (10, 30, 50).  Gives the solution with prices  1/7  and  3.  There are  (25, 16)  solutions.

Dell'Abbaco.  c1370.  Prob. 101, pp. 85‑87.  (10, 30, 50).  Gives the solution with prices  3  and  1/7.

Provençale Arithmétique.  c1430.  Op. cit. in 7.E.  F. 116v, pp. 62-63.  (10, 30, 50).  Gives the solution with prices  3  and  1/7  and then gives a solution with three prices!

Chuquet.  1484.  English in FHM 227-228.  Prob. 145.  (10, 30, 50).  Same two solutions as in Provençale Arithmétique.  FHM says it appears in one of Dudeney's books, where he expresses "grave dissatisfaction with the answer".

HB.XI.22.  1488.  P. 54 (Rath 247).  Rath doesn't give the numbers and says it is similar to Munich 14684 and p. 73v of Cod. Vindob. 3029.  Glaisher dates Cod. Vindob. 3029 as c1480.

Johann Widman.  Op. cit. in 7.G.1.  1489.  F. 134v+.  ??NYS _ discussed by Glaisher, pp. 1‑18.  (10, 30, 50)  _ one solution with prices  3  and  1/7.  He then generalises this example to construct single solutions for other examples:  (30, 56, 82),  (17, 68, 119, 170),  (305, 454, 603, 752, 901),  which have  (225, 196),  (45, 35),  (11552, 11400)  solutions respectively.  Glaisher then describes several examples that Widman might have constructed.

Pacioli.  De Viribus.  c1500.  Prob. 65: Dun mercante ch' a  3  factori et a tutti manda a uno mercato con perli.  (10, 20, 30).  Gives the solution with prices  1  and  1/6,  which has a part of zero.  There are  (25, 16)  solutions.

Anon.  Demandes joyeuses en manière de quodlibets.  End of 15C.  ??NYS.  Selected and translated as: The Demaundes Joyous.  Wynken de Worde, London, 1511.  [The French had 87 demandes, but the English has 54.  This is the oldest riddle collection printed in England, surviving in a single example in Cambridge Univ. Library.  Often attributed to de Worde.  Santi 9 uses Yoyous and Wynkyn and list de Worde as author.]  Facsimile with transcription and commentary by John Wardroper, Gordon Fraser Gallery, London, 1971, reprinted 1976.  Prob. 50, pp. 6 of the facsimile, 26-27 of the transcription.  (10, 30, 50)  apples.  One solution with prices  3  and  1/7.

Blasius.  1513.  F. F.iii.r: Decimaquarta regula.  Selling eggs _  (8, 17, 26).   There are  (16, 9)  solutions.  He give one with prices  2  and  1/5  and each sold as many batches of  5  as possible.  Discussed by Glaisher.

Tagliente.  Libro de Abaco.  (1515).  1541.  Prob. 115, ff. 57r-57v.  Women selling eggs _  (10, 30, 50).  One solution with prices  1/7  and  3.  Glaisher, below and in op. cit. in 7.G.1, cites  Hieronymus Tagliente and says the 1515 & 1527 editions give  (10, 30, 50)  with solution at prices  1/7  &  3,  and the 1525 ed. has  (20, 40, 60)  with solution at prices  3  and  1/7.  The latter has  (100, 81)  solutions.

Ghaligai.  Practica D'Arithmetica.  1521. 

Prob. 21, ff. 65r-65v.  (10, 30, 50).  One solution, with prices  1/7  and  3.  Glaisher says it is p. 66 (misprinted 64) in the 1548 ed.  (H&S 53 gives Italian from 1552 ed., but no solution.)  Ghaligai says the problem was known to Benedetto (da Firenze, flourished 1470s) and Ghaligai's teacher Giovanni del Sodo, as a problem outside of any rule, and Ghaligai labels it  "Ragione apostata" (Exceptional problem).

Prob. 22, f. 65v.  (10, 50)  with first making twice the second.  Solution with prices  13  and  1/7.

Tartaglia.  General Trattato, 1556, art. 136‑139, pp. 256r‑256v.

136.  (10, 30, 50)  yielding  10  each.  He gives the solution with prices  3  and  1/7.  There are  (25, 16)  solutions.

137.  (11, 33, 55)  yielding  11  each.  He gives the solution with prices  2  and  1/6.  There are  (30, 20)  solutions.

138.  (16, 48, 80)  yielding  16.  He gives the solution with prices  3  and  1/11.  There are  (64, 49)  solutions.

139.  (10, 20, ..., 90),  yielding  100.  He gives the unique positive solution.  There are  (3, 1)  solutions.

Bachet.  Problemes.  1612.  Prob. XXI, 1612: 106-115.  Prob. XXIV, 1624: 178-186;  1884: 122‑126.  He gives a new "general and infallible rule", which is fairly general _ Glaisher says it produces a selection of the solutions.  He applies the idea to  (20, 30, 40),  exhibiting  4  solutions.  It has  (100, 81)  solutions.  Cf. van Etten.  1612 also does  (20, 25, 32).

                    In the 5th ed., the general material is dropped and replaced by some vague algebra.  Labosne gives two solutions for  (18, 40),  but one of them uses fractions.  It has  (171, 36)  solutions.  He then considers  (18, 40, 50)  and gives one fractional solution _ there are  (3, 1)  solutions.  He then makes some discussion of  (10, 12, 15)  (which has  (7, 4)  solutions) and  (31, 32, 37)  (which has  (70, 60)  solutions).

van Etten.  1624.  Prob. 69 (62), pp. 64‑65 (90‑91).  (20, 30, 40).  Gives one solution with prices  3  and  1.  There are  (100, 81)  solutions.  Cf. Bachet.  Henrion's 1630 Notte, p. 22, states that Bachet found many other solutions and gives a solution with prices  2  &  7. 

Ozanam.  1694.  Prob. 24, 1696: 77-80;  1708: 68‑70.  Prob. 28, 1725: 201‑210.  Prob. 12, 1778: 199-204;  1803: 196-201;  1814: 170-174;  1840: 88-90.  (10, 25, 30).  Glaisher describes the material in the 1696 ed. and says that Ozanam first considers the same general form that Bachet considered and then applies it to the example.  Glaisher indicates that Ozanam's and Bachet's methods are essentially the same, but Ozanam certainly gets all solutions, while I am not sure that Bachet can do so.  1696 gives two solutions at prices  7  and  2  and at prices  6  and  1.  1725 et seq. gives a general method and finds all  10  solutions of the specific problem.  (Glaisher notes that the Remarques on pp. 203‑210 are new to the 1723 ed.  They give an algebraic form of the solution.)  1725 refers to the second part of Arithmétique Universelle, p. 456 (more specifically identified as by M. de Lagny [1660-1734] in 1778 et seq.), where  6  solutions are found.  1725 says there are  10  solutions and 1778 says de Lagny is mistaken _ but in fact, there are  (10, 6)  solutions and de Lagny probably meant just the positive ones.  1778 drops the preliminary general general form.

Amusement for Winter Evenings.  A New and Improved Hocus Pocus; or Art of Legerdemain: Explaining in a Clear and Comprehensive Manner Those Apparently Wonderful and Surprising Tricks That are performed by Slight of Hand and Manual Dexterity: Including Several Curious Philosophical Experiments.  M. C. Springsguth, London, nd [c1800 _ HPL], 36pp.  Pp. 22-23: Of three sisters.  (10, 16, 22)  sold at  7 a penny  and then a penny apiece, i.e. prices  1/7  and  1,  each earning  4.

Bestelmeier.  1801.  Item 718: Das Eyerverkauf.  Three women sell different numbers of eggs and make the same.  Further details not given.

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions.  No. 14, pp. 17 & 74.  (10, 30, 50).  Gives the solution with prices  3  and  1/7.

John Badcock.  Domestic Amusements, or Philosophical Recreations.  Op. cit. in 6.BH.  [1823].  P. 155, no. 192: Trick in reasoning.  (10, 30, 50)  _ gives the solution with prices  3  and  1/7.

Boy's Treasury.  1844.  Amusements in arithmetic: Three country-women and eggs, p. 299.  (10, 30, 50)  _ gives the solution with prices  3  and  1/7.

Leske.  Illustriertes Spielbuch für Mädchen.  1864?  Prob. 564-21, pp. 254 & 395-396: Die Blumenmädchen.  (27, 29, 33).  This has  (117, 100)  solutions _ she gives one with prices  _  and  1,  each earning  13.

Hugh Rowley.  More Puniana; or, Thoughts Wise and Other-Why's.  Chatto & Windus, London, 1875.  P. 179.  Man with three daughters and lots of apples _  (10, 30, 50).  Solution with prices  3  and  1/7.

Hoffmann.  1893.  Chap. IV.

No. 65: The three market‑women, pp. 160 & 215.  (27, 29, 33).  c= Leske.

No. 66: The farmer and his three daughters, pp. 160 & 216.  (10, 30, 50).  He gives the solution with prices  3  and  1/7.

Dudeney??  Breakfast Table Problems No. 328:  "How were the oranges sold".  Daily Mail (27  &  28 Jan 1905) both p. 7.  (20, 30, 40, 50, 60).  Gives the solution with prices  1  and  1/11.  There are  (45, 36)  solutions.

Pearson.  1907.  Part II, no. 37, pp. 121 & 199.  (27, 29, 33)  c= Leske.

J. W. L. Glaisher.  On certain puzzle-questions occurring in early arithmetical writings and the general partition problems with which they are connected.  Messenger of Mathematics 53 (1923-24) 1-131.  Discusses the versions in Blasius, Widman, Tagliente  and attempts to explain the methods used.  Unfortunately he is rather prolix and I often get lost in the many examples and special cases, but he seems to have general solutions.  On p. 12, he mentions that Tagliente's problem could be extended to  (20, 40, ..., 140)  _ cf. Gould below.  On p. 77, he says more results will appear in a later paper _ check index of Messenger??

A. A. Krishnaswami Ayyangar.  A classical Indian puzzle-problem.  J. Indian Math. Soc. 15 (1923-24) 214-223.  Responding to Glaisher, he analyses the Indian version, obtaining a simple complete solution with two parameters having an infinite range and a third parameter being bounded.  I found this a bit confusing since he sometimes uses price to mean the number of items per unit cost.  He says many of the solutions are not in Glaisher's system given on p. 19, but I can't tell if Glaisher intends this to be a complete solution.

Rupert T. Gould.  The Stargazer Talks.  Geoffrey Bles, London, 1944.  A Few Puzzles _ write up of a BBC talk on 10 Jan 1939, pp. 106-113.  Seven applewomen with  20, 40, 60, 80, 100, 120, 140.  One solution with prices  3  and  1/7  _ cf. Glaisher.  There are  (27, 21)  solutions.

McKay.  At Home Tonight.  1940.  Prob. 16: Extraordinary sales, pp. 65 & 81.  (7, 18, 29)  eggs.  This has  (12, 6)  solutions.  He asks for a solution where all make  10d.  He gives one solution with prices  1/4  and  3  and selling as many eggs in batches of  4  as possible.

W. T. Williams & G. H. Savage.  The Penguin Problems Book.  Penguin, 1940.  No. 75: A falling market, pp. 43 & 128.  Cauliflowers  (19, 25, 27)  with each making  85d,  both prices being integral and each sells some at the lower price, not to be less than  2d.  Actually there is only one solution with integral prices and each making  85d.

Philip E. Bath.  Fun with Figures.  Op. cit. in 5.C.  1959.  No. 45: Potatoes for sale, pp. 19 & 47.  (60, 63, 66).  This has  (900, 841)  solutions.  He wants solutions where all make  9/6 = 114d.  There are two solutions with integral prices, but he gives the solution with prices  2d  and  12/7 d,  i.e.  7 for 1s,  but he doesn't require maximum numbers of batches of  7  to be sold.

David Singmaster.  Some diophantine recreations.  In:  Papers Presented to Martin Gardner on the occasion of the opening of the exhibition: Puzzles: Beyond the Borders of the Mind at the Atlanta International Museum of Art and Design; ed. by Scott Kim, 16 Jan 1993, pp. 343-356  AND  The Mathemagician and Pied Piper  A Collection in Tribute to Martin Gardner;  ed. by Elwyn R. Berlekamp & Tom Rodgers;.  A. K. Peters, Natick, Massachusetts, 1999, HB, pp. 219-235.  Sketches some history, gives complete solutions for the Western and Indian cases (filling a gap in Ayyangar) and finds a new simple formula for the number of solutions in the Western case.

 

          7.P.6. CONJUNCTION OF PLANETS, ETC.

 

          See Tropfke 642. 

          Some overtaking problems in 10.A take place on a circular track and are related to or even identical to these problems.  In particular, if two persons start around an island of circumference  D, from the same point and in the same direction at rates  a, b,  this is the same as  O-(a, b; D)  of Section 10.A.  This is easily adapted to deailing with different starting points and going in opposite directions (which gives a meeting problem).  Clock problems, 10.R, are also related to these.

 

Sun Zi.  Sun Zi Suan Ching.  Op. cit. in 7.P.2.  4C.  ??NYS.  Sisters come home every  5, 4, 3  days, when do they all come together?.  (Mikami 33 gives English.)

Zhang Qiujian.  Zhang Qiujian Suan Jing.  Op. cit. in 7.E.  468,  ??NYS _ translated on p. 139 of:  Shen Kangsheng; Mutual‑subtraction algorithm and its applications in ancient China; HM 15 (1988) 135‑147.  Circular route of length 325.  Three persons start with speeds  150, 120, 90.  When do they all meet at the start?

Brahmagupta.  Brahma‑sphuta‑siddhanta.  628.  Several of the sections of Chap. XVIII discuss astronomical versions of this problem, but with complicated values and unclear exposition.

Gherardi.  Libro di ragioni.  1327.  P. 47.  Two men start going in a circuit.  One can do it in  4  days, the other in  5½  days.  When do they meet again?  Same as  O-(1/4, 1/5½),  D = 1,  in the notation of Section 10.A.

AR.  c1450.  Prob. 148, pp. 72, 164‑165, 214.  Though titled 'De planetis' and described as conjunction by Vogel, this is really just an overtaking problem _ see 10.A.

H&S 74‑75 says sun and moon problems are in van der Hoecke and Trenchant (1556).

Cardan.  Practica Arithmetice.  1539.  Chap. 66, sections 20-24, ff. CC.vii.v - DD.i.r (pp. 142‑143).  Several versions concerning conjunctions of planets, including irrational ratios and three planets.  Examples with periods:  Ö7, Ö5;  Ö18, Ö30;  Ö8, Ö20;  1000, 999;  and  Ö5, Ö4, Ö3.  In section 23, he gives periods of Saturn and Jupiter as  30  &  12  years and periods of Jupiter and Mars as  144  &  23  months.  (H&S 75 gives English and some of the Latin.)

Cardan.  Opus Novum de Proportionibus Numerorum.  Henricpetrina, Basil, 1570, ??NYS.  = Opera Omnia, vol. IV, pp. 482-486.  General discussion and examples, e.g. with periods  2, 3, 7.

Vyse.  Tutor's Guide.  1771?  Prob. 20, p. 85 & Key p. 110-111.  Island  73  in circumference;  three persons set out in the same direction at speeds  5, 8, 10.  When do they all meet again?

Bonnycastle.  Algebra.  1782.  P. 86, no. 23.  Identical to Vyse.

Pike.  Arithmetic.  1788.  P. 353, no. 31.  Island  50  in circumference.  Three walkers start in the same direction at speeds  7, 8, 9.  When and where do they meet again?

Hutton.  A Course of Mathematics.  1798?  Prob. 37,  1833: 223;  1857: 227.  Identical to Vyse.

Kaida Anmuyo.  c1800.  Problem given on pp. 139‑140 of Shen Kangsheng, loc. cit. under Zhang Qiujian above.  Assume  365¼  degrees in a circle.  Five stars are in a line and travel at speeds of  28 13/16,  19 1/4,  13 5/12,  11 1/7,  2 7/9  degrees per day.  When do they meet at the starting point again?

Anonymous.  A Treatise on Arithmetic in Theory and Practice: for the use of The Irish National Schools.  3rd ed., 1850.  Op. cit. in 7.H.  P. 360, no. 48.  Three walkers start to circle an island of circumference  73,  at rates  6, 10, 16.  When do they meet again?

James B. Thomson.  Higher Arithmetic; or the Science and Application of Numbers; ....  Designed for Advanced Classes in Schools and Academies.  120th ed., Ivison, Phinney & Co, New York, (and nine copublishers), 1862.  Prob. 94, p. 308 & 422.  Same as Anon: Treatise.

Lemon.  1890.  The Maltese cross, no. 483, pp. 63 & 115.  Walkers complete  6, 9, 12, 15  circuits per hour _ when are they all again at start?

Hoffmann.  1893.  Chap. IV.

No. 38: When will they get it?, pp. 152 & 202.  Guests come to restaurant with periods  1, 2, ..., 7  days.  When do they all meet again?

No. 47: The walking match, pp. 154 & 207.  Four men walk around a track of length  1  with speeds  5, 4, 3, 2  per hour.  When do all meet at start again?

Dudeny.  "The Captain" puzzle corner.  The Captain 3:2 (May 1900) 97 & 179  &  3:4 (Jul 1900) 303.  No. 2: The seven money boxes.  Boy puts a penny in  i-th box on  i-th day, where day  1  is  1 Jan 1900.  When he has to put in seven pennies, he will then open them all up.  When is this and how much will he have?  Answer:  420 days  =  24 Feb 1901  and  £4 10s 9d.

Depew.  Cokesbury Game Book.  1939.  Bicycle racers, p. 221.  One can travel around the track in  6  minutes, the other in  9  minutes.  When are they together again?

 

          7.P.7. ROBBING AND RESTORING

 

          Men keep money together and divide it into amounts   x₁,  x₂,  ...   _ usually by robbing the common fund.  They put fractions  a_ix_i  into a pool and divide the pool in proportion  b₁ : b₂ : ....  They then have money in the proportion  c₁ : c₂ : ...,  or actual amounts  d₁, d₂, ....

          I use  A  for  (a₁, a₂, a₃),  etc.

 

Kurt Vogel.  Ein unbestimmtes Problem al-Kara__s in Rechenbüchern des Abendlands.  Sudhoffs Archiv 61 (1977) 66-74.  Gives the history of this problem, particularly the transmission to Fibonacci via John of Palermo and the different methods of solving it _ Fibonacci gives three methods.  He mentions all the entries below except Gherardi and Calandri, which had not been published when he wrote, and Pacioli, which is sad because Pacioli is not very clear!

Jacques Sesiano, op. cit. under 7.R, 1985, discusses this problem along with the problems in 7.R.  He calls it "The disloyal partners".  He cites Ab_ K_mil's Algebra, ff. 100r-101r, ??NYS, and says it is solved two ways there.  Martin Levey's 1966 edition of the Algebra does not give the problem and states that the unique Arabic MS ends on f. 67r, and uses a 'better' Hebrew text.  The Arabic MS actually continues and the third part of the book, ff. 79r-111r, was treated by Schub & Levey in 1968 & 1970.  Sesiano; Les méthodes d'analyse indéterminée chez Ab_ K_mil; Centauraus 21 (1977) 89-105 is scathing about the work of Levey and of Schub & Levey, saying the Hebrew MS is third-rate, and the translators have made serious mathematical and philological errors.  Sesiano studies some of Abu Kamil's problems in this article, but unfortunately the problem of this section is not among them.

 

al‑Karkhi.  c1010.  Sect I, no. 45 & 47;  sect III, no. 6;  pp. 80‑81 & 90.

I‑45: Two men have  d₁ = 40  and  d₂ = 60.  From the common sum, they take  x₁  and  x₂  =  100 ‑ x₁.  The first gives  a₁ = 1/4  of what he took to the second and the second gives  a₂ = 1/5  of what he took to the first.  Then they have correct amounts.  Answer:  (400, 700)/11.

I‑47: Usual version with two people.  A = (_, ¼),   B = (1, 2),   D = (30, 70).  Answer:  (156, 344)/5.

III‑6: Usual version:  A = (1/2, 1/3, 1/6),   B = (1, 1, 1),   C = (3, 2, 1).  Answer:  33, 13, 1.

Fibonacci.  Flos.  c1225.  Pp. 234-236: De tribus hominibus pecuniam comunem habentibus.  = Fibonacci, below, pp. 293-294.  = al‑Karkhi's III‑6.  Problem II in Picutti, pp. 310‑312. 

                    Cantor; op. cit. under Fibonacci in 7.P.1; 1863; p. 345, discusses the contest between Fibonacci and John of Palermo before Frederick II at Pisa in 1225 (or 1226) and says this problem is the third and last of the contest problems.  However both Flos and Picutti have this as the second problem.  Licks, op. cit. in 5.A, says it was problem 5 in the contest.  Vogel doesn't mention which problem it was.

                    Leonardo says he later found three further methods of solution, which are "in libro uestro, quem de numero composui, patenter inserui".  Leonardo is here addressing the Emperor, so Vogel interprets 'libro uestro' as a book dedicated to the Emperor.  Vogel interprets this as referring to the material in Liber Abbaci, so I have now dated the next entry as 1228 rather than 1202, although there is no mention of the contest or the Emperor in Liber Abbaci.

                    The solution here is different than below and Vogel calls this Fibonacci's third method, the shortest and cleverest, and which Fibonacci described as "exceedingly beautiful"'.  Vogel notes the remarkable hybrid notations:  XXX3 for 33;  XXIII 1/1 for 23½;  X 1/1 for 10½  in this 15C MS.

Fibonacci.  1228 _ see above entry.  Pp. 293‑297.  Several versions.  He often notes that the values  x_i  can be multiplied through by any value.

Pp. 293‑294.  A = (1/2, 1/3, 1/6),   B = (1, 1, 1),   C = (3, 2, 1).  Answer:  33, 13, 1.  (= Al‑Karkhi III‑6.)  Vogel calls this Fibonacci's first method and notes a minor variation, which Fibonacci may have intended as another method.  Vogel notes some typographical errors.

P. 294.  Same  A,  with   B = (3, 2, 1) = C.  Answer:  30, 15, 6.

P. 295.  A = (1/3, 1/4, 1/5),   B = (1, 1, 1),   C = (3, 2, 1).  Answer:  177, 92, 25.

P. 295.  Same  A,  with  B = (3, 2, 1) = C.  Answer:  162, 96, 45.

Pp. 295‑296.  Same  A,  with  B = (5, 4, 1),   C = (3, 2, 1).  Answer:  543, 296, 175.  In this problem he computes  360 - 360  as  0  and  0/2  as  0.

Pp. 296‑297.  A = (1/2, 1/3, 1/6),   B = (1, 1, 1),  C = (5, 4, 1).  Answer:  326, 174, ‑30.  "Quare hec questio non potest solui, nisi solvatur cum aliqua propria pecunia tercii hominis ..."  He treats  ‑30  as a debt of the fund to the third man who steals nothing, thereby losing  30.  See Sesiano.

P. 297.  Four men.  A = (1/2, 1/3, 1/4, 1/5),   B = (1, 1, 1, 1),   C = (10, 9, 6, 5).  Answer:  1034, 666, 300, 190.

Pp. 335‑336.  Problem on pp. 293‑294 done by false position.  Vogel calls this Fibonacci's second method, but there is a typographical error citing p. 235f.

Gherardi. Libro di ragioni.  1327.  Pp. 54‑56.  A = (½, _, ¼),   B = (1,1,1),   C = (6, 4, 3).  Answer is  240 : 93 : 44.  He gives  4/29  of these values by starting with  x₁ + x₂ + x₃ = 52.

Columbia Algorism.  c1370.  Prob. 5, pp. 34‑35.  (= al‑Karkhi III‑6.)  Gives only the answer with no explanation.  Vogel's Introduction, p. 22, sketches the history.

Lucca 1754.  c1390.  Ff. 60v‑61r, pp. 138‑139.  (= al‑Karkhi III‑6.)  Gives some explanation, but Vogel says only the beginning makes sense.

Pacioli.  Summa.  1494.  Gives a number of versions.

F. 105r, prob. 11.  A = (½, _),   B = (1, 1),   D = (100, 100).  Solution is  (600, 800)/7.

F. 157v, prob. 77.  = al-Karkhi III-6 with total  12.  He then sketches the solution with total  60, but immediately has wrong values.  He also seems to have changed some of the parameters _ his answers don't add up to  60,  and he gives final values in the ratio  6 : 4 : 3.

Ff. 157v-158r, prob. 78.  A = (1/4, 1/6, 1/5),   B = (9, 7, 4),   C = (6, 4, 3),  with total  12.  His working is correct until he has  2 158/1651  instead of  2 958/1651  for his unknown and then the erroneous value is used in the final steps.  Answer should be  (9639, 5526, 4650)/1651.

F. 158v, prob. 82' (the second problem numbered 82).  Seems to be a discussion of modification of prob. 78 to have   B = (1, 1, 1),   C = (1/2, 3/10, 1/7),  but he never uses the values in  C  and winds up giving the answers of prob. 77 for total of  564,  namely  396, 156, 12.

F. 158v, prob. 83.  A = (_, ¼),   B = (1, 1),   D = (15, 15).

Calandri, Raccolta.  c1500.  Same as Fibonacci, pp. 296‑297.  Calandri simply says it is "insolubile".

Tonstall.  De Arte Supputandi.  1522.  Pp. 244-245.  Same as Pacioli, prob. 11.

Cardan.  Practica Arithmetice.  1539.  Chap. 66. 

Section 90, ff. GG.vii.v - GG.viii.v (pp. 164-165).  Same as Fibonacci, p. 295, first problem.  Answer:  354, 184, 50.

Section 91, ff. GG.viii.v - HH.i.v (pp. 165-166).  A somewhat similar situation, where the first two take money leaving the third with  5.  Friend says the first is to give  10  and  _  of what he has left to the second and the second is then to give  7  and  ¼  of what he has left to the third to make  C = (3, 2, 1).  This is determinate.  Answer is  x = 172,  y = 39  and the total sum is  216.

Buteo.  Logistica.  1559. 

Prob. 7, pp. 335-336.  A = (½, ¼),   B = (1, 1),   C = (1, 1).  He assumes total is  500,  then answer is  300, 200.

Prob. 8, pp. 336-337.  A = (½, _, ¼),   B = (1, 1, 1),   D = (116, 116, 116).  Answer:  144, 108, 96.

Vogel says that Clavius; Epitome Arithmeticae; Rome, 1595, pp. 249-252, gives a simple example with two persons and that then the problem vanishes from the literature.

 

          7.Q.   BLIND ABBESS AND HER NUNS _ REARRANGEMENT ALONG SIDES

                                        OF A  3 x 3  SQUARE TO CONSERVE SIDE TOTALS

 

          This is a kind of magic figure, except that here we generally have repeated values. 

          There are three trick versions of 6.AO which might be classified here or in 7.Q.2. 

(12,   4, 5) _ Trick version of a hollow  3 x 3  square with doubled corners, as in 7.Q:  Family Friend (1858),  Illustrated Boy's Own Treasury,  Secret Out.

 

Shihâbaddîn Abû’l‑‘Abbâs Ahmad ibn Yahya ibn Abî Hajala at‑Tilimsâni alH‑anbalî.  Kitâb ’anmûdhaj al‑qitâl fi la‘b ash‑shatranj  (Book of the examples of warfare in the game of chess).  c1370.  Copied by  Muhammed ibn ‘Ali ibn Muhammed al‑Arzagî  in 1446.

                    This is the second of Dr. Lee's MSS, described in 5.F.1, denoted Man. by Murray.  Described by Bland and Forbes, loc. cit. in 5.F.1 under Persian MS 211, and by Murray 207‑219.

                    Murray 280 says No. 46‑49 give the problems of arranging  32, 36, 40, 44  men along the walls and corners so the total along each edge is  12.

Pacioli.  De Viribus.  c1500.  Prob. 71: Un quadro quale  3  per ogni verso, Diametro e lati et giontovi  3  doventa  4  per ogni verso.  Agostini says to start with one object in each cell, then add one to the cells along a diagonal to get  4  in each line.  (This gets  6  on that diagonal however??)

van Etten.  English ed., 1653, prob. 72: Of the game of square formes, pp. 124‑125.  24  men on sides of a fort, becoming  28  and  20.  Discusses case of  12  men making  3,  4  or  5  on a side.

Anon.  Schau‑Platz der Betrieger: Entworfen in vielen List‑ und Lustigen Welt‑Händeln.  Hamburg & Frankfurt, 1687, pp. 543‑545.  ??NYS  (A&N, p. 5.)

Ozanam.  1694.  Prob. 1, 1696: 1-2;  1708: 1‑2;  1725: 1-3.  Prob. 20, 1778: 172-174;  1803: 172-174;  1814: 151-153.  Prob. 19, 1840: 77‑78.  Blind abbess and  24  nuns with  9  on a side.  1696 gives three arrangments with  24, 28, 20  on a side.  1725 adds another arrangement with  32.  1778 says Ozanam has presented this in a rather indecent manner to excite the curiosity of his readers and adds arrangements with  36  and  18  on a side.  The last has  5  and  4  in the corners and none in the side cells, but can be done in other ways.  1803 drops the 'indecent' reference.

Dilworth.  Schoolmaster's Assistant.  1743.  Part IV: Questions: A short Collection of pleasant and diverting Questions, p. 168.  Problem 1.  General stationing guards around a castle, wanting  18  on a side, starting with  48  men and changing to  56,  then  40.

Les Amusemens.  1749.  Prob. 11, p. 131: Les rangs de Neuf.  Wine merchant with  32  bottles,  9  on a side, reduced to  28, 24, 20.

Catel.  Kunst-Cabinet.  Vol. 2, 1793.  Die Nonnenlist (The nuns' strategem), pp. 15-16 & fig. 251 on plate XII.  The diagram shows the eight outside cells with  5  spots in the form of a  5  on a die and one spot in the centre.  However, the text says there are  25  cones or pieces and one must read the instructions to learn the game.  The number of pieces seems peculiar and I'm not entirely sure this is our problem, despite its name.

Bestelmeier.  1801.  Item 191: Die Nonnenlist.  Picture is an obscure copy of Catel.  Text is copied from part of Catel, but says there are  15  pieces!

Badcock.  Philosophical Recreations, or, Winter Amusements.  [1820].  Pp. 10-11, no. 20: The blind abbess and her nuns.  9  on a side, starts with  24  and changes to  28,  then  20.

Jackson.  Rational Amusement.  1821.  Arithmetical Puzzles.

No. 22, pp. 6 & 57.  18  men on a side of a castle, total =  48, 56, 40.

No. 29, pp. 7 & 58.  Blind abbess and nuns,  9  on a side, total ranging from  20  to  32.

No. 32, pp. 8 & 59.  Wine merchant and casks,  9  on a side, total of  32,  diminished to  20.

No. 36, pp. 9 & 59-60.  Blind abbess and nuns, equal numbers on each side, then with four extras, then with four gone away.  Solution starts with  24.

John Badcock.  Domestic Amusements, or Philosophical Recreations.  Op. cit. in 6.BH.  [1823].  Pp. 156-157, no. 194: Dishonest contrivance.  32  sheep, with  12  on a side, reduced to  28.

Manuel des Sorciers.  1825.  Pp. 75-78, art. 38.  ??NX  Blind abbess.  Gets totals of  32,  28,  24,  20.

Endless Amusement II.  1826? 

Pp. 108-109: Curious arithmetical question.  9  on each side, changed from a total of  24  to  20, 28, 32.

Prob. 32, pp. 209-210.  Uses a novel figure _ take a  3 x 3  square and draw the  1 x 1  cells in each corner, then diagonally connect the interior vertices of these to form an  X  in the central cell.  This gives  8  cells _ the four squares at the corners and four pentagonal shapes along the edges.  (See Wehman, p. 22.)  15  on each side, beginning with  40,  reduced to  36.

The Boy's Own Book.   The wine merchant and his clerk.  1828: 412;  1828‑2: 418;  1829 (US): 211;  1855: 565;  1868: 669.  32  bottles, reducing to  20.

The Riddler.  1835.  The wine merchant and his clerk, pp. 4-5.  Identical to Boy's Own Book.

Crambrook.  1843.  P. 10, no. 23: The Blind Abbess and her Nuns, a laughable trick.

Magician's Own Book.  1857.

The square of Gotham, pp. 229-230.  24  scholars,  9  on a side, changing to  20, 28, 32.  = Boy's Own Conjuring Book.

Prob. 24: The nuns, pp. 274 & 297.  24  nuns.  =  Book of 500 Puzzles, prob. 24.  = Boy's Own Conjuring Book, prob. 23.  c= Illustrated Boy's Own Treasury, prob. 29.

Landells.  Boy's Own Toy-Maker.  1858.  Pp. 149-150.  c= Magician's Own Book, prob. 24.

The Sociable.  1858.  Prob. 21: The blind abbot and the monks, pp. 292-294 & 309.  24  monks,  9  on a side, changed to  20, 28, 32, 36, 18.  = Book of 500 Puzzles, prob. 21. 

Book of 500 Puzzles.  1859. 

Prob. 21: The blind abbot and the monks, pp. 10-12 & 27.  As in The Sociable.

Prob. 24: The nuns, pp. 88 & 111.  Identical to Magician's Own Book, prob. 24.

Boy's Own Conjuring book.  1860.

The square of Gotham, pp. 199‑200.  Identical to Magician's Own Book.

Prob. 23: The nuns, pp. 236 & 261.  Identical to Magician's Own Book, prob. 24.

Illustrated Boy's Own Treasury.  1860.  Prob. 29, pp. 429 & 434.  Very similar to Magician's Own Book, prob. 24.

Leske.  Illustriertes Spielbuch für Mädchen.  1864?

Prob. 585-4, pp. 296 & 409: Der Kasten des Juweilers.  32  rings reduced to  28.

Prob. 585-6, pp. 296 & 409: Des Müllers Säcke.  32  sacks reduced to  28.

The Secret Out.  Op. cit. in 4.A.1.  1871?  Both of the following are presented with cards.

The unfaithful knave, pp. 4-5.  32  wine bottles,  9  on a side, reduced to  28, 24, 20.

The blind abbot and his monks, pp. 5-6.  24  monks,  9  on a side, changing to  20, 28, 32, 18.

Bachet-Labosne.  Problemes.  3rd ed., 1874.  Supp. prob. VI, 1884: 189-191.  60  bottles diminished to  44.

Kamp.  Op. cit. in 5.B.  1877.  No. 16, pp. 325‑326.  Nine wine bottles on each side on the square.  Servant steals some.

Cassell's.  1881.  Pp. 98‑99: The twenty‑four monks.  = Manson, 1911, pp. 249-250.

Handy Book for Boys and Girls.  Op. cit. in 6.F.3.  1892.  Pp. 38-39: The counter puzzle.  "In an old book published over half a century ago, I came across this puzzle ...."  Rearranges  24  as  20, 28, 32. 

Somerville Gibney.  So simple!  V. _ A batch of match tricks.  The Boy's Own Paper 20 (No. 988) (18 Dec 1897) 188-189.  24  changed to  25, 20, 28, 32, 30.

Dudeney.  Problem 70: The well and the eight villas _ No. 70: The eight villas.  Tit‑Bits 33 (5 Mar 1898) 432  &  34 (2 Apr 1898) 8.  How many ways can numbers be placed in the  8  cells to make  9  along each side?  Answer is  2035.  He gives a general formula.

Dudeney.  The monk's puzzle.  London Mag. 9 (No. 49) (Aug 1902) 89‑91  &  9 (No. 50) (Sep 1902) 219.  (= CP, prob. 17, pp. 39‑40 & 172‑173.)  How many ways can numbers be placed in the  8  cells to make  10  along each side?

Benson.  1904.  The dishonest servant puzzle, p. 228.  28  bottles,  9  on each edge, reduced to  24,  then  20.

Wehman.  New Book of 200 Puzzles.  1908. 

Pp. 20-21: The blind abbot and the monks.  = The Sociable.

P. 22: Fifteen "square" puzzle.  Uses the unusual figure of Endless Amusement II, prob. 32.  Starts with  5  marks in each cell so that it adds to  15  each way.  Remove four marks.  Solution is a bit unclear. 

Adams.  Indoor Games.  1912.  The cook and the jam, pp. 353-354.  36  jars of jam.

Blyth.  Match-Stick Magic.  1921.  Escaping from Germany, pp. 75-76.  32  with  9  on a side _ arranged  1, 7, 1  _ reduced to  28,  25  and  24.

Will Blyth.  Money Magic.  C. Arthur Pearson, London, 1926.  The stolen tarts, pp. 91-95.  9  on a side, start with  32,  reduce to  28, 24, 20.

Rohrbough.  Brain Resters and Testers.  c1935.  Ba Gwa, p. 18 (= pp. 18‑19 of 1940s?).  3 x 3  frame.  Two player game.  Start with  7  in each corner and  1  in each edge.  A player places a extra counter in the frame and the other tries to rearrange to preserve  15  in each outside row.  It says you can get  56  men on the board.  [I can get  60  if the corners can be empty.]

Jeffrey J. F. Robinson.  Musings on a problem.  MTg 37 (1966) 23‑24.  A farmer has  41  cows and wants to see  15  along each side of his house which is in the centre of a  3 x 3  array.  How many solutions are there if only  1, 2, ..., 8  different values can be used?  He finds all the solutions in some cases.

J. A. Dixon & Class 3T.  Number squares.  MTg 57 (1971) 38‑40.  Use the digits  1 ‑ 8  so the four side totals are the same.  They find that the sum can only be:  12, 13, 14, 15,  with  1, 2, 2, 1  solutions.  They then use  8  digits from  1 ‑ 9  and find  35  solutions.

 

          7.Q.1. REARRANGEMENT ON A CROSS

 

          The counts from the base to the top and to the end of each arm remain constant though some (usually  2)  of the pearls or diamonds have been removed.  Trick versions with doubling up are in 7.Q.2.

 

Pacioli.  De Viribus.  c1500.  Prob. 70: Un prete ch'inpegno la borscia del corporale con la croci de perle al Giudeo.

Thomas Hyde.  Historia Nerdiludii, hoc est dicere, Trunculorum; ....  (= Vol. 2 of De Ludis Orientalibus, see 7.B for vol. 1.)  From the Sheldonian Theatre (i.e. OUP), Oxford, 1694.  De Ludo dicto Magister & Servus, pp. 234-236.  Cross with  28  thalers with  16  from the base to the end of each arm.  He points out that the picture has an error causing the count to the ends of the side arms to be  17.  Discusses general solution.  Says it is known to the Arabs of the Holy Land.

Les Amusemens.  1749.  P. xxvi.  13  markers reduced to  11.

Manuel des Sorciers.  1825.  Pp. 136-138, art. 16.  ??NX  Cross of coins reduced from  13  to  11.

Endless Amusement II.  1826?  Prob. 11, pp. 195-196.  15  diamonds reduced to  13.

The Boy's Own Book.  The curious cross.  1828: 414;  1828-2: 420;  1829 (US): 213;  1855: 568;  1868: 628.  13  markers, reducing to  11.

Nuts to Crack III (1834), no. 80.  The curious cross.  Almost identical to Boy's Own Book.

The Riddler.  1835.  The curious cross, p. 6.  Identical to Boy's Own Book.

Young Man's Book.  1839.  P. 60.  Easy Method of Purloining without Discovery.  Identical to Endless Amusement II, except that has no title.

The Sociable.  1858.  Prob. 25: The dishonest jeweller, pp. 295 & 310.  15  diamonds, reducing to  13.  = Book of 500 Puzzles, 1859, prob. 25, pp. 13 & 28.  = Wehman, New Book of 200 Puzzles, 1908, p. 7.

Boy's Own Conjuring Book.  1860.  Easy method of purloining without discovery, p. 295.  15  diamonds, reducing to  13.  Very similar to The Sociable.

Leske.  Illustriertes Spielbuch für Mädchen.  1864?  Prob. 585-5, pp. 296 & 409: Juwelenkreuz.  18  jewels reduced to  16.

Mittenzwey.  1879?  Prob. 252 & 253, pp. 42 & 90.

Lucas.  La croix de perles.  RM2, 1883, pp. 134‑135.  c= Lucas; L'Arithmétique Amusante; 1895; pp. 10-11.  15  reduced to  13  and discussion.

Lemon.  1890.  The puzzling pearls, no. 535, pp. 69 & 117.  15  reduced to  13.

Don Lemon.  Everybody's Pocket Cyclopedia.  Revised 8th ed., 1890.  Op. cit. in 5.A.  P. 136, no. 7.  Diamond cross reduced from  15  to  13.  No solution.

É. Ducret.  Récréations Mathématiques.  Op. cit. in 4.A.1.  1892?  P. 104: Les croix de jetons.  Rearrange a cross of  13  to  11.

Adams.  Indoor Games.  1912  The dishonest steward, pp. 22 & 24.  Rearrangement of a  Y.

Dudeney.  AM.  1917.  Prob. 423: The ruby brooch, pp. 144-145 & 249.  Complicated version.  Brooch is a circle with  8  radii and the owner used to count from the centre to the circle, along  _  of the circumference and then back in, getting  8  each time.  Dishonest jeweller reduces stones from  45  to  41  in a symmetric pattern of one in the middle, two on each arm and three on each arc.  What was it before?

Blyth.  Match-Stick Magic.  1921.  Counting the cross, p. 33.  Rearrange a cross of  15  to  13.

 

          7.Q.2. REARRANGE A CROSS OF SIX TO MAKE TWO LINES OF FOUR,

                                        ETC.

 

          For the standard version, one has to put one marker on top of the one at the crossing.  See 6.AO and 7.Q for some similar trick versions.  The one which is closest to this section is:

(12,   7, 4) _ Trick version of a  3 x 3  square with doubled diagonal:  Hoffmann (1876),  Mittenzwey,  Hoffmann (1893), no. 8.

          See also in 6.AO, Hoffmann (1893), no. 9.

 

Les Amusemens.  1749.  P. xxv. 

Blyth.  Match-Stick Magic.  1921.  Five by count, pp. 33-34.  A cross of seven _ five crossing three _ rearranged to count five both ways.

J. F. Orrin.  Easy Magic for Evening Parties.  Jarrolds, London, nd [1930s??].  The five puzzle, pp. 66-67.  As in Blyth.

Sid G. Hedges.  More Indoor and Community Games.  Methuen, Lonodn, 1937.  Penny puzzle, pp. 51-52.  Four pennies in the shape of a  T  or  Y  tetromino.  Make two rows of three.  Put one from an end of the row of three on the crossing.

Depew.  Cokesbury Game Book.  1939.  Seven coins, p. 223.  As in Blyth.

 

          7.R.    "IF I HAD ONE FROM YOU, I'D HAVE TWICE YOU"

 

Jacques Sesiano.  The appearance of negative solutions in mediaeval mathematics.  Archive for the History of the Exact Sciences 32 (1985) 105-150.  In this, he discusses problems of the types given here and in 7.P.1, 7.R.1, 7.R.2 and 7.P.7.  He pays particular attention to whether the author discusses the problems in general or recognizes conditions for positivity or consistency, covering this in more detail than I do here.

 

          See Tropfke 609.

          NOTATION.

(a, b;  c, d) denotes the general form for two people.

          "If I had  a  from you, I'd have  b  times you."

          "And if I had  c  from you, I'd have  d  times you."

 I‑(a, b;  c, d;  ...) denotes the case for more people where 'you' means all the others.

II‑(a, b;  c, d;  ...) denotes the same where 'you' means just the next person, taken cyclically.

          With two people, there is no need to distinguish these cases.

          See Alcuin for an example where the second statement is interpreted as occurring after the first is carried out. 

          See Hall for a version where the first equation is  x + a = by.

          See Ghaligai, Cardan for versions with  "If I had  a  times yours from you, I'd have  b  times you", i.e.  x + ay = b(y - ay),  ....

          Sometimes  a = 0  _ see: Kelland (1839).  If  a = c = ... = 0,  this can interpreted as a form of 7.R or 7.R.1 _ see: Dodson (1775).

          Let  T  be the total of the amounts.  Then  I‑(a₁,b₁; a₂,b₂; ...)  with  n  people has  n  equations   x_i + a_i  =  b_i(T ‑ x_i - a_i),   which can be rewritten as   x_i + a_i(1+b_i)  =  b_i(T - x_i),   so we see that this is the same problem as discussed in 7.R.1 below where men find a purse, but with variable known purses,  p_i = a_i(1+b_i).  We get  x_i  =  b_iT/(1+b_i) - a_i.   Adding these for all  i  gives one equation in the one unknown  T,   T [Σ {b_i/(1+b_i} - 1]  =  Σ a_i.

          For  II‑(a₁,b₁; a₂,b₂; ...),  systematic elimination in the  n  equations  x_i + a_i = b_i (x_i+1 ‑ a_i)  leads to   x₁ [b₁b₂...b_n ‑ 1]   =  a₁(b₁+1) + a₂b₁(b₂+1) + a₃b₁b₂(b₃+1) + ....  ,  and any other value can be found by shifting the starting point of the cycle.

 

Euclid.  c-325.  Opera.  Ed. by J. L. Heiberg & H. Menge, Teubner, Leipzig, 1916.  Vol. VIII, pp. 286‑287.  Ass and mule in Greek and Latin verse.  (1, 2;  1, 1).  (Sanford 207 gives English of Clavius's 1605 version.  Cf Wingate/Kersey.)

Diophantos.  Arithmetica.  c250.  Book I.

No. 15. pp. 134‑135.  "To find two numbers such that each after receiving from the other may bear to the remainder a given ratio."  Does  (30, 2;  50, 3).

No. 18, pp. 135‑136.  "To find three numbers such that the sum of any pair exceeds the third by a given number."  E.g. "If I had 20 more, I'd have as much as you two."  Does with values  20, 30, 40.  This is like finding several purses _ see 7.R.1.

No. 19, pp. 136.  Same as no. 18, with  4  people.  Does with values  20, 30, 40, 50.

Metrodorus.  c510.  Art. 145‑146, p. 105.  (10, 3;  10, 5);  (2, 2;  2, 4).

Alcuin.  9C.  Problem 16: Propositio de duobus homines boves ducentibus.  (2, 1;  2, 2),  but the second statement in the problem is interpreted as happening after the first is actually carried out.  If a problem with parameters  (a, b;  c, d)  is interpreted this way, it is the same as our usual problem with parameters  (a, b;  c-a, d).

Mahavira.  850.  Chap. VI, v. 251‑258, pp. 158‑159.

253.  I‑(9, 2;  10, 3;  11, 5).

256.  I‑(25, 3;  23, 5;  22, 7).

al‑Karkhi.  c1010.  Sect. III, no. 5, p. 90.  II‑(1, 2;  2, 3;  3, 4;  4, 5).

Tabari.  Mift_h al-mu‘_mal_t.  c1075.  P. 145f., no. 7.  ??NYS _ cited by Tropfke 611.

Bhaskara II.  Bijaganita.  1150.  Chap. IV, v. 106.  In Colebrooke, p. 191.  (100, 2;  10, 6).

Fibonacci.  1202.  Pp. 190‑203.  Numerous versions, getting up to five people, some inconsistent examples and types where the second clause is  "I'd have  b₁  times you plus  b₂  more." 

Pp. 190.  (1, 1;  1, 10).

Pp. 190‑191.  (7, 5;  5, 7)  and general rules.

P. 192.  (7, 5, 1;  5, 7, 1)  _ denoting the extended type where the first says  "If I had  7  from you, I'd have  5  times you plus  1  more".  I.e. the first equation is  x + 7  =  5 (y ‑ 7) + 1.

Sesiano analyses the extended type on pp. 192-198 in detail.

Pp. 197‑198.  (7, 5, ‑1;  5, 7, ‑3).

Pp. 198‑199.  I‑(7, 5;  9, 7;  11, 7),  but he solves  I‑(7, 5;  9, 6;  11, 7).

Pp. 199‑200.  I‑(7, 5, 1;  9, 7, 1;  11, 7, 1),  but he solves as though the middle  7  is a  6.

Pp. 200‑201.  Types giving  x + y + 7  =  5 (z ‑ 7)  etc.

P. 201.  De eodem inter quattuor homines questio insolubilis.  This gives equations like  w + x + 7  =  3 (y + z - 7)  with coefficients   7, 3;  8, 4;  9, 5;  11, 6.  This is indeed inconsistent.

Pp. 202‑203.  I‑(7, 2;  8, 3;  9, 4;  10, 5;  11, 7),  but he solves with the last  7  as a  6.

Pp. 325‑326.  Problem of pp. 190‑191 done by false position.

Pp. 332‑333.  I‑(7, 4;  9, 5;  11, 6).

Pp. 344‑346.  I‑(7, 3;  9, 4;  11, 5)  done in two ways.

Munich 14684.  14C.  Prob. XV, p. 80.   (1, 1;  2, 2)  and  (n, 1;  n, 2).

Dell'Abbaco.  c1370.

Prob. 69, pp. 63‑65 with simple plate on p. 64.  (8, 2;  10, 3).

Prob. 126, pp. 100‑102 with simple plate on p. 101.  (3, 2;  5, 3).

Lucca 1754.  c1390.  Ff. 29r‑30v, pp. 68‑70.  (12, 2;  17, 3).  II‑(15, 2;  18, 3;  21, 5).

Giovanni di Bartolo.  Op. cit. in 7.H.  c1400.  He gives complex examples in probs. 10‑14, 54, 56, 57 on pp. 18‑27, 101‑107.  E.g. prob. 10, pp. 18‑21.  "If I had the square root of your money, I'd have  3  times you."  "And if I had the square root of your money, I'd have  4  times you."

Provençale Arithmétique.  c1430.  Op. cit. in 7.E.

F. 114r, p. 60.  (1, 1;  1, 2).

F. 114v,  p, 61.  (1, 2;  1, 4).  This has non-integral answers.

F. 115r, p. 61.  II-(1, 1;  1, 2;  1, 3).  Non-integral answers.

AR.  c1450.  Prob. 138‑147, 220, 334‑336.  Pp. 70‑71, 102, 146‑147, 169‑171, 218.

138.  Regula augmentationis:  (1, 2;  1, 3);   (10, 2;  10, 3).

139‑147.  (3, 4;  4, 3);   (3, 3;  4, 4);   (3, 2;  5, 3);   (5, 2;  3, 3);   (1, 1;  1, 3);   (15, 2;  15, 3);   (3, 3;  3, 5) _ erroneous answer;   (1, 10;  1, 20);   (1, 1;  1, 2).

220.  same as 144.

334.  I‑(3, 2;  3, 4;  3, 10;  3, 100)  phrased as "If you each give me  1, ..."

335.  I‑(3, 2;  4, 4;  6, 7;  5, 9).

336.  I‑(2, 2;  2, 4;  2, 10). 

He then says that  I‑(1, 2;  1, 3;  1, 5)  and  I‑(1, 2;  1, 3;  1, 4)  are impossible.  However, I find solutions in both cases, though each person has  £ 1  which may be why AR is unhappy.  Vogel finds the same solutions that I do, but doubled because he reads the problems as  I‑(2, 2;  2, 3;  2, 5)  as in 334.  I don't read them that way, but the text and numerical layout are a bit inconsistent.

Muscarello.  1478.  Ff. 81v-82r, p. 198.  (2, 2;  2, 1).

P. M. Calandri.  c1480.  Pp. 153.  (60, 6;  50, 13);   II‑(10, 2;  19, 4;  15, 9);   I‑(16, 2;  30, 8;  26, 7/2).

Chuquet.  1484.  Prob. 57, 58, 59, 60.  Prob. 61‑79 extend in various ways.  English of 69, 70 78 in FHM 210-212.  78 is indeterminate.

57.  (7, 2;  9, 6).

58.  (20, 2;  30, 3).

59.  (1, 1;  1, 2).

60.  (2, 1;  3, 4).

61.  II-(3, 2;  4, 3;  5, 5).

62.  II-(2, 2;  2, 3;  2, 4).

63.  I-(7, 5;  9, 6;  11, 7).

64-76 lead to equations like   x + 7  =  5 (y - 7) + 1   or   5 (y + z - 7) + 1.

77.  II-(-9, 1/6;  -11, 1/7;  -7, 1/5) - i.e. the first equation is  y + z + 9  =  6 (x - 9).

78.  This has equations like   w + x + 100  =  3 (y + z - 100)   with coefficients  (3, 100;  4, 106;  5, 145;  6, 170), which is indeterminate.  "Thus it appears that such problems have a necessary answer for two by two, but for one by one they have whatever answer one desires."

79.  This has equations like   v + w + x + 7  =  2 (y + z - 7)   with coefficients  (7, 2;  8, 3;  9, 4;  10, 5;  11, 6).

Calandri.  Arimethrica.  1491.  F. 66r.  (20, 2; 30, 3).  This has the unusual feature that  x = y  and I do not recall any other such example.  The condition for  x = y  is more complex than one might expect:   a (b+1)/(b-1)  =  c (d+1)/(d-1).

Pacioli.  Summa.  1494.  He has numerous problems, sometimes mixing amounts and parts and sometimes mixing this topic with 7.R.1 and 7.R.2, often saying "that part of yours that 12 is to mine", i.e.  y(12/x)  {cf. 7.R.1}, and he often continues into problems where one gives the square root of what one has or says something about the square of an amount.

F. 105v, prob. 19.  Two men find two purses of values  p+10, p,  giving equations:  x + p+10 + 10  =  4 (y - 10),   y + p + 20  =  5 (x - 20).  He assumes the purses are worth  100  in total, so  p = 45,  p+10 = 55.  Answer:  (765, 690)/19.

Ff. 189r-189v, prob. 12.  x + 12  =  2 (y - 12),   y + x(12/y)  =  3 {x ‑ x(12/y)}

F. 191r-191v, prob. 24 & 25.  x + y(20/x)  =  y - y(20/x) + 28,   y + x(30/y)  =  x ‑ x(30/y) + 70.  Answer:  100, 120.  Prob. 25 is an alternative way to solve the problem.

F. 192v, prob. 29.   x + _ (y + z)  =  96,   y + 60  =  2 (z + x - 60) - 4,   z + ¼ (x  + y) + 5  =  3 {¾ (x + y)} - 5.   Answer:  -79, 236, 289.  His algebra leads to  79 + x  =  0. 

F. 193v, prob. 34.   x + 6  =  2 (y + z - 6),   y + _ (z + x)  =  3 {_ (z+x)},   z + ¾ (x + y)  =  4 {¼ (z + x)}.   Answer:  (198, 90, 72)/7.

Calandri, Raccolta.  c1500. 

Prob. 23, pp. 22‑23.  (20, 2;  30, 3).

Prob. 44, pp. 40‑41.  Three people _ first two as  (12, 2;  20, 3),  third says  "If I had  24  from you two, I'd have  3  times you plus the square root of what you have."

Hans Sachs (attrib.).  Useful Table-talk, or Something for all; that is the Happy Thoughts, good and bad, expelling Melancholy and cheering Spirits, of Hilarius Wish-wash, Master-tiler at Kielenhausen.  No publisher, place or cover, 1517, ??NYS _ discussed and quoted in: Sabine Baring-Gould; Strange Survivals  Some Chapters in the History of Man; (1892), 3rd ed., Methuen, 1905, pp. 220-223.  [Not in Santi.]  Baring-Gould, p. 221 has  (1, 2;  1, 1).

Ghaligai.  Practica D'Arithmetica.  1521.  He has a series of problems of this type, of increasing complexity, all involving men and money.  I omit the more complex cases.  He also uses parts as in Pacioli.

Prob. 1, f. 100r.  (10, 1;  20, 2).

Prob. 2, f. 100r.  (20, 2;  30, 3).

Prob. 3, f. 100v.  x + ¼y  =  y - ¼y,   y + ½x  =  4 (x - ½x) + 2.

Prob. 5, f. 101r.  x + 10  =  y - 10,   y + x(20/y)  =  3 {x - x(20/y)}.

Prob. 6, f. 101r.  x + 12  =  2 (y - 12),   y + x(12/y)  =  3{x - x(12/y)}.  = Pacioli 12.

Prob. 7, f. 101v.  x + y(6/x)  =  21,   y + x(3/y)  =  20,   given that  y(6/x) + x(3/y)  =  11.  [Without the extra condition, this gives a fourth order equation with solutions   x = 0, 12, 15 ± Ö6;   y = 0, 18, 11 _ 3Ö6,  though  0, 0  is indeterminate in the original equations.

Prob. 9, f. 102r.  Same as prob. 7, but the extra condition is replaced by   x + y  =  30.

Prob. 10, f. 102r-102v.  x + ry  =  2 (y - ry),   y + rx  =  5 (x - ax),  where  r  is an unspecified ratio.  This seems to be a unique version of this problem and both forms I and II lead to interesting solutions _  r  is determined by the coefficients  2, 5.  Cardan has a related version, but he gives a value of  r  which is inconsistent. 

                    For the 'all others' (type I) version, we let  T = Σ x_i.  Then the equations are:   x_i + r(T - x_i)  =  a_i(1-r)(T - x_i).  Since   x_i + r(T - x_i) + (1-r)(T - x_i)  =  T,   we see that   x_i + r(T - x_i)  =  a_iT/(1+a_i),   so   T  =  (1+a_i)[x_i + r(T‑x_i)]/a_i =  (1+a_i)(1‑r)(T-x_i)  or   T - x_i  =  T/(1+a_i)(1-r),  assuming  1-r ¹ 0.  Adding these last equations gives  (n‑1)T  =  [T/(1‑r)] Σ 1/(1+a_i).  Assuming  T ¹ 0  gives us   1‑r  =  [1/(n‑1)] Σ 1/(1+a_i).  Hence  r  is determined by the  a_i's,  or else  T = 0  and/or  1-r = 0.  In fact  T = 0  holds if and only if  1-r = 0  or  all x_i = 0.  When  1-r = 0,  the  x_i  are arbitrary.  In either of these degenerate cases, the  a_i  are arbitrary.

                    For the 'next one' (type II) version, the equations are:   x_i + rx_i+1  =  a_i(1‑r)x_i+1,   or   x_i  =  [(1-r)(1+a_i) - 1] x_i+1.   Multiplying these together, we find that the product of the factors must be  1  and this gives an  n-th order polynomial for  1-r.  Set   P(x)  =  Π [(1+a_i)x - 1].   If we assume all  1+a_i > 0,  then this has  n  positive roots and hence  P(x) = 1  occurs for some  x = 1-r  greater than the largest term  1/(1+a_i).  If we further assume the  a_i  are not too small, namely that  Π a_i > 1,  then this product is  P(1)  and we hence know there is a point with  P(x) = 1  for some positive  x  less than  1,  so the corresponding  r = 1-x  is also between  0  and  1.  There may well be other suitable roots.  If some  1+a_i < 0,  the situation is more complex and there need not be any positive roots.  Even if all  a_i  are positive,  P(x) = 1  may only occur for  x > 1,  and hence  r < 0.  When  n  is even, the constant terms in the equation cancel and one can factor out  1-r  (since  1-r = 0  leads to  r = 1,  and an easy solution or inconsistency).  E.g., for  n = 2,  we get   1‑r  =  1/(1+a₁) + 1/(1+a₂).

Riese.  Rechnung.  1522.  1544 ed. _ p. 93;  1574 ed. _ pp. 62v‑63r.  (1, 1;  1, 3).

Riese.  Die Coss.  1524.  No. 25‑30, p. 44 & No. 63‑64, p. 49.

No. 25.  (1, 1;  1, 2).

No. 26.  (1, 1;  1, 3).

No. 27.  (1, 10;  1, 20).

No. 28.  (1, 2;  1, 5).

No. 29.  II‑(1, 1;  1, 2;  1, 3).

No. 30.  II‑(1, 1;  2, 2;  3, 3).

No. 63.  (1, 1;  4, 2).

No. 64.  (1, ½;  5, 3).

Cardan.  Practica Arithmetice.  1539.  Chap. 61.

Section 3, f. S.viii.v (p. 110).  (5, 4;  4, 4).

Section 4, f. S.viii.v (p. 110).  Variation leading to:   x + ½y  =  3 (y - ½y),   y + ½x  =  7 (x - ½x),   which he rightly states is impossible (unless  x = y = 0).

Section 7, f. T.i.v (p. 111).  Variation giving equations:   x + ½y + 2  =  9 (y - ½y - 2),   y + _x + 3  =  3 (x - _x - 3).

Tartaglia.  General Trattato.  1556. 

Book 16, art. 12‑15, pp. 240v‑241v.  (6, 1;  9, 2).   (7, 2;  13, 3).   (30, 1;  30, 2).  I‑(16, 1;  24, 2;  33, 3).

Book 17, art. 23, 34, 35,  pp. 271v-272r  &  273v-274v.  II‑(32, 2;  38, 3;  50, 4;  76, 7).   (24, 2;  42, 3).   II-(34, 2;  52, 3;  80, 5).

Buteo.  Logistica.  1559.  Prob. 59, p. 264.  i-th  says  "I have  a_i  times as much as the rest of you."  with   (a_i)  =  (1, 1/2, 1/5).   This could be considered as  I-(0, 1;  0, 1/2;  0, 1/5).  This is indeterminate with general solution proportional to  (3, 2, 1).  He assumes  x = 24  and gets  y = 16,  z = 8.

van Etten.  1624.  Prob. 83 (76), parts a & c, pp. 90‑92 (134‑136).  Ass & mule _  (1, 2;  1, 1)  = Euclid.   (10, 3;  10, 5),   (2, 2;  2, 4)  = Metrodorus.

Wingate/Kersey.  1678?.  Quest. 39, pp. 502-503.  Ass and mule in Latin verse - cf Euclid.  (1, 2;  1, 1)

Edward Cocker.  Arithmetic.  Ed. by John Hawkins.  T. Passinger & T. Lacy, London, 1678.  [De Morgan states "I am perfectly satisfied that Cocker's Arithmetic is a forgery of Hawkins" and then spends several pages detailing this charge and showing that the book is a rather poor compilation from several better books.  However Ruth Wallis [Ruth Wallis; Edward Cocker (1632?-1676) and his Arithmetick: De Morgan demolished; Annals of Science 54 (1997) 507-522] has argued that De Morgan is wrong.  Inspection of a 1st ed. at the Graves collection and a 3rd ed., 1680, at Keele shows no noticeable difference in the texts other than resetting which makes the book smaller with time _ all the editions seen have the same 32 chapters.  The 1st and 3rd eds. seem to have identical pagination so I will not cite the 1680.]  1st ed., 1678 & 3rd ed., 1680, both T. Passinger & T. Lacy, London.  = 33rd ed., Eben. Tracy, London, 1715.  = Revised by John Mair; James & Matthew Robertson, Glasgow, 1787.  Chap. 32, quest. 4.  1678: p. 333;  1715: p. 215;  1787: p. 186.  (1, 5;  1, 1).

Ozanam.  1725.  De l'asne et du mulet, prob. 24, question 3, 1725: 176‑178.  Prob. 6, 1778: 189-190;  1803: 186-188;  1814: 162-163;  1840: 84-85.  (1, 2;  1, 1).  1725 gives two versions and solutions in Latin verse.  1778 et seq. gives just one version and solution, but with slight differences, and refers to the Metrodorus problems in Bachet's Diophantus, though these are not the numbers in Metrodorus.

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790), prob. XXIII, p. 93 (not in 1790 ed.).  A  says to  B,  if I had  a  of your money, I'd have as much as you together with half of  C,  etc., giving:   x + a  =  y-a + ½z,   y + b  =  z-b + _x,  z + c  =  x-c + ¼y.  Finds general solution and does case  a = b = c = 5.

Walkingame.  Tutor's Assistant.  1751.  1777: p. 175, prob. 98;  1860: p. 184, prob. 97.  (4, 1;  4, 2),  flocks of sheep.

Euler.  Algebra.  1770.  I.IV.IV.612: Question 3, pp. 208‑209.  Ass and mule:  (1, 2;  1, 3).

Vyse.  Tutor's Guide.  1771?  Prob. 11, p. 138 & Key p. 183.  (1, 1;  1, 2)  (= Euclid).

Dodson.  Math. Repository.  1775.

P. 8, Quest. XIX.  (1, 1;  1, 2)  (= Euclid).

P. 19, Quest. L.  Find  x/y  such that  (x+1)/y  =  1/3,  x/(y+1)  =  1/4.  [(x+1)/y  =  1/n;  x/(y+1)  =  1/(n+1)  has solution  x = n+1,  y = (n+1)² - 1.  In general,  (x+1)/y  =  a/b,  x/(y+1)  =  c/d  gives  x = c(a+b)/(ad-bc),  y = b(c+d)/(ad-bc).  One would normally assume  a/b > c/d.  Cf Wolff in 7.R.3 for a different phrasing of the same problem.]

P. 31, Quest. LXXVI.  Find  x/y  such that  (x+4)/(y+4)  =  4/3,  (x-4)/(y-4)  =  3/2.  [In general  (x+A)/(y+A)  =  a/b,  (x-A)/(y-A)  =  c/d  has solution  x = A (2ac‑bc‑ad)/(bc-ad),  y = A (ad+bc-2bd)/(bc-ad).  One would normally assume  a/b < c/d.]

P. 46, Quest. XCIX.  w = (x+y+z)/2,  x = (w+y+z)/3,  y = (w+x+z)/4,  x = 14 + z. 

Eadon.  Repository.  1794.  P. 296, no. 8.  (5, 1;  5, 3).

Augustus De Morgan.  Arithmetic and Algebra.  (1831 ?).  Reprinted as the second, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  Art. 121, p. 32.  (10, 2;  10, 3).

Unger.  Arithmetische Unterhaltungen.  1838.  Pp. 117-119 & 256-257, nos. 443‑450.  (5½, 2;  6½, 5),  (13½, 7;  16½, 3),  (10, 4;  7½, 2),  (10, 4;  12, 3),  (20_, 3;  12_, 4),  (9½, 4;  15, 2),  (33½, 2;  16½, 3),  (6, 5;  4 2/5, 8).

Philip Kelland.  The Elements of Algebra.  A. & C. Black, Edinburgh, et al., 1839.  ??NX.  P. 134:  "A's  money or debt is  a  times  B's;  if  A  lose  £10  to  B,  it will be  b  times  B's."  (Also entered in 7.X.)

The Family Friend (1856) 376,  Enigmas, Charades, &c.  176  Arithmetical Puzzle.  Standard  (1, 2;  1, 1)  given in a four stanza poem involving two costermongers with barrows of apples.  Signed  G. M. F. G.

Thomas Grainger Hall.  The Elements of Algebra: Chiefly Intended for Schools, and the Junior Classes in Colleges.  Second Edition: Altered and Enlarged.  John W. Parker, London, 1846.  P. 132, ex. 7.  Appears to be  (50, 2;  50, 1),  but it reads: "A says to B, if you give me £50, I shall have twice as much as you had; but if I give you £50, each will have the same sum."  The use of 'had' means the first equation is  x + 50  = 2y,  while the second equation is the usual  x - 50  =  y + 50.  Answer:  250, 150.

Magician's Own Book.  1857.  The two drovers, p. 246.  (1, 1;  1, 2).  = Book of 500 Puzzles, 1859, p. 60.  = Boy's Own Conjuring Book, 1860, p. 218.

Leske.  Illustriertes Spielbuch für Mädchen.  1864?  Prob. 564-33, pp. 255 & 396.  (1, 1;  1, 2).  Notes that the solution to  (a, 1;  a, 2)  is just  a  times the solution of the original.

(Beeton's) Boy's Own Magazine 3:4 (Apr 1889) 175  &  3:6 (Jun 1889) 255.  (This is undoubtedly reprinted from Boy's Own magazine 1 (1863).)  Mathematical question 34.  (5, 1;  10, 2)  with postage stamps.

William J. Milne.  The Inductive Algebra ....  1881.  Op. cit. in 7.E.

No. 21, pp. 156 & 333.  (20, 4;  20, 3/2).

No. 11, pp. 163 & 334.  II-(20, 4;  40, 4/9;  60, 7/4).  Answer:  (5620, 1880, 6700)/19

Fireside Amusements _ A Book of Indoor Games.  Op. cit. in 7.L.1.  1890?  P. 96, no. 2:  (1, 2;  1, 1)   between market women with eggs.

Hoffmann.  1893.  Chap. IV, no. 26: A rejected proposal, pp. 150 & 195.  (1, 3;  2, 1),  but the first person also says he already has twice the second, so this is an overdetermined problem.  (Hoffmann's other example is not overdetermined.)

Sullivan.  Unusual.  1947.  Prob. 36: A problem old enough to be considered new.  (1, 1;  1, 2).

David Singmaster.  Some diophantine recreations.  Op. cit. in 7.P.5.  1993.  Sketches some history; finds condition for integer data to produce an integer solution.

 

          7.R.1. MEN FIND A PURSE AND 'BLOOM' OF THYMARIDES

 

          See Tropfke 604 & 606. 

          Algebraically, 7.R.1 and 7.R.2 differ only in signs.

          NOTATION.  Finding a purse has two forms.

I‑(a₁, a₂, ..., a_n)  _  i-th says "If I had the purse, I'd have  a_i  times the rest of you".

II‑(a₁, a₂, ..., a_n)  _  i-th says "If I had the purse, I'd have  a_i  times the  i+1-st person".

          There are two related problems which I call forms  III  and  IV.

III‑(a₁, a₂, ..., a_n)  _ the sum of all amounts except the  i‑th is  a_i.  See the discussion under

          Iamblichus.

IV-(a₁, a₂, ..., a_n)  _  x_i + x_i+1 = a_i.  (This is determinate only if  n  is odd.  For  n = 3,  types III and  IV  are the same, though the constants or the variables are taken in a different order, so that  III‑(a, b, c) = IV-(c, a, b)  if we keep the variables in the same order.

          I give answers as a list of the amounts, in order;  then the purse.

          Let  p  be the value of the purse and let  T  be the total of the amounts.  Then  I‑(a₁,a₂,...)  with  n  people has  n  equations   x_i + p  =  a_i(T ‑ x_i),   so we see that this is the same problem as discussed in 7.R.2 below where men buy a horse, but with the value of the horse and the multipliers all negative, which makes this version have fewer sign complications in its solution.  Thus we get  x_i  =  (a_iT-p)/(1+a_i).   Adding these for all  i  gives one equation in the two unknowns  T  and  p.  However, letting  C = T + p  leads to the simplest equation:   (n‑1)T  =  [Σ  1/(1+a_i)] C.

          For  II‑(a₁,a₂,...),  systematic elimination in the  n  equations  x_i + p = a_ix_i+1  leads to   x₁ [a₁a₂...a_n - 1]  =  p [1 + a₁ + a₁a₂ + ... + a₁a₂...a_n],  and any other value can be found by shifting the starting point of the cycle.

          In either case, the solution can be adapted to variable purses _ see 7.R.  In some problems, gaining the purse is replaced by paying out, so the purse can be treated as having a negative value _ see Unger, 1838.

          Some of the cistern problems in 7.H are of type III

Diophantos.  Arithmetica.  c250.  Book I.

No. 16, p. 135.  "To find three numbers such that the sums of pairs are given numbers."  He does  III‑(20, 30, 40).

No. 17, p. 135 is the same for four numbers.  He does  III‑(22, 24, 27, 20).

No. 18 & 19 are like 3 and 4 men finding 3 and 4 purses _ see under 7.R.

No. 20, pp. 136‑137.  This is Type I with a purse of 0, i.e.  "I have  a_i  times the rest of you".  He does this as  I‑(3, 4, a₃)  since  a₃  is determined by  a₁  and  a₂.

Iamblichus.  On Nicomachus's 'Introductio Arithmetica'.  c325.  Pp. 62‑63, ??NYS.  Partly given in SIHGM 1, pp. 138‑141.  Describes the 'bloom' which has  n+1  unknowns  x,  x₁,  ...,  x_n  and we know   x + x_i  =  a_i   and   x + x₁ + ... + x_n  =  s.   Then  x  =  (a₁ + ... + a_n ‑ s)/(n‑1).  (Iamblichus uses  n  unknowns.)  Heath (HGM I 94‑96) says Iamblichus continues and applies the Bloom to  I‑(a₁, ..., a_n),  with integral  a_i,  by letting  x  be the value of the purse and  s  =  (a₁+1)...(a_n+1),  which yields   x + x_i  =  sa_i/(a_i+1).  (We can take  n-1  times the value of  s  to insure integer solutions.)  For rational  a_i,  we let  s  be the  LCM  of the denominators of  a_i/(a_i+1).  Iamblichus gives the problems  I‑(2, 3, 4)  and  I‑(3/2, 4/3, 5/4).  See Chuquet for an indeterminate version of the Bloom.

                    This is closely related to problems like the following:   x + y = a,  y + z = b,  z + x = c,   i.e.  III‑(b, c, a) = IV-(a, b, c).  We set  T = x + y + z  and so   T ‑ z = a,  T ‑ y = b,  T ‑ x = c.  This is a case of the 'bloom' for  n = 3,  with   x = T,  x₁ = ‑z,  a₁ = a,  etc., and  s = 0.  In general, this gives us   x  =  T  =  (a₁ + ... + a_n)/(n‑1).

Aryabhata.  499.  Chap. II, v. 29, pp. 71-72.  (Clark edition: pp. 40‑41.)  III‑(a₁, ..., a_n).  Gives   T  =  (a₁+...+a_n)/(n‑1).

Bakhshali MS.  c7C.  Kaye I 39-42, sections 78-79  and  Datta, pp. 45‑46  discuss two types of related systems for  n  amounts  x₁, x₂, ..., x_n.  See the discussion under Iamblichus.

IV-(a₁, ..., a_n).  When  n = 3,  this is equivalent to type III.  Kaye notes that n is always odd and says the following occur:  IV-(13, 14, 15) (Kaye III 166, f. 29r);  IV‑(16, 17, 18, 19, 20) (Kaye III 166-167, ff. 29v & 27v);  while the following are implied:  IV‑(9, 5, 8);  IV‑(70, 52, 66);  IV-(1860, 1634, 1722);  and possibly  IV‑(36, 42, 48, 54, 60).  Kaye's concordance (I 38-39) implies these examples should be in the same area of the text, but I can't find them in his Part III _ ??.  Also  III-(317, 347, 357, 362, 365) (Kaye I 40 (omitting the fourth equation); III 168‑169, ff. 1v-2r).

T ‑ x_i  =  c ‑ d_ix_i.   These are variations of Type III problems or of the Present of Gems problem, section 7.P.4.

Bhaskara I.  629.  Commentary to Aryabhata, chap. II, v. 29.  Sanskrit is on pp. 125-127; English version of the examples is on pp. 307-308.

                    Ex. 1:  III-(30, 36, 49, 50).

                    Ex. 2:  III-(28, 27, 26, 25, 24, 23, 21).

Mahavira.  850.  Chap. VI, v. 159, 233‑250, pp. 136‑137, 153‑158.

V. 159, pp. 136‑137.  III‑(22, 23, 24, 27).

V. 236, p. 155.  I‑(2, 3, 5).   Answer:  1, 3, 5; 15.

V. 239, p. 156.  i-th says  "If I had  b_i  of the purse, I'd have  3  times the rest of you", with  B = (1/6, 1/7, 1/9, 1/8, 1/10).  Answer:  261, 921, 1416, 1801, 2109; 110880.

V. 242, p. 157.  i-th says  "If I had  b_i  of the purse, I'd have  a_i  times the rest of you", with  a₁, b₁;  a₂, b₂  =  2, ½;  3, _.   Answer:  11, 13; 30

V. 244, p. 157.  I‑(2, 3).   Answer:  3, 4; 5.

V. 245, p. 157.  I‑(8, 9, 10, 11).   Answer:  103, 169, 223, 268; 5177.

V. 248, pp. 157‑158.  As in v. 242, with four men and   a₁, b₁;  ...  =  2, 1/5;  3, 1/4;  5, 1/2;  4, 1/3.   Answer:  356, 585, 445, 624; 14760.

V. 249.  As in v. 242, with   2, ¼;  3, _;  4, ½.   Answer:  55, 71, 66; 876.

al‑Karkhi.  c1010.  Sect. III, no. 24‑25 & 29‑30, pp. 95 & 98.

24:  III‑(20, 30, 40).   (= Diophantos I 16.)

25:  III‑(30, 45, 40, 35).

29:  Three men find purses  (30, 40, 20).  i-th says:  "If I had the  i‑th purse, I'd have as much as all of you."

30:  same as 29 with four men and purses  (20, 30, 40, 50).

Tabari.  Mift_h al-mu‘_mal_t.  c1075.  P. 129, no. 46.  ??NYS _ Hermelink, op. cit. in 3.A, says this is a problem with two persons.  Tropfke 607 cites this with no details.

Fibonacci.  1202.  Pp. 212‑228, Chap. XII, part 4: De inventione bursarum.  Many problems, going up to five men, 4 purses and an example with a negative solution.

Pp. 212‑213.  I‑(3, 4).   Answer:  4, 5; 11.

Pp. 213‑214.  I‑(2, 3, 4).   Answer:  7, 17, 23; 73.   (= Iamblichus's 1st.)

Pp. 214‑215.  I‑(3, 4, 5, 6).   Answer:  4, 67, 109, 139; 941.

Pp. 215‑216.  I‑(5/2, 10/3, 17/4, 26/5, 37/6).   Answer:  ‑49154, 30826, 89478, 131962, 163630, 1088894.   "... aut positio huius questionis indissolubilis est; aut primus homo debitum habebit ...."  See Sesiano.

P. 216.  II‑(2, 3, 4).   Answer:  9, 16, 13; 23.

P. 217.  II‑(5/2, 10/3, 17/4).   He doesn't find the answer which is  284, 444, 381; 826.

Pp. 218‑220.  II‑(2, 3, 4, 5).   Answer:  33, 76, 65, 46; 119.

Pp. 220‑223.  Four men find four purses of values:   p₁, p₂, p₃, p₄  =  p, p+3, p+7, p+13.  i-th says  "If I had the  i-th purse, I'd have  a_i  times the rest of you",  with  A  =  (a₁, a₂, a₃, a₄)  =  (2, 3, 4, 5).

P. 223.  Two men & purses of values:  p, p+13,  with  A = (2, 3).

Pp. 223‑224.  Three men & purses:  p, p+10, p+13,  with  A = (2, 3, 4).

Pp. 224‑225.  Four men & purses:  p, p+10, p+13, p+19,  with  A = (2, 3, 4, 5).

Pp. 225‑226.  Four men, one purse, giving  w + x + p  =  2y,  etc.

P. 227.  Three men, one purse, giving  x + y + p  =  2z,  etc.

P. 227.  Four men, one purse, giving  w + x + p  =  2y + z,  etc.

Pp. 227‑228.  Five men, one purse, giving  v + w + p  =  2 (x + y + z),  etc., with constants  2, 3, 4, 5, 6.   Answer:  22, -9, 57, 12, 71; 267  (the text has  7  instead of  71).  "... quare hec questio est insolubilis, nisi ponamus, secundum hominem habere debitum  9, ...."  See Sesiano.

P. 284.  III‑(31, 34, 37, 27).

P. 285.  III‑(31, 34, 37, 39, 27).  On pp. 284-285 & 286, he also considers problems like  IV-(a, b, c, d)  (cf. Bakhshali) and notes that some are inconsistent (e.g. for  a, b, c, d  =  27, 31, 34, 37)  and others are indeterminate.

Pp. 302‑303.  Four men, one purse, giving equations   w + x + p  =  3/2 y,  etc., with constants  3/2, 9/4, 16/5, 25/6.   Answer:  8665, 5682, 12718, 10280; 4730.

Pp. 326‑327.  Problem of pp. 218‑220 done by false position.

Pp. 330-331.  III-(75, 70, 67, 64, 54, 50). 

P. 333.  Problem of pp. 214‑216 done by false position.

Pp. 346‑347.  Three men & purses of values  18, 16, 20,  giving   x' + 18  =  3y',   y' + 16  =  4z',   z' + 20  =  5x'.  By setting   x' = x ‑ 11,  y' = y ‑ 7,  x' = z ‑ 9,  he converts to the ass & mule problem (7.R) on pp. 344‑346.

Pp. 349‑352.  Four men, one purse, giving   w + p  =  2 (x + y),   etc., with constants  2, 3, 4, 5.   Answer:  ‑1, 4, 1, 4; 11.  "... hec questio insolubilis est, nisi concedatur, primum hominem habere debitum."  See Sesiano.

Fibonacci.  Flos.  c1225.  In Picutti, pp. 316-319, numbers IV-V.

Pp. 238-239: De quatuor hominibus et bursa ab eis reperta, questio notabilis.  Described as the second of the problems that Fibonacci sent to Frederick II.  Same as Fibonacci, pp. 349‑352, though he doesn't cite this.  "... hanc quidem questionem insolubilem esse monstrabo, nisi concedatur, primum hominem habere debitum: ...."  See Sesiano.

Pp. 239-240: De eadem re.  Does the same problem with multipliers  4, 5, 6, 7.   Answer:  -1, 6, 1, 6; 29.  Implies a general solution for multipliers   k ‑ 2, k ‑ 1, k, k + 1   is   -1, k, 1, k;  k² - k - 1.

Jordanus de Nemore.  De Numeris Datis.  c1225.  Critical edition and translation by Barnabas Hughes.  Univ. of Calif. Press, Berkeley, 1981.  Prob. II‑24, pp. 150‑151.  General version of type  I  with purse given.  Example:  I‑(1/9, 1/3, 3/5, 1)  with purse  6.  Answer:  2, 14, 24, 34.

Ibn Badr  = Abenbéder  = Abu ‘Abdallah Muhammad ibn ‘Umar ibn Muhammad.  c1225.  Arabic text with Spanish translation by José A. Sánchez Pérez as:  Compendio de Álgebra de Abenbéder; Centro de Estudios Históricos, Madrid, 1916.  Tercer problema análogo, pp. 109-111 (pp. 70-71 of the Arabic).   (4, 7).   Answer:  (8, 5; 27)/9.

BR.  c1305.  No. 61, pp. 84‑87.  (7, 11).  Answer given is  12, 8; 76,  but should be  8, 12; 76.

Gherardi.  Libro di ragioni.  1327.  P. 53.  3 men find a purse.  I‑(2, 3, 4).  Answer:  7, 17, 23; 73.

Dell'Abbaco.  c1370.  Prob. 125, p. 100.  I‑(2, 3, 4).  Answer:  7, 17, 23; 73.  (= Iamblichus' 1st problem.)

Lucca 1754.  c1390.  F61r, pp.139‑140.  II‑(2, 3, 4, 5).  Answer:  33, 76, 65, 46; 119.  (= Fibonacci, pp. 218‑220.)

Bartoli.  Memoriale.  c1420.  Prob. 4, f. 75r (= Sesiano, pp. 136-137 & 147.  I‑(2, 3, 4).   Answer:  7, 17, 23; 73.   (= Iamblichus's 1st.)

AR.  c1450.  Prob. 113, 159, 227.  Pp. 63‑64, 76, 168‑169, 217.

113:  I‑(1, 2, 3),  answer:  2, 10, 14; 22.

159:  I‑(4, 10),  answer:  5, 11; 39.

227:  I‑(1, 2, 3),  answer:  1, 5, 7; 11.  Cf. 113.

Correspondence of Johannes Regiomontanus, 1463?-1465.  Op. cit. in 7.P.1.

P. 238, letter from Bianchini, 5 Feb 1464.  Query 8:  III‑(42, 54, 30).

Pp. 259 & 291, letter to Bianchini, nd [presumably 1464].  P. 259 gives the answer:  21, 9, 33.   P. 291 is part of Regiomontanus's working for this letter where he solves the problem by assuming the first has  x,  so the second has  30 - x  and the third has  54 - x,  hence   84 - 2x  =  42.

Muscarello.  1478.  Ff. 65r-65v, pp. 172-173.  Men find a purse,  I‑(3, 4, 5).  Answer:  7, 13, 17; 83.

P. M. Calandri.  c1480.

P. 67:  III‑(12, 14, 10).

P. 68:  III‑(10, 8, 15, 12).

   Chap. 22:  "... huomini che ànno danari et trovano borse di danari", pp. 181‑183.

I‑(2, 3),  answer:  3, 4; 5.  (= Mahavira v. 244.)

II‑(2, 3, 4),  answer:  9, 16, 13; 23.  (= Fibonacci, p. 216.)

I‑(2, 3, 4),  answer:  7, 17, 23; 73.   (= Iamblichus's first.)

Chuquet.  1484. 

   Triparty, part I.  FHM  80-81 & 83-85.  Sesiano cites pp. 642 & 646 of Chuquet, ??NYS. 

III-(120, 90, 80, 75).   Answer:  (5, 95, 125, 140)/3.   Not mentioned by Sesiano; English in FHM 80, where the next two cases are just given as formulae with solutions.

III-(120, 90, 80, 75, 72).   Answer:  (-43, 77, 117, 137, 149)/4.   Sesiano notes that this is the same as the solution of the problem in 7.R.2 in the Provençale Arithmétique, c1430, and in Pellos, 1492.

III-(120, 180, 240, 300, 360).   Answer:  180, 120, 60, 0, -60.   Sesiano notes these numbers occur in the solution of the problem on p. 641 _ see 7.R.2.

FHM 83 gives the next four just by formulae with solutions.

II-(2, 3, 4)  with purse worth  40.  Answer:  (280, 680, 920)/73.

I-(3, 4, 5, 6)  with purse worth  50.  Answer:  (200, 3350, 5450, 6950)/941.

I-(2, 3, 4, 5)  with purse worth  26.  Answer:  (‑78, 312, 546, 702)/123.  FHM puts down  +78  and hence misses this interesting example which Sesiano discusses.

Version giving equations like   w + z  =  2 (x + y)  =  26_,   but this is a bit far away from the problems of this section.

FHM 83-84: Indeterminate version of the Bloom of Thymarides:   w + x  =  17,  w + y  =  18,  w + z  =  19.   He solves by setting  w = 12,  and says any other value less than  17  can be used "wherefor such calculations may have innumerable responses."

   Appendice.

Prob. 81.  III-(19, 23, 30).   Answer:  17, 13, 6.

Prob. 82, English in FHM 212-213.  (3, 5).   Answers:  4, 6; 14  and  (60, 90; 210)/7.  Says "such questions do not have a necessary answer."

Borghi.  Arithmetica.  1491? _ this material is additional to the 1484 ed. and Rara indicates that the first ed. to have 100ff is the 4th of 1491.  The folio numbers are from the 1509 ed.

Ff. 99r-99v.  Men find a purse,  I-(3, 4).   = Fibonacci p. 212.

Ff. 99v-100r.  Men find a purse,  I-(2, 3, 4).   = Fibonacci p. 213.  = Iamblichus' 1st.

F. 100r.  I-(3, 4)  with purse worth  16.

F. 100r.  I-(2, 3, 4)  with purse worth  30.

Calandri.  Arimethrica.  1491. 

F. 65v.  III-(12, 14, 10).

F. 65v.  I-(6, 40).  Answer:  7, 41; 239.

Pacioli.  Summa.  1494.  Some of his problems mix this with 7.R.2.

F. 105v, prob. 19.  Two men find two purses of values  p+10, p,  giving equations:  x + p+10 + 10  =  4 (y - 10),   y + p + 20  =  5 (x - 20).  He assumes the purses are worth  100  in total, so  p = 45,  p+10 = 55.   Answer:  (765, 690)/19.   = Tonstall, pp. 245-246.

F. 190v, prob. 22.  I-(3, 4).   Answer:  4, 5; 11.

Ff. 190v-191r, prob. 23.   x + p  =  2y + 2,   y + p  =  4x + 2.   Answer:  12, 20; 30.

Ff. 192r-192v, prob. 28.  I-(2, 3, 4).   He assumes  p = 10  and gets answer:  (70, 170, 230; 730)/73.

F. 192v, prob. 30.  I-(3/2, 7/3, 15/4).  Assumes  p = 1  and gets answer:  (63, 177, 279; 621)/621.

F. 192v, prob. 31.  I-(2, 3, 4).  Doesn't observe that this  = prob. 28.  Answer:  (7, 17, 23; 73)/73.

F. 193v, prob. 39 & 40.  III-(35, 32, 27).   III-(122, 114, 106, 96).   Says one can deal similarly with more people.  Prob. 40 discusses the point further.

Ff. 193v-194r, prob. 41.  3  men find a purse and want to buy a horse, giving:  x + y + p/3  =  h,   y + z + p/5  =  h,   z + x + p/4  =  h.  If   T  =  x + y + z,   one gets   T + 47p/60  =  3h   and the solution space is actually two dimensional.  He assumes  p = 60,  h = 47  and this problem reduces to prob. 39.

Ghaligai.  Practica D'Arithmetica.  1521.  He gives several versions, of increasing complexity _ the later ones involve various numbers in geometric progressions or using roots and I omit these.

Prob. 13, f. 103r.  I-(6, 10).   Answer:  (42, 66; 354)/7.   He arbitrarily sets the first man's money at  6.

Prob. 14, ff. 103r-103v.  I-(4, 6),  but with total money  = 100.  Answer:  (100, 140, 460)/7.

Prob. 16, ff. 103v-104r.  Two men find purses worth  p  and  p+13,  with coefficients  2  and  3,   i.e.   x + p  =  2y,   y + p+13  =  3x.   Answer:  (339, 400; 461)/12.

Tonstall.  De Arte Supputandi.  1522. 

Quest. 42,  pp. 172-173.  III-(150, 240, 326).  Considers four person case and sketches general solution.

Pp. 245-246.  Same as Pacioli's prob. 19, but clearly says the two purses are worth  100.

Riese.  Die Coss.  1524.

No. 45, p. 46.  2 men find a purse,  (2, 5).

No. 65, p. 49.  2 men find a purse, leading to:   x + p + 1  =  b ‑ 1,   y + p + 4  =  3 (x ‑ 4).   He assumes  p = 2,  but the general solution is  b  =  2a ‑ 7,  p  =  a ‑ 9.

Apianus.  Kauffmanss Rechnung.  1527. 

F. M.iii.v.  I-(½, 3)  with  p = 30.  Answer:  60, 160.

F. M.iv.r.  III-(44, 36, 30).  Answer:  11, 19, 25.

Cardan.  Practica Arithmetice.  1539.  Chap. 66.

Section 97, ff. HH.v.r - HH.vi.r (p. 169).  III-(34, 73, 72, 88).

Section 98, ff. HH.vi.r - HH.vi.v (p. 169).  Men find a purse and buy a horse, giving:   x + y + p/2  =  y + z + p/5  =  x + z + p/3  =  h.   Answer:  6, 10, 15; 30, 31.

Tartaglia.  General Trattato.  1556.  Book 16, art. 28‑35 & 40, pp. 243r‑245r.

Art. 28.  I‑(2, 3, 4).   Answer:  7, 17, 23; 73.  (= Iamblichus' 1st.)

Art. 29.  I‑(3, 4, 5).   Answer:  7, 13, 17; 83.

Art. 30.  I‑(4, 6, 8).   Answer:  17, 53, 73; 487.

Art. 31.  Same as 30 with purse given as  100.

Art. 32.  Same as 30 with total given as  1200.

Art. 33.  II‑(2, 3, 4).   Answer:  9, 16, 13; 23.  (= Fibonacci, p. 216.)

Art. 34.  II‑(4, 5, 6).   Answer:  25, 36, 31; 119.

Art. 35.  II‑(2, 3, 4, 5).   Answer:  33, 76, 65, 46; 119.  (= Fibonacci, pp. 218‑220  &  Lucca 1754.)

Art. 40.  III‑(24, 28, 32, 36).

Buteo.  Logistica.  1559.  Prob. 9, p. 209-210.  III-(4900, 3760, 4660).   Remarks on the case with four people.

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790), prob. V, pp. 79-80 (1790: prob. VII, p. 80).  III-(17, 16, 15).   1745 gives two methods of solution;  1790 gives one.

Vyse.  Tutor's Guide.  1771?  Prob. 27, pp. 61-62 & Key p. 67.   III‑(33000, 30000, 32000, 28000, 25000).

Dodson.  Math. Repository.  1775.

P. 6, Quest. XV.  III-(30, 36, 40).

P. 38, Quest LXXXIX.  I-(1, 2, 3)  with purse worth 55.  The context is three horses and a saddle.  The first horse, with the saddle, is worth as much as the other two horses, etc.  Answer:  5, 25, 35.

P. 64, Quest. CXX.  Multiplicative form of type III.  xy = 12,  xz = 18,  yz = 24.  He doesn't multiply the equations to get  (xyz)² = 72².  Answer:  3, 4, 6.

P. 75, Quest. CXL.  Like the previous, with four variables.  wxy = 252,  wxz = 432,  wyz = 756,  xyz = 336.  Here he does multiply them together to find  wxyz = 3024.  Answer:  9, 4, 7, 12.

Hutton.  A Course of Mathematics.  1798?  Prob. 30,  1833: 222;  1857: 226.  Two horses with saddle.  I‑(2, 3)  with saddle worth  50.  Answer:  30, 40; 50.  (=  10 times Mahavira v. 244.)

Unger.  Arithmetische Unterhaltungen.  1838.  Pp. 115-117 & 256, nos. 431-442.  All of these have different purses except no. 435, which is  I-(½, _)  with a purse of  -10,  i.e.  10  is taken away rather than gained.  No. 432 uses the context of ages.

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Arithmetical puzzles, no. 2, pp. 172-173 (1868: 184).  I-(1, 2, 3)  with purse of  11.  Answer:  1, 5, 7; 11.  See AR 113 & 227.

Ken Russell & Philip Carter.  Intelligent Puzzles.  Foulsham, Slough, 1992.  Prob. 57, pp. 45 & Answer 75, p. 164: Money.  Jack has 75p and ¾ of what Jill has;  Jill has 50p and ½ of what Jack has.  This is a straightforward problem, but it is also a concealed version of  I‑(4/3, 2)  with a purse of 100.

 

          7.R.2. "IF I HAD  1/3  OF YOUR MONEY, I COULD BUY THE HORSE"

 

          See Tropfke 608.

          For related problems, see 7.H.4.

          NOTATION.  Again there are two forms.

I‑(a₁, a₂, ..., a_n)  _ i-th says "If I had  a_i  of what the rest of you have, then I could buy the

          horse".

II‑(a₁, a₂, ..., a_n)  _ i-th says "If I had  a_i  of what the i+1-st has, then I could buy the horse".

 

          The problem often replaces horse by house, ship, etc.  In some cases, the value of the horse, house, etc. is given.  In some cases the value is given with no reference to anything bought and I say  "with  h = ..."  to indicate the value.  Some problems have different values of horses.  If values are simply given, I say  "with  h₁, h₂ = ..."  or  "with  h_i = ...".  See:  Fibonacci,  Gherardi,  Lucca 1754,  AR,  P. M. Calandri,  Riese,  Tartaglia,  Simpson  for examples where there are different values of horses.

          Solution notation as in 7.R.1 with the horse last.

          Let  h  be the value of the horse and let  T  be the total of the amounts.  Then  I‑(a₁,a₂,...)  with  n  people has  n  equations   x_i + a_i(T ‑ x_i)  =  h,   so   x_i  =  (h - a_iT)/(1-a_i).   Adding these for all  i  gives one equation in the two unknowns  T  and  h.  However, letting  C = T - h  leads to the simplest equation:   (n‑1)T  =  [Σ 1/(1‑a_i)] C.

          For  II‑(a₁,a₂,...),  systematic elimination in the  n  equations  x_i + a_ix_i+1 = h  leads to   x₁ [1 + (-1)ⁿ⁺¹a₁a₂...a_n]  =  h [1 - a₁ + a₁a₂ - ... + (-1)ⁿa₁a₂...a_n],  and any other value can be found by shifting the starting point of the cycle.

          In either case, the solution can be adapted to variable purses _ see 7.R.  Taking negative values of  h  and all  a_i  converts this into 7.R.1 _ men find a purse, which is slightly easier to deal with.

 

Chiu Chang Suan Ching.  c‑150.  Chap. VIII.

Prob. 10, p. 86.  I‑(½, _)  with  h = 50.

Prob. 12, p. 87.  Equivalent to  II‑(3, 2, 1)  with  h = 40.

Prob. 13, pp. 87‑88.  Equivalent to  II‑(6, 5, 4, 3, 2).  Answer given is  265, 76, 129, 148, 191; 721  and this is the least integral solution.

Diophantos.  Arithmetica.  c250.  Book I.

No. 22, p. 138:  "To find three numbers such that, if each give to the next following a given fraction of itself, in order, the results after each has given and taken may be equal."  However, he says that the equal amounts are the results after both giving and taking, i.e.   (1 ‑ b) y + ax  =  (1 ‑ c) z + by  =  (1 ‑ a) x + cz.   He does  a, b, c  =  1/3, 1/4, 1/5.   Answer:  6, 4, 5.

No. 23, pp. 138‑139 is the same with four people and numbers  1/3, 1/4, 1/5, 1/6.  Answer:  150, 92, 120, 114.

No. 24, p. 139:  "To find three numbers such that, if each receives a given fraction of the sum of the other two, the results are all equal."  Does  I‑(1/3, 1/4, 1/5).   Answer:  13, 17, 19; 25.

No. 25, pp. 139‑140.  Same with four numbers.  Does  I‑(1/3, 1/4, 1/5, 1/6).   Answer:  47, 77, 92, 101; 137.

Sun Zi.  Sun Zi Suan Ching.  Op. cit. in 7.P.2.  4C.  ??NYS.  Chap. III, no. 28.   I-(½, _)  with  h = 48  _ like Chiu Chang Suan Ching, prob. 10.  (English in Lam & Shen, HM 16 (1989) 117.)

See 7.P.4 _ Bakhshali MS for a problem which is related, leading to   x₁/2 + x₂ + x₃ + x₄ + x₅  =  h,  etc.,  where  h  is the price of a jewel.  Solution: 120, 90, 80, 75, 72; 377.  Also an example with three values and diagonal coefficients   -7/12,  -3/4,  -5/6   and solution:  924, 836, 798; 1095.

Sesiano cites Ab_ K_mil's Algebra and al-Karaj_'s K_f_, ??NYS.  Hermelink, op. cit. in 3.A, cites K_f_  &  Beha-Eddin, ??NYS

al‑Karkhi.  c1010.

   Sect I, no. 42‑43, p. 80.

42:  x + _y  =  20  =  y + ¼x.

43:  x + _y + 5  =  20  =  y + ¼x + 6.

   Sect. III, no. 26‑27, 32‑35, pp. 95‑100.

26:  I‑(1/3, 1/4, 1/5)  with horse worth  20.   Answer:  (52, 68, 76)/5.   See Diophantos I 24.

27:  I‑(1/3, 1/4, 1/5, 1/6).   Answer:  47, 77, 92, 101; 137.   (= Diophantos I 25.)

32:  x  =  y + _z,   y  =  z + _x,   z  =  x + _y.

33:  II‑(1/3, 1/5, 1/4).   Answer:  (44, 51, 50; 61)/11.

34:  = Diophantos I 22.

35:  = Diophantos I 23, but with answer divided by  23,  making  y = 4.

Tabari.  Mift_h al-mu‘_mal_t.  c1075.  P. 133, no. 52.  ??NYS _ Hermelink, op. cit. in 3.A, mentions this without details.  Tropfke 609 cites this problem and also p. 150f., no. 13, as buying a horse.

Fibonacci.  1202.  Pp. 228‑258: chap. 12, part 5: De emptione equorum inter consocios secundum datam proportionem.  Many examples, getting up to seven men, up to five horses, an inconsistent example and negative solutions.  I have omitted some of the variations and some of the more complex problems.

                    See:  K. Vogel; Zur Geschichte der linearen Gleichungen mit mehreren Unbekannten; Deutsche Mathematik 5 (1940) 217‑240;  for a thorough study of the problems on pp. 228‑258.  Some further versions occur on pp. 327‑349.  Vogel cites several sources, ??NYS:  Codex lat. Monacensis 14908 (1456)  and, for a simpler type of problem,  Mich. Pap. 620.

P. 228.  II‑(_, ¼).   Answer:  8, 9; 11.

P. 229.  II‑(1/3, 1/4, 1/5).   Answer:  45, 48, 52; 61.

P. 231.  II‑(2/3, 4/7, 5/9).   Answer:  135, 141, 154; 229.

Pp. 231‑232.  II‑(1/3, 1/4, 1/5, 1/6).   Answer:  264, 285, 296, 315; 359.

Pp. 234.  II‑(1/3, 1/4, 1/5, 1/6, 1/7).   Answer:  1855, 1998, 2092, 2145, 2156; 2521.  The margin has  1815  for  1855  and  2256  for  2156.

Pp. 234‑235.  II‑(2/3, 4/7, 5/11, 6/13, 8/19).   Answer:  35435, 35313, 41712, 38643, 44057; 58977.

Pp. 235.  II‑(1/4 + 1/3,  1/5 + 1/4,  1/6 + 1/5,  1/7 + 1/6).  Answer:  176274, 200772, 205820, 238830; 293391.

Pp. 235‑236.  Two men, two horses of values  h₁, h₂  =  h, h+2.   i‑th says  "If I had  a_i  of what the  i+1-st has, then I could buy the  i-th horse",  with  A = (_, ¼).  Gives the first two solutions:  8, 12; 12  and  16, 21; 23.  Varies to  h₂ = h+3  and gives the solution  20, 27; 29,   which is the third positive solution of the problem.

Pp. 236‑240.  Versions with  n  men and  n  horses,  n = 3, 4, 5.

Pp. 240‑242.  Four men, one horse, giving   w + _ (x + y)  =  ...  =  h,   with constants  1/3, 1/4, 1/5, 1/6.   Answer:  187, 209, 247, 273; 339.

Pp. 242‑243.  Three men, one horse, giving   x + y + _ z  =  ...  =  h,   with constants  1/3, 1/4, 1/5.   Answer:  16, 15, 18; 37.

P. 243.  Same with four men and constants  1/3, 1/4, 1/5, 1/6.  Answer:  15, 18, 15, 20; 38.   The margin has  28  for  38.

Pp. 243‑244.  Five men, one horse, giving   v + w + x + ¼ y  =  ...  =  h.   Answer:  1218, 1295, 1200, 1260, 1365; 4028.   The margin has  1269  for  1260.

P. 245.  I‑(1/3, 1/4, 1/5).   Answer:  13, 17, 19; 25.   (= Diophantos I 24.)

Pp. 245‑248.  I‑(1/2, 1/3, 1/4, 1/5).   Answer:  1, 19, 25, 28; 37.

Pp. 248‑249.  I‑(2/5, 3/8, 4/11, 6/19).   Answer:  1774, 2047, 2164, 2614; 4504.   Margin has  2164  for  2614.  States a variation:   I‑(1/3 + 1/4,  1/4 + 1/5,  1/5 + 1/6,  1/6 + 1/7)   with answer:  1376, 54272, 76022, 87902; 128657.

Pp. 249‑250: Questio nobis proposita a peritissimo magistro musco constantinopolitano in constantinopoli.  Buying a ship.  I‑(1/5 + 2/3,  1/480 + 1/6 + 2/3,  1/638 + 1/6 + 2/3,  1/420 + 1/7 + 2/3,  1/810 + 1/27 + 1/10 + 2/3).  Answer:  3, 228, 231, 348, 378; 1030.

Pp. 250‑251.  Five men, giving   v + w + ½ (x + y + z)  =  ...  =  h,   with constants  1/2, 1/3, 1/4, 1/5, 1/6.   Answer:  58, 19, 148, 49; 163.

Pp. 251‑252: Questio insolubilis.  Four men, giving equations   w + x + ½ (y + z)  =  ...  =  h,   with constants  1/2, 1/3, 1/4, 1/5.  This is inconsistent unless  h = 0.  He then varies the constants to  1/2, 3/7, 3/11, 5/13,   which gives answer:  5, 6, 7, 9; 19.

Pp. 252‑253.  Seven men, giving   t + u + v + ½ (w + x + y + z)  = ...  =  h,   with constants  ½, _, ..., _.   Answer:  507, 171, ‑9, 1347, 451, 131, 1431; 2349.  "... quare tercius homo habet debitum ipsos  9,  vel hec questio est insolubilis: sit itaque solubilis cum debito tercii hominis; ..."  He then varies the constants to  1/3, 1/4, ..., 1/9,  with answer:  1077, 717, 489, 1637, 997, 657, 1749; 3963.  See Sesiano.

Pp. 253‑254.  Two men and two horses of values  h₁, h₂ = h, h+2.  i-th says  "If I had  a_i  of what the rest of you have, then I could buy the  i-th horse",  with constants  _, ¼.  (This is the same as on pp. 235‑236 above, but only because  n = 2.)

P. 254.  Same with  3  men,  3  horses worth  h, h+2, h+4  and constants  1/3, 1/4, 1/5.  Answer:  7, 13, 17; 17.

Pp. 254‑257.  Same with  4  men,  4  horses worth  h, h+3, h+7, h+12  and constants  1/3, 1/4, 1/5, 1/6.   Answer:  (‑4, 13, 27, 41; 23)/2.   "Unde hec questio cum hiis iiii‑or positis residuis solui non potest, nisi primus homo haberet debitum."  He later gives another solution:  (82, 193, 265, 325; 343)/6,   by choosing a larger value for  h.  The margin has  32 1/2  for  32 1/6  which I have converted to  193/6.  See Sesiano.

Pp. 257‑258.  Four men, one horse, giving   w + x/2 + y/3 + z/4  =  x + y/3 + z/4 + w/5  =  ...  =  h.   Answer:  105, 168, 210, 240; 319.

Pp. 327‑329.  II‑(1/2, 1/3, 1/4, 1/5, 1/6)  done by false position.  Answer:  456, 530, 573, 592, 645; 721  _ the text has  529  instead of  592.  Sesiano notes that negatives are used in one of the false positions.

Pp. 334‑335.  Problem of pp. 245‑248 done by false position.

Pp. 336-338.  (_, ¼)  with  h_i = 14, 17.   Answer:  (100, 162)/11.

P. 338.  II-(1/3, 1/4, 1/5)  with  h_i = 14, 17, 19,  done by false position.   Answer:  (595, 777, 1040)/61.

Pp. 338-339.  I-(1/3, 1/4, 1/5)  with  h_i = 14, 17, 19,  done by false position.   Answer:  (241, 594, 783)/50.

Pp. 347‑349.  Three men, one horse, giving   x + y/2 + z/3  =  y + z/4 + x/5  =  z + x/6 + y/7  =  h.  Answer:  1530, 3038, 3540; 4229.

P. 349.  Four men, one horse, giving   w + x/2 + y/3 + z/4  =  x + y/4 + z/5 + w/6  =  y + z/6 + w/7 + x/8  =  z + w/8 + x/9 + z/10  =  h.  Answer:  8569848, 21741336, 26955060, 29657460; 35839901.

Fibonacci.  Flos.  c1225.  In:  Picutti, pp. 312-316 & 320-326, numbers III, VI & VII.

Pp. 236-238: De quinque numeris reperiendis ex proportionibus datis.  I‑(1/2, 1/3, 1/4, 1/5, 1/6).   Answer:  7, 10, 19, 25, 28; 34.

Pp. 240-242: No heading _ paragraph begins:  Item de mode predicto extraxi ....  Three men, three horses, type I with constants  1/3, 1/4, 1/5  and horses worth  14, 17, 19.   Answer:  (241, 594, 783)/50.

Pp. 242-243: De quatuor hominibus bizantios habentibus.  He refers to Liber Abaci, apparently to pp. 338-339.  I‑(1/2, 1/3, 1/4, 1/5)  with horses worth  33, 35, 36, 37.   Answer:  ‑3, 18, 25, 29.   "... hanc insolubilem esse sub posita conditione."  On p. 243, he states that if the values of the horses are  181, 183, 184, 185,   then the answer is  1, 94, 125, 141.  See Sesiano, who notes that for  h_i = h, h+2, h+3, h+4,  the smallest positive integral solution is that given by Fibonacci.

Fibonacci.  Epistola.  c1225.  In Picutti, pp. 338-340, numbers XV & XVI.

Pp. 250-251: Modus alius solvendi similes questiones.  I‑(1/2, 1/3, 1/4, 1/5, 1/6)  with horses worth  12, 15, 18, 20, 23.   Answer:  (4938, 7428, 10161, 11268, 15760)/721.

Pp. 251-252: Investigatio unde procedat inventio suprascripsit.  II‑(1/2, 1/3, 1/4, 1/5, 1/6)  with horses worth  12, 15, 18, 20, 23.   Answer:  (‑5316, 1479, 4532, 6157, 7920)/394.   He says  "tunc questio esset insolubilis, nisi concederetur, primum habere debitum; quod debitum esset [5316/394]."  See Sesiano.

Jordanus de Nemore.  c1225.  Op. cit. in 7.R.1.

Prob. II‑25, p. 151.  General version of type II with value of horse given.  Example:  II‑(1/2, 1/3, 1/4, 1/5)  with horse worth  119.  Answer:  75, 88, 93, 104.

Prob. II‑26/28, pp. 152‑155.  General version of type I with value of horse given.  Example in II‑26:  I‑(3, 13/4, 25/7, 4)  with horse worth  28.  Answer:  1, 2, 3, 4.  Example in II‑27:  I‑(1/2, 1/3, 1/4, 1/5)  with horse worth  37.  Answer:  1, 19, 25, 25.   II‑28 is II‑26  done in a different way.

BR.  c1305.

No. 6, pp. 26‑27.  Like Fibonacci's pp. 242‑243 with constants  1/5, 1/7, 1/9  and  h = 100.  Method is wrong, but the answer is right:  (2700, 2800, 3000)/61.

No. 7, pp. 26‑29.  Buying a ship,  (_, ¼)  with ship worth  100.  Answer:  (800, 900)/11.

No. 8, pp. 28‑29.  Buying a business,  (2/3, 3/5)  with business worth  100.  Answer:  (215, 258)/9.

No. 50, pp. 66‑69.  Same as no. 6 with constants  (1/3, 1/4, 1/5)  and  h = 11.  Answer:  45, 48, 54.

No. 55, pp. 72‑75.  Buying a business worth  20,  (4/9, 12/35).  Answer:  (10500, 12420)/801.

No. 56, pp. 74‑77.  Same,  (12/35, 13/42),  with  h = 72.  Answer:  (69552, 73080)/1314.

No. 57, pp. 78‑81.  Same,  (71/105, 37/60),  with  h = 50.  Answer:  (102000, 120750)/3673.

No. 58, pp. 80‑83.  A  gives  B  7/12  of what  A  has, then  B  returns  9/20  of what he has, then both have  12.   Answer:  (288, 1032)/55.

No. 72, pp. 94‑97.  Same with constants  1/7, 1/4,  and both finish with  36.  Answer:  28, 44.

No. 114, pp. 130‑131.  (= Fibonacci, p. 229.)

Gherardi.  Libro di ragioni.  1327.

P. 42.  Same as Fibonacci, pp. 235‑236, with  h_i = 10, 12,  A = (_, ¼).   Answer:  (72, 114)/11.

Pp. 45‑46.  Chopera.  II‑(1/2, 1/3, 1/4, 1/5).   Assumes house is worth  60.   Answer is  60/119  times Jordanus's II‑25.

Pp. 46‑47.  I‑(½, _, ¼).   Answer:  5, 11, 13; 17.

Pp. 59‑60.  Three men and three horses:  I‑(1/3, 1/4, 1/5)  with  h_i = 40, 47, 55.   Answer:  (702, 1602, 2289)/50.

Munich 14684.  14C.  Prob. XVII, p. 80.  I‑(½, _, ¼).   Answer:  10, 22, 26; 34.

Lucca 1754.  c1390.

F. 58r, p. 131.  II‑(½, _, ¼).

F. 58r, pp. 131‑132.  I‑(½, _, ¼).

F. 58v, p.132.  Two men.  First says:  "If you give me  _  of your money, then I can buy  20  horses."   Second says:  "If you give me  ¼  of your money, then I can buy  21  horses."

Ff. 61r‑61v, p. 141.  Three men and a friend.  i-th says:  "If I had  a_i  of our friend's money, I could buy the horse",  with  (a_i) = (½, _, ¼).  This has a two dimensional solution space.  He gives only  1/2,  5/2,  7/2  with friend having  12  and horse worth  13/2.

Provençale Arithmétique.  c1430.  Op. cit. in 7.E.  F. 100r-101v, pp. 49-53.

Three men buy a horse:  I‑(½, _, ¼).   Answer:  (5, 11, 13; 17).

Four men buy a horse:  I‑(1/2, 1/3, 1/4, 1/5)  (= Fibonacci, p. 245).  The answer here takes  C = 60,  which gives:  (5, 95, 125, 140; 185)/3

Five men buy a piece of cloth:  I‑(1/2, 1/3, 1/4, 1/5, 1/6).   Answer:  (‑43, 77, 117, 137, 149; 197)/4.  Sesiano, loc. cit. in 7.E, asserts  "Tel est le plus ancien exemple de l'acceptation d'une solution négative dans un texte mathématique, ...."  The text says simply  "restan  10  et  ¾  mens de non res."  Sesiano, loc. cit. at beginning of 7.R, gives this in English:  "... the first work in which a negative result is admitted without any restriction, ....  No interpretation whatsoever is given of the negative result.  The only hint at its peculiarity is the _ exceptional _ verification of the results: ...."

AR.  c1450.  Prob. 156, 157, 171‑181, 186, 221, 223, 224, 338‑341.  Pp. 74‑75, 82‑85, 87, 102‑103, 149‑150, 171‑173, 218‑219.

156:  Regula posicionum:  II‑(1/3, 1/4, 1/5)  with horse worth  100.

157:  II‑(1/3, 1/4, 1/5).   Answer:  45, 48, 52; 61.

171:  I‑(_, ¼).   Answer:  8, 9, 11.   Vogel says this occurs in Ibn al‑Haitham, ??NYS.

172‑175:  same as 171, with various values of horse:  31, 100, 81, 1.

176:  I‑(_, _)  with horse worth  30.

177:  173 redone.

178:  Pferd mit  3  an namen (Horse with  3  [men] without price).   I‑(½, _, ¼).   Answer:  10, 22, 26; 34.

179:  same heading,  II‑(½, _, ¼).   Answer:  16, 18, 21; 25.

180:  Pferd mit  3  [und] mit namen (horse with  3  [men and] price).   II‑(½, _, ¾)  with horse worth  100.   Answer:  (200, 200, 150)/3.

181:  3  men buy a fish, leading to   x + ¼ (y + z)   =   ½y + ¼ (x + z)  =  _z + ¼ (x + y)  =  cost.   Answer:  1, 3, 9; 4  and its multiples.

186:  3  men want to buy horses worth  10,  20,  30,  as in Fibonacci, pp. 235‑240, with  A = (½, _, ¼).   Answer:  (24, 52, 144)/5.

221:  same as 173.

223:  II‑(½, _, ¼)  with horse worth  51.  See 179.

224:  same as 178, with answer:  5, 11, 13; 17.

338:  same as 157.

339:  same as 179, with comment that multiplying the solution is again a solution.

340:  same as 178.

341:  I‑(1/3, 1/4, 1/5).   Answer:  (65, 85, 95; 125)/2.

Dresden MS C80, 15C, has some of these problems.  ??NYS _ mentioned in BR, p. 157.

Muscarello.  1478.  Ff 56v-58r, pp. 158-161.  Four men buying a house worth  100,  II‑(2/3, 5/8, 4/5, 7/10).   Answer should be:  (1250, 1575, 1160, 1425)/23,  but there are two copying errors in the MS.  The second answer is given as  66 11/23  instead of  68 11/23  and the fourth answer is given as  61 12/23  instead of  61 22/23.

P. M. Calandri.  c1480.

   Pp. 155‑157. 

(½, _)  with  h_i = 60, 80.

I-(½, _, ¼)  with  h_i = 60, 50, 163/5. 

II-(½, _, ¼)  with  h_i = 60, 50, 40.

   Chap. 20:  "... uomini che ànno d. et vogliono chompare chavagli," pp. 168‑171.

I‑(1/6, 1/7).  Answer:  35, 36; 41.

II‑(1/4, 1/6, 1/8).  Answer:  152, 164, 174; 193.  Illustration on p. 169.

I‑(½, _, ¼).  Answer:  5, 11, 13; 17.

Chuquet.  1484.  Triparty, part 1.  Sesiano cites p. 641, ??NYS

I-(½, _, ¼)  with object worth  30.  English in FHM 79.  Answer:  (150, 324, 390)/17.

FHM 79-80 then simply states and discusses the next two, which are discussed by Sesiano.

I-(2/3, 3/4, 4/5, 5/6)  with horse worth  40.   Answer:  24, 16, 8, 0.

I-(1/2, 2/3, 3/4, 4/5, 5/6)  with horse worth  40.   Answer:  0, 20, 10, 0, ‑10.  Chuquet then explains how to deal with negatives and zero.

FHM 81 gives the English of a version with equations   x + y + ½z  =  _x + y + z  =  x + ¼y + z  =  20.   Answer:  (180, 160, 240)/23.

FHM 81-83 then gives the next two by formulae with solutions, then then gives the English of the next.

II-(½, _, ¼)  with object worth  20.   Answer:  (64, 72, 84)/5.

II-(½, _, ¾)  with object worth  30.   Answer:  (600, 540, 495, 500)/29.

II-(1/3, 1/4, 1/5)  with object worth  20.   Answer:  (900, 960, 1040)/61.

Borghi.  Arithmetica.  1484.  Ff. 113v-116v (1509: ff. 95v-98r).  I‑(½, _, ¼)  with  h = 20.   Answer:  (100, 220, 260)/17.

Calandri.  Arimethrica.  1491.  F. 667v.  Two men buy a lamprey.  I-(1/3, 1/5).  Takes  h = 60  rather arbitrarily and gets answer:  (300, 360)/7.

Francesco Pellos.  Compendion de lo Abaco.  Turin, 1492.  ??NYS _ see Rara 50-52  &  both Sesiano papers mentioned at Provençale Arithmétique, c1430, above.  Sesiano says ff. 64v-65r translates the three problems in the Provençale Arithmétique and is the first printed problem with a negative solution.  He says it may have been composed c1460.  Smith doesn't mention either of these points.

Pacioli.  Summa.  1494.

Ff. 105v-106r, prob. 23.  I-(½, _, ¼)  with each saying  "If you ..., I will have  20".  Answer:  (100, 220, 260)/17.

F. 190v, prob. 21.  Two men find a purse, but leads to:   x + _y  =  p,   y + ¼x  =  p,   so this is  I-(_, ¼).  He assumes   x + y + p  =  100   and gets answer:  (200, 225; 275)/7.

Ff. 191v-192r,  prob. 26.   x + ½ (y + z)  =  90,   y + _ (z + x)  =  84,   z + ¼ (x + y)  =  81.   Answer:  (576, 900, 1008)/17.

F. 192r, prob. 27.  Same as prob. 26 with all values equal to  50.  Answer:  (250, 550, 650)/17.

F. 192v, prob. 31.  I-(½, _, ¼).   Notes that the common value,  h,   can be set, as in prob. 23 (cf. 7.R.1) to  50,  but when fractions appear, he converts to answer:  (5, 11, 13; 17).

F. 193r, prob. 35.   x + (y + z)/3 + 1  =  14,   y + (z + x)/4 ‑ 2  =  17,   z + (x + y)/5 + 3  =  19.   Answer:  (197, 748, 611)/50.

F. 193v, prob. 37.   x + y/2 + z/3  =  12,   y + z/3 + x/4  =  15,   z + x/4 + y/5  =  20.   Answer:  (164, 810, 1677)/94.

Ff. 193v-194r, prob. 41.  2  men find a purse and want to buy a horse, giving:   x + y + p/3  =  h,   y + z + p/5  =  h,   z + x + p/4  =  h.   If  T = x + y + z,  one gets   T + 47p/60  =  3h   and the solution space is actually two dimensional.  He assumes  p = 60,  h = 47  and then this problem reduces to prob. 39.

Calandri, Raccolta.  c1500.  Prob. 25, pp. 24‑25.  I‑(1/3, 1/5)  with horse worth  60.  Answer:  (300, 160)/7.

Riese.  Rechnung.  1522.  1544 ed. _ pp. 91‑92 & 94‑95;  1574 ed. _ pp. 61v‑62v & 63v‑64r.

I‑(_, ¼)  with horse worth  15.   Answer:  (120, 135)/11.

I‑(_, ¾)  with horse worth  39.  Answer:  (52, 39)/2.

II‑(½, _, ¼)  with cow(?) worth  200.   Answer:  64, 72, 84.

Tonstall.  De Arte Supputandi.  1522.  Pp. 246-248.  I-(½, _, ¼)  with common value  20.

Riese.  Die Coss.  1524.  Many examples, including the following.

No. 31, pp. 44‑45.  3  men buy a horse worth  100,  II‑(½, _, ¼).

No. 47, pp. 46‑47.  Same with horse worth  17,  I‑(½, _, ¼).

No. 48, p. 47.  3  men buy  3  horses leading to:   a + b + c/5  =  12,   a/2 + b + c  =  18,   a + b/3 + c  =  16.

No. 120, p. 56.  Same as no. 31.

No. 121, p. 56.  3  men buy horse worth  100,  II‑(3/4, 4/5, 5/6).   Answer:  (510, 520, 475)/9.

No. 122, p. 56.  Same,  I‑(½, _, ¼).   Says his friend Hans Conrad learned this from a Dominican(?) monk named Aquinas.

No. 123, p. 56.  Same with horse worth  204,  I‑(½, _, ¼).

No. 126, p. 58.  4  men buy a horse worth  37,  I‑(1/2, 1/3, 1/4, 1/5).   Answer:  1, 19, 25, 28.

No. 124 & 125, pp. 56‑58.  Complex formulations leading to the same problem with horse worth  2701  and  14800.   No. 125 has the order of the constants reversed.

No. 140, pp. 60‑61.  7  men buy a horse worth  100,  II‑(2/3, 3/4, ..., 8/9).  Answer:  (27630, 27855, 24460, 27175, 22830, 27265, 21640)/462.

No. 143, pp. 61‑62.  4  men buy a horse worth  100  leading to  II‑(1/3, 1/5, 1/6, 1/8)  except the last man borrows from all the others.  Answer:  (9747, 11058, 11875, 9348)/707.

Apianus.  Kauffmanss Rechnung.  1527.  Ff. M.iii.r - M.iii.v.  x + y/3  =  y + z/2  =  z + 2y/3  =  30.  Answer:  (45, 45, 60)/2.  Because  x = y,  this is actually the same as  II-(1/3, 1/2, 2/3).

Georg von Peurbach.  Elementa arithmetices, algorithmus ....  Joseph Klug, Wittenberg, 1534, ??NYS.  (There were several previous editions back to 1492, with variant titles.  Rara 53‑54.  Glaisher, op. cit. in 7.G.1 under Widman, describes this extensively and gives the following on p. 97.  This edition is substantially better than previous ones, but Peurbach died in 1461!)

F. D.iii.verso.  I-(½, _)  with horse worth  10.

F. E.i.verso.  II-(½, _, ¼)  with horse worth  100.

Cardan.  Practica Arithmetice.  1539.  Chap. 66, section 98, ff. HH.vi.r - HH.vi.v (p. 169).  Men find a purse and buy a horse, giving:   x + y + p/2  =  y + z + p/5  =  x + z + p/3  =  h.   Answer:  6, 10, 15; 30, 31.

Tartaglia.  General Trattato.  1556.  Book 17, art. 8, 13, 14, 22, 32, 40, 41, pp. 268v, 269v, 271v, 273v, 275v-277r.

Art. 8.  (_, ¼)  with  h_i = 14, 17.   Answer:  (100, 162)/11.

Art. 13.  II-(½, _, ¼)  with  h_i = 40, 52, 62.   Answer:  (584, 832, 1404)/25.

Art. 14.  II-(2/3, 5/8, 4/5, 7/10)  with house worth  200.  Answer:  (2500, 3150, 2320, 2850)/23.

Art. 22.  II-(½, _, ¾).   Answer:  48, 48, 36; 72,   which is not in lowest terms.

Art. 32.  (_, ¼)  with  h_i = 32, 42.   Answer:  (216, 408)/11.

Art. 40.  I-(½, _, ¼)  with  h = 40.   Answer:  (200, 440, 520)/17.

Art. 41.  I-(½, _, ¼)  with  h = 20.   Takes about two pages to finally get half of the preceeding answer.

Buteo.  Logistica.  1559. 

Pp. 189-190.  I-(½, _)  with amounts  30, 20.   Answer:  24, 12.

Pp. 190-192.  I-(1/3, 1/4, 1/5)  with amounts  14, 8, 8.  I find this remarkable in that he uses three unknowns _ A, B, C _ and solves by systematic elimination.   Answer:  11, 4, 5.

Pp. 192-193.  Same with amounts assumed equal.  He assumes the amount is  17  and gets:  5, 11, 13; 17.

Pp. 193-196.  I-(1/2, 1/3, 1/4, 1/6)  with amounts  17, 12, 13, 13.   Answer:  6, 4, 8, 10.

Prob. 81, pp. 289-291.  Problem with soldiers, equivalent to  I‑(1/2, 1/3, 1/4)  with amount  14280.   Answer:  4200, 9240, 10920.

Prob. 30, pp. 357-358.  Amounts desired from others are variable, giving   x + y/2 + z/3  =  14,   x/3 + y + z/4  =  13,   x/6 + y/8 + z  =  14.   Answer:  6, 8, 12.

"A Lover of the Mathematics."  A Mathematical Miscellany in Four Parts.  2nd ed., S. Fuller, Dublin, 1735.  Part III, no. 45, p. 94.  Men buying a house worth 1200.  I-(_, ¾).   Answer:  800, 600.

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790), prob. XXV, pp. 95-96 (1790: prob. 34, pp. 94‑95).   II‑(1/2, 1/3, 1/4, 1/5)  with  h_i = a, b, c, d.   Does example with values  357, 476, 595, 714  and answer:  190, 334, 426, 676.   1745 gives a second method.

Les Amusemens.  1749.  Prob. 172, pp. 319-320.  I‑(½, _, ¼)  with  h = 40.   Solves with a general  h.

Euler.  Algebra.  1770.  I.IV.

III: Question for practice, no. 19, p. 205.  I‑(½, _, ¼)  with horse worth  34.

IV.618: Question 5, pp. 212‑213.  I‑(_, ¾)  paying a debt of  29.

IV.619‑620: Question 6, pp. 213‑214.  II‑(½, _, ¼)  buying a vineyard worth  100.

IV.621, pp. 214‑215.  Gives general form of solution of type II problem, using  4  person case as an example.

IV.622: Question 7, pp. 215‑216.  Problem equivalent to  I‑(½, _, ¼)  with  h = 901,  i.e.  53/2  times the problem on p. 205.

Vyse.  Tutor's Guide.  1771?  Prob. 13, p. 138 & Key pp. 183-184.  I‑(_, ¾)  buying a horse worth  1200.

Pearson.  1907.  Part II, no. 139: The money‑boxes, pp. 141 & 218.  I‑(1/2, 1/3, 1/4, 1/5)  with  h = 740.

Adams.  Puzzles That Everyone Can Do.  1931.  Prob. 101, pp. 43 & 144: A garaging problem.  A variation on type IV.  a + b + c + d + e = 100;  a + b = 52;  b + c = 43;  c + d = 34;  d + e = 30.

 

          7.R.3. SISTERS AND BROTHERS

 

          New section.

          NOTATION:  (a, b)  means each boy has  a  times as many sisters as brothers, while each girl has  b  times as many brothers as sisters.  This only has integer solutions for the integer pairs:

   (a, b)                   = (1, 2),  (1, 3),  (2, 1),  (2, 2),  (3, 1),   with solutions: 

(Boys, Girls)  = (4, 3),  (3, 2),  (3, 4),  (2, 2),  (2, 3).

          See Wolff for a related problem.

 

Peano.  Giochi.  1924.  Prob. 55, p. 15.  (1, 2).

King.  Best 100.  1927.  No. 39, pp. 19 & 47.  = Foulsham's, no. 11, pp. 7 & 11.  (2, 1).

Dr. Th. Wolff.  Die lächelnde Sphinx.  Academia Verlagsbuchhandlung, Prague, 1937.  Prob. 28, pp. 193 & 203.  A man in an office says he has four times as many male colleagues as female colleagues.  One of the women answers that she has five times as many male colleagues as female colleagues.  In general, this would give  m - 1 = af,  m = b (f - 1),  which has solution  f = (b + 1)/(b - a).  See Dodson in 7.R for a different phrasing of the same problem.

Depew.  Cokesbury Game Book.  1939.  Sisters and brothers, p. 218.  (2, 1).

Owen Grant.  Popular Party Games.  Universal, London, nd [1940s?].  Prob. 8, pp. 36 & 50.  (2, 1).

John Henry Cutler.  Dr. Quizzler's Mind Teasers.  Greenberg, NY, 1944.  ??NYS _ excerpted in: Dr. Quizzler's mind teasers; Games Magazine 16:3 (No. 109) (Jun 1992) 47 & 43, prob. 27.  "Mr. and Mrs. Twichell have six daughters.  Each of the daughters has a brother.  How many persons are there in the entire family?"  Only  9  (counting the parents).

Liz Allen.  Brain Sharpeners.  Op. cit. in 5.B.  1991.  The sum of the siblings, pp. 31 & 110.  (2, 1).

 

          7.R.4. "IF I SOLD YOUR EGGS AT MY PRICE, I'D GET ...."

 

          New section.

 

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790), prob. XL, pp. 106-107 (1790: prob. LIII, p. 108).  Market women bring  x  and  y  eggs to market, with  x + y = 100 [c]  and sell at prices  A  and  B  such that they get the same, i.e.  Ax = By.  First says to the second:  "Had I brought as many eggs as you I should have received  18 [a]  Pence for them".  The other responds:  "Had I brought no more than you, I should have received only  8 [b]  Pence for mine".  I.e.  Ay = a,  Bx = b.  This gives   (x/y)²  =  b/a,   x  =  cÖb/[Öa+Öb],   y  =  cÖa/[Öa+Öb].

McKay.  At Home Tonight.  1940.  Prob. 11: Buying cows, pp. 64 & 78.  Farmers  A  and  B  each buy  £350  worth of cows.  If  A  had bought at  B's  price, he would have paid  £250.  What would  B  have paid if he bought at  A's  price?  Letting  x, y  be the numbers of cows and  A, B  be the prices, we have   Ax = By = 350,   Bx = 250,  and we want  Ay.  Then   y/x  =  7/5   and   Ay  =  Ax(7/5)  =  490.

 

          7.S.    DILUTION AND MIXING PROBLEMS

 

          See Tropfke 569.

          There are a number of problems of this sort.  One type is the same as the Hundred Fowls problem (7.P.1) where the solutions need not be integers.  Lucca 1754, c1390, has a number of these.  Here I consider only some of special interest and the following.

 

Robert Recorde.  .  Reginald Wolfe, London, 1542.  ??NYS.  1579 ed., f. Y.3.  "It hath great use in composition of medicines, and also in myxtures of metalles, and some use it hath in myxtures of wines. but I wshe it were lesse used therein than it is now a daies."  (Quoted in H&S 83.)  (I haven't found this in the 1542 ed., which has gatherings only up to P.)

 

          7.S.1.  DISHONEST BUTLER DRINKING SOME AND REPLACING WITH

                                        WATER

 

          Dodson and Clark are problems to determine the amount taken off each time.

 

Papyrus Rhind, op. cit. in 7.C.  Prob. 71, p. 108 of vol. 1 (1927) (= p. 57 of 1978 ed.).  ¼  is poured off & replaced, what is the strength?  (H&S 85 quotes Peet's version.)

Bakhshali MS.  c7C.  Kaye I 48; III 201-202, ff. 12r-12v.  Man has bottle holding  4  prasthas of wine.  (Drinks  ¼  and refills with water)  four times.  How much wine is left?

Cardan.  Practica Arithmetice.  1539.  Chap. 66, sections 36 & 37, ff. DD.v.r - DD.v.v (p. 146).  Drink three pitchers and replace with water four times leaving wine of half strength.  Then the same for three times.

Tartaglia.  Quesiti, et Inventioni Diverse, 1546, op. cit. in 7.E.1, Book 9, quest. 18, pp. 102v‑103r.  (Remove 2 and replace)  thrice to halve strength.

Buteo.  Logistica.  1559.  Prob. 85, pp. 296-298.  Butler drinking some and replacing with water.  (7/8)⁵.  (H&S 85)

Trenchant.  Op. cit. in 7.L, 1566.  1578 ed., p. 297.  ??NYS.  (Remove  1/12th  and replace)  six times.  (H&S 85 gives French and English.  Sanford 209 gives English.)

Les Amusemens.  1749.  Prob. 176, pp. 326-327.  Sommelier drinks  6  pints from a cask of  360  and replaces with water three times _ how much wine has he drunk?

Dodson.  Math. Repository.  1775.  P. 76, Quest. CXLI.  Cask of 81 gallons.  (x  is drawn off and replaced by water) four times, leaving 16 gallons of wine in the mixture.  Gives a general solution.

Ozanam‑Montucla.  1778.  Prob. 21, 1778: 212-214;  1803: 207-209.  Prob. 20, 1814: 179‑181;  1840: 93.  Dishonest butler  (removes  1/100th  and replaces)  30  times.  Notes that it is easier to use logarithms.

Bullen.  Op. cit. in 7.G.1.  1789.  Chap. 38, prob. 33, p. 243.  Cask of  500  gallons;  (remove  1/10th  and replace with water)  five times.

Louis Pierre Marie Bourdon (1779-1854).  Élémens d'Algèbre.  7th ed., Bachelier, Paris, 1834.  (The 6th ed., was 1831, but I haven't yet looked for the early editions _ ??)  Art. 222, last problem, p. 364.  Barrel of  100  pints of wine.  One pint is drunk and replaced by water each day.  How much wine is left after  50  days?  When is the wine diluted to half its strength?  One third?  One quarter?

Anonymous.  A Treatise on Arithmetic in Theory and Practice: for the use of The Irish National Schools.  3rd ed., 1850.  Op. cit. in 7.H.  P. 358, no. 35.  Same as Bullen.

M. Ph. André.  Éléments d'Arithmétique (N^o 3) a l'usage de toutes les institutions ....  3rd ed., Librairie Classique de F.-E. André-Guédon, Paris, 1876.  Prob. 98, p. 62.  Barrel of wine holding  210.  Remove  45  and replace with water three times.  Determine the amount of wine and water.

Lucas.  L'Arithmétique Amusante.  1895.  Prob. XLI: Le tonneau inépuisable, p. 183.  (Remove  1/100  and add water)  20  times _ how much is left?  Gives general solution.

Dudeney.  Weekly Dispatch (8 Feb 1903) 13.  (Remove  1/100th  and replace)  30  times.

Clark.  Mental Nuts.  1904: no. 39.  Find capacity of the keg.  (Fill a keg from a  20 gallon cask and then replace with water) three times to dilute the cask to half-strength.  How big is the keg?

 

          7.S.2.  WATER IN WINE VERSUS WINE IN WATER

 

          Two containers, one of wine and one of water.  One puts an amount of water in the wine, stirs and then transfers the same amount of the mixture back to the water.  Is there now more water in the wine or wine in the water?

 

M. Ph. André.  Éléments d'Arithmétique (N^o 3) a l'usage de toutes les institutions ....  3rd ed., Librairie Classique de F.-E. André-Guédon, Paris, 1876.  A predecessor of this problem _ prob. 92, p. 61.  Vase  A  has  12  litres of wine and  4  litres of water.  Vase  B  has  8  litres of wine and  3  litres of water.  Take off  4  litres from each and then put the  4  litres from  A  into  B  and the four litres from  B  into  A.  Determine contents in each vase.

Viscount Simon.  [Memory of Lewis Carroll.]  IN:  Appendix A of Derek Hudson; Lewis Carroll, An Illustrated Biography; Constable, (1954);  new ed., 1976, pp. 248‑249.  50  spoonfuls of brandy and  50  spoonfuls of water _ transfer a spoonful back and forth.  He says Carroll posed this.

Ball.  MRE, 3rd ed., 1896, pp. 26‑27.  Water in wine versus wine in water.  He says this is a question "which I have often propounded in past years".  Not in the 1st ed.

W. P. Workman.  The Tutorial Arithmetic, op. cit. in 7.H.1.  1902.  Section IX, examples CXLV, prob. 35, pp. 427 & 544  (= 433 & 577 in c1928 ed.).  A  contains  11  pt water and  7  pt wine,  B  contains  5  pt water and  13  pt wine.  Transfer  2  pt from  A  to  B  and back.  Find changes of water and wine in both  A  &  B.  This is a precursor of the puzzle idea.

Pearson.  1907.  Part II, no. 18, pp. 117 & 194‑195.  Butter in lard versus lard in butter.

Loyd.  Cyclopedia, 1914, pp. 287 & 378.  Forty quarts milk and forty quarts water with a quart poured back and forth.  He says the ratios of milk to water are then  1 : 40  and  40 : 1,  which is correct, but isn't the usual question.

H. E. Licks.  Op. cit. in 5.A.  1917.  Art. 20, pp. 16‑17.  Water and wine.

Ahrens.  A&N, 1918, p. 89.

T. O'Conor Sloane.  Rapid Arithmetic.  Quick and Special Methods in Arithmetical Calculation Together with a Collection of Puzzles and Curiosities of Numbers.  Van Nostrand, 1922.  [Combined into: T. O'Conor Sloane, J. E. Thompson & H. E. Licks; Speed and Fun With Figures; Van Nostrand, 1939.]  Wine and water paradox, pp. 168‑169.

F. & V. Meynell.  The Week‑End Book.  Op. cit. in 7.E.  1924.  Prob. two: 2nd ed., p. 274;  5th?? ed., p. 407.  Find "proportion of the amount of water in  A  to the amount of milk in  B."

Peano.  Giochi.  1924.  Prob. 24, p. 7.  Water and wine.

Loyd Jr.  SLAHP.  1928.  Cheating the babies, pp. 40 & 98.  Two large cans with  10  gallons of milk and  10  of water.  Pour  3  gallons back and forth.  "Have I more milk in the water can than I have water in the milk can?"  He works out that each can has the proportion  7 9/13 : 2 4/13  [=  100 : 30  =  10 : 3].

Phillips.  Week‑End.  1932.  Time tests of intelligence, no. 25, pp. 17 & 190.  Whisky and water in equal amounts.  He asks about proportions rather than amounts.

Phillips.  The Playtime Omnibus.  Op. cit. in 6.AF.  1933.  Section XVII, prob. 17: The two vessels, pp. 56 & 238.  Same as in his Week‑End.

Abraham.  1933.  Prob. 26 _ Whiskey and water, pp. 10 & 24 (7 & 112).

Perelman.  FMP.  c1935?  Water and wine, pp. 215 & 218.

Phillips.  Brush.  1936.  Prob. D.5: Whisky and water, pp. 12 & 82.  Same as in his Week‑End.

Dr. Th. Wolff.  Die lächelnde Sphinx.  Academia Verlagsbuchhandlung, Prague, 1937.  Prob. 36, pp. 194 & 204.  Man takes a cup of black coffee.  He drinks  1/6  of it and fills it up with milk.  He then drinks  1/3  of it and fills it up with milk.  He then drinks half of it and fills it up with milk.  Then he drinks the whole cup.  Has he drunk more milk than coffess or vice-versa?

McKay.  Party Night.  1940.  No. 6, pp. 176-177.  Water and wine.  "I have seen, after dinner, parties discuss the problem for a long time, ...."

Sullivan.  Unusual.  1943.  Prob. 21: Mixtures.  Gasoline and alcohol.

Northrop.  Riddles in Mathematics.  1944.  1944: 14-16;  1945: 13‑15;  1961: 23‑25.  Water and milk.

Gamow & Stern.  1958.  Gin and tonic.  Pp. 103‑104.

 

          7.T.    FOUR NUMBER GAME

 

          The game takes  (a, b, c, d)  to  (½a ‑ b½, ½b - c½, ½c ‑ d½, ½d ‑ a½).  See 7.BB for other iterated functions.

 

C. Ciamberlini & A. Marengoni.  Su una interessante curiosità numerica.  Period. Mat. (4) 17 (1937) 25‑30.  They attribute the problem to E. Ducci.

G [= J. Ginsburg].  Curiosa 30.  An interesting observation.  SM 5 (1938) 135.  Brief report on above article.

Benedict Freedman.  The four number game.  SM 14 (1948) 35‑47.  No references.  Obtains basic results for  n‑number version.

S. P. Mohanty.  On a problem of S. J. Bezuszka & M. J. Kenney on cyclic difference of pairs of integers.  Fibonacci Quarterly 19 (1981) 314‑318.  (This and three of its references are not in Meyers.)

Leroy F. Meyers.  Ducci's four‑number problem:  a short bibliography.  CM 8 (1982) 262‑266.  See Ludington-Young, below, for nine additional references.

M. Gardner.  Riddles of the Sphinx and Other Mathematical Puzzle Tales.  New Math. Library, MAA, 1987.  Prob. 29: Hustle off to Buffalo, parts 2‑5, pp. 134‑136, 151‑152, 160‑163.  Gives a proof that most quadruples converge to all zeroes and finds the quadruples that cycle.

Joseph W. Creely.  The length of a three-number game.  Fibonacci Quarterly 26:2 (May 1988) 141-143.  Solves three and two number versions.

Stanley Bezuszka with Lou D'Angelo & Margaret J. Kenny.  The Wonder Square.  Booklet 2, Boston College Math. Inst.  Motivated Math Project Activity.  Boston College Press, Chestnut Hill, Mass., 1976.  32pp.  Studies the process with various special numbers, e.g. progressions, Fibonacci, Tribonacci and figurate numbers.  The Tribonacci case produces starting sequences with length  n  for any  n.

Anne Ludington-Young.  The length of the  n-number game.  Fibonacci Quarterly 28:3 (Aug 1990) 259-265.  Obtains a bound and solves some cases.  Cites Meyers and 9 additional references.

 

          7.U.   POSTAGE STAMP PROBLEM

 

          What integers are non‑negative integral combinations of  a, b, ...?  In particular, what is the largest integer which is not such a combination?  This is well known for the case of two values, but remains unknown for more values.  From about 1960 onward, the case with two values frequently occurs in number theory texts and as a puzzle problem, but I haven't entered such appearances.

 

Dickson, vol. II, chap. II is uncharacteristically obscure about this.  It is generally attributed to Frobenius (1849-1917).

J. J. Sylvester.  Math. Quest. Educ. Times 41 (1884) 21.  ??NYS.  Solves the problem for two values.

Alfred Brauer.  On a problem of partitions.  Amer. J. Math. 64 (1942) 299‑312.  On p. 301, he says that some of the results are due to I. Schur in a lecture in 1935, others are due to himself and others are joint work.  He cites Sylvester, but doesn't mention Frobenius.

Alfred Brauer & B. M. Seelbinder.  On a problem of partitions II.  Amer. J. Math. 76 (1954) 343‑346.  "A number of years ago, the first of the authors studied together with I. Schur the following problem of Frobenius: ..."

Alfred Brauer & James E. Shockley.  On a problem of Frobenius.  J. reine angew. Math. 210 (1962) 215‑220.  "G. Frobenius, in his lectures, raised the following problem repeatedly ....  No result was obtained for many years.  In 1935, I. Schur proved in his last lecture in Berlin the following result ...."

Ernest S. Selmer.  On the linear diophantine problem of Frobenius.  J. reine angew. Math. 293/294 (1977) 1‑17.  Gives 25 references which he believes to be a complete list.  Cites Sylvester, but the next oldest are the 1942 and 1954 papers above.  The 1962 paper above is the first to mention Frobenius in the title.

 

          7.V.    X^Y  =  Y^X  AND ITERATED EXPONENTIALS

 

D. Bernoulli.  Letter to C. Goldbach, 29 Jun 1728.  In:  P. H. Fuss, ed. op. cit. in 5.F.1, vol. 2, p. 262.  x^y = y^x  in integers.

Goldbach.  Letter to D. Bernoulli, 31 Jan 1729.  Ibid, pp. 280‑283.  Reply to the above.  Setting  y = ax,  he gives an easy proof for the only integer solutions.  He says the fractional solutions are  (f/g)^g/(f‑g).

L. Euler.  Introductio in Analysin Infinitorum.  Bousquet, Lausanne, 1748.  Vol. 2, § 519, pp. 295-296 & Tab. XXV, fig. 103.  = Introduction to the Analysis of the Infinite; trans. by John D. Blanton; Springer, NY, 1988-1990; Book II, pp. 339-340 & 489.  Gets  x  =  (1 + 1/n)ⁿ;   y  =  (1 + 1/n)ⁿ⁺¹.

L. Euler.  De formulis exponentialibus replicatus.  Acta Acad. Sci. Petropol. 1 (1777(1778)) 38‑60.  = Opera Omnia (1) 15 (1927) 268‑297.  Iterated exponentials.

Ahrens.  A&N, 1918, pp. 76‑78, discusses the problem and notes that Goldbach's fractional solution is rational if and only if  g/(f‑g)  is integral, say  n,  which gives   x  =  (1 + 1/n)ⁿ.

R. L. Goodstein.  Note 1725:  The equation  a^b = b^a.  MG 28 (No. 279) (May 1944) 76.  Quick derivation of general solution and the rational solutions.

R. A. Knoebel.  Exponentials reiterated.  AMM 88 (1981) 235‑252.  Extensive history and bibliography.

F. Gerrish.  Note 76.25:  a^b = b^a:  the positive integer solution.  MG 76 (No. 477) (Nov 1992) 403.  Short note on the integer case with two recent references.

R. F. Churchhouse.  Solutions of the equation  x^y = y^x.  Bull. Inst. Math. Appl. 31:7/8 (Jul/Aug 1995) 106.  Easily finds the real solutions, then the rational and integer solutions.  Notes that  x = i,  y = -i  is a solution!

 

          7.W.  CARD PILING OVER A CLIFF

 

          Identical cards (or dominoes) of length  1  can be stacked to reach out from the edge of a cliff.  The simplest analysis shows that  n  cards can reach out  1/2 + 1/4 + 1/6 + ... + 1/2n  _  (log n)/2.  Some authors consider real dominoes which can be piled in several orientations.

 

J. G. Coffin, proposer.  Problem 3009.  AMM 30 (1923) 76.  Asks for maximum overhang for  n  cards.  (Never solved!)

A. S. Ramsey.  Statics.  2nd ed., CUP, 1941, example 4.68, pp. 47‑48.  Discusses equal spacing with a support at the outer end, e.g. a staircase.  (Is this in the 1st ed. of 1934??  My source indicated that Ramsey cited a Tripos exam.)

R. T. Sharp, proposer;  C. W. Trigg, solver.  Problem 52.  Pi Mu Epsilon J. ??  &  (April 1954) 411‑412.  Shows overhang approaches infinity, but the proposal asks for the largest overhang for  n  dominoes, which is not answered.  Notes that the dominoes can be angled so the diagonal is perpendicular to the cliff edge.  This is also in Trigg; op. cit. in 5.Q; Quickie 52: Piled dominoes, pp. 17 & 99, but it still doesn't answer the proposal.

P. B. Johnson, proposer;  Michael Goldberg, Albert Wilansky, solvers.  Problem E1122 _ Stacking cards.  AMM 61 (1954) 423  &  62 (1955) 123‑124.  Both show overhang can go to infinity.

Paul B. Johnson.  Leaning tower of lire.  Amer. J. Physics 23 (Apr 1955) 240.  Claims harmonic series gives greatest overhang!!

P. J. Clarke.  Note 2622:  Statical absurdity.  MG 40 (No. 333) (Oct 1956) 213‑215.  Considers homogeneous bricks with weight bounded above and below and length bounded below.  Then one can take such bricks in any order to achieve an arbitrarily large overhang.

Gamow & Stern.  1958.  Building blocks.  Pp. 90‑93.

D. St.P. Barnard.  Problem in The Observer, 1962.  ??NYS.

D. St.P. Barnard.  Adventures in Mathematics.  Chap. 8 _ The Domino Story.  (1965);  Funk & Wagnalls, NY, 1968, pp. 109‑122.  Gets  3.969  for  13  dominoes.

Birtwistle.  Math. Puzzles & Perplexities.  1971.  Overbalance, pp. 122-124.  Standard piling.  Shows the harmonic series diverges.

See A. K. Austin, 1972, in 5.N for a connection with this problem.

Stephen Ainley.  Letter:  Finely balanced.  MG 63 (No. 426) (1979) 272.  Gets  1.1679  for  4  cards.

[S. N. Collings].  Puzzle no. 47.  Bull. Inst. Math. Appl. 15 (1979) 268 & 312.  Shows that a simple counterbalancing scheme gets  m/2  for  2^m ‑ 1  dominoes, so the overhang for  n  dominoes is at least  ½ log₂(n+1).

R. E. Scraton.  Letter:  A giant leap.  MG 64 (No. 429) (1980) 202‑203.  Discusses some history, especially Barnard's problem.

Nick Lord, proposer;  uncertain solver.  Problem 71.E.  MG 71 (No. 457) (Oct 1987) 236  &  72 (No. 459) (Mar 1988) 54‑55.  Overhang can diverge even if the lengths converge to zero.

Jeremy Humphries, proposer;  various solvers.  Prob. 129.5 _ Planks.  M500 129 (Oct 1992) 18  &  131 (Feb 1993) 18.  Uniform planks of lengths  2, 3, 4.  Find maximum overhang.  Best answer is  3 5/18.

Unnamed solver, probably Jeremy Humphries.  Prob. 145.2 _ Overhanging dominoes.  M500 145 (?? 1995) ??  & 146 (Sep 1995) 17.  He uses approximately real dominoes of size  .7 x 2.2 x 4.4 cm.  Given three dominoes, how far out can you reach?  And how far away from the edge can one domino be?  Answer is basically independent of the shape.  Place one domino on its large face with its centre of gravity at the cliff edge and its diagonal going straightout.  For maximum reach, similarly place another domino with its centre of gravity at the far corner of the first domino.  To get a domino maximally far from the edge, place one on edge so its largest face is parallel to the cliff edge and with its centre of gravity at ther far corner of the first domino.  In both cases, counterbalance by putting the other domino with its centre of gravity at the inner corner of the first domino.  So the answer to the first problem is the longest face diagonal and the answer to the second problem is half the longest face diagonal minus half the thickness.

 

          7.X.    HOW OLD IS ANN? AND OTHER AGE PROBLEMS

 

          The simplest age problems are 'aha' problems, like 'Diophantos's age', going back at least to Metrodorus and will not be considered here _ see Tropfke 575 for these.  More complicated, but still relatively straightforward age problems appear in various 16-19C arithmetic and algebra works, e.g.  Schooten;  Record;  Baker;  Cocker;  Pike;  American Tutor's Assistant;  Ainsworth & Yeats, 1854, op. cit. in 7.H.4;  Colenso.  I have included only a sample of these to show the background and a few earlier versions.  Simple problems of Form III then appear from the mid 18C and later in standard arithmetic books, and later in puzzle books like Hoffmann;  Pearson;  Loyd;  Dudeney  and  Loyd Jr.  These usually lead to two equations in two unknowns, a bit like 7.R, or one equation in one unknown, depending on how one sets up the algebra.  In the 19C, this problem was popular in discussions of algebra as the problem can have a negative solution, which means that the second time is before the first rather than after, and so the problem was used in discussions of the existence and meaning of negative numbers _ see:  Hutton, 1798?;  Kelland;  De Morgan (1831, 1836, 1840).  About 1900, we get the  "How old is Ann (or Mary)"  versions, forms I and II below.

          Form I:  "The combined ages of Mary and Ann are  44  years, and Mary is twice as old as Ann was when Mary was half as old as Ann will be when Ann is three times as old as Mary was when Mary was three times as old as Ann.  How old is Ann?"  Answer is  16½;  Mary is  27½.  See:  Pearson;  Kinnaird;  Loyd;  Bain;  White;  Dudeney;  Loyd Jr;  Grant;  Ransom;  Doubleday-II.

          Form II:  "Mary is  24.  She is twice as old as Ann was when Mary was as old as Ann is now.  How old is Ann?"  Answer is  18.  See:  Menninger;  Ransom;  Doubleday-II.

          Form III‑(a, b, c).  X  is now  a  times as old as  Y;  after  b  years,  X  is  c  times as old as  Y.  I.e.   X = aY,   X + b  =  c (Y + b).   This can be rephrased depending on the time of narration _ e.g.  X  is now  c  times as old as  Y,  but  b  years ago,  X  was only  a  times as old as  Y.  See Clark for an equivalent problem with candles burning.

          Form IV-(a, b, c).  X  is now  a;  Y  is now  b;  when will  X  be  c  times as old as  Y?  I.e.  a + x  =  c (b + x).  See:  Hutton, 1798?;  Manuel des Sorciers;  De Morgan (1831, 1836, 1840);  Kelland;  Colenso;  Young;  Hoffmann;  Dudeny;  Perelman; 

          General solution of Form III occurs in:  Milne;  Singmaster.

          General solution of Form IV occurs in:  De Morgan (1831).

 

Examples of type III.

 

       10/7            2        4/3              Boy's Own Mag.

            2          12        8/5              Milne

            2          30        7/5              Dilworth

            2          60        5/4              Dilworth

        7/3            6       13/9              Milne

            3       10½            2              Walkingame, 1751

            3          15            2              Mair;  Amer. Tutor's Asst.;  Eadon;  Bonnycastle, 1815;  Child;  Walkingame, 1835;  Charades, etc.;  Pearson; 

            3          18            2              Dudeney

            4           -2            6              Unger

            4            4            3              Tate

            4            5            3              Young

            4          10        5/2              Tate

            4          14            2              Simpson

            5            5            3              Dudeney

            5            5            4              Hutton, c1780?;  Hutton, 1798?;  Treatise; 

            6            3            4              Unger;  Milne

            7            3            4              Berkeley & Rowland

            7          15            2              Unger

            9            3            3              Adams

          14          10          5_              Unger

 

Robert Record.  The Whetstone of Witte.  John Kyngstone, London, 1557.  Facsimile by Da Capo Press, NY  &  Theatrum Orbis Terrarum, Amsterdam, 1969.  A question of ages, ff. Gg.i.v - Gg.ii.r.  (The gathering number at the bottom of folio ii is misprinted G.)  Father and two sons.   B = A + 2.   C = A + B + 4.   A + B + C = 96.

Baker.  Well Spring of Sciences.  1562?  Prob. 1,  1580?: ff. 189r‑190r;  1646: pp. 297-299;  1670: pp. 340-341.  A  is  120.  B  says if he were twice his present age, he would be as much older than  A  as  A  is older than  B  is now.  C  says the same with three times his age,  D  with four and  E  with five.

Frans van Schooten Sr.  MS algebra text, Groningen Univ. Library, Hs. 443, c1624, f. 54r.  ??NYS _ reproduced, translated and described by Jan van Maanen; The 'double_meaning' method for dating mathematical texts;  IN: Vestigia Mathematica; ed. by M. Folkerts & J. P. Hogendijk; Rodopi, Amsterdam, 1993, pp. 253_263.  "A man, wife and child are together  96  years, that is to say the man and child together  2  years more than the wife and the wife with the child together  14  years more than the man.  I ask for the years of each."  Van Maanen feels that this problem probably describes the van Schooten family and it is consistent with the family situation for a period in 1623_1624, when the man was  41,  the wife  47  and the child  8,  thereby giving a reasonable date for the undated MS.  Van Schooten's two sons also used the problem in their works, but reversed the role of man and wife because it was not common for the wife to be so much older.

Edward Cocker.  Arithmetic.  Op. cit. in 7.R.  1678.  Several problems, of which the following are the most interesting.

Chap. 31, quest. 2.  1678: p. 325 (misprinted 305 in 1678);  1715: p. 210;  1787: p. 182.  B  says he is  3/2  A.  C  says he is twice  B.  A  says the sum of their ages is  165.

Chap. 32, quest. 2.  1678: pp. 332-333;  1715: p. 214;  1787: p. 186.  A  says he is  18.  B  says he is  A + C/2.  C  says he is  A + B.

Ladies' Diary.  1708.  Question V.  ??NYS _ reproduced in The Diarian Repository, 1774, Wallis 341.5 DIA, ??NX.  III‑(3, 15, 2).

                    When first the marriage knot was tied

                              Betwixt my wife and me,

                    My age to her's we found agreed

                              As nine doth unto three;

                    But after ten and half ten years,

                              We man and wife had been,

                    Her age came up as near to mine,

                              As eight is to sixteen.

                    Now tell me if you can, I pray,

                              What was our age o' th' wedding day?

Dilworth.  Schoolmaster's Assistant.  1743. 

P. 92, no. 4.  A  is  20;  B  is  A + C/2;  C = A + B.

P. 167, no. 104.  III-(2, 30, 7/5).  After another  30  years, we have  III-(2, 60, 5/4),  when they died.  "I demand ... also the reason why the lady's age, which was continually gaining upon her husband's, should, notwithstanding, be never able to overtake it."

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790), prob. XVI, p. 87 (1790: prob. XXII, pp. 85-86).  Essentially  III‑(4, 14, 2),  though the present time is halfway between the times involved in the problem _ i.e.  7  years ago, the father was  4  times the son's age;  in  7  years, he will be  2  times.

Les Amusemens.  1749.  Prob. 97, p. 239.  Father is three times the age of his son, when will he be only twice as old? _ i.e.  III-(3, b, 2)  _ with answer  b = Y,  the son's present age.

Walkingame.  Tutor's Assistant.  1751. 

1777, p. 82, prob. 5;  1835: p. 91, prob. 4;  1860: p. 11, prob. 5.  H  is  30.  K = H + L/4.  L = H + K.

1777, p. 177, prob. 120;  1860: p. 185, prob. 113.  (The 1835 edition differs _ see under 1835 below.)  III-(3, 10½, 2).

                    When first the marriage knot was ty'd,

                              Between my wife and me,

                    Her age did mine as far exceed,

                              As three times three does three;

                    But when seven years, and half seven years,

                              We man and wife had been,

                    My age came then as near to her's,

                              As eight is to sixteen.

          Quest.  What was each of our ages when we married?

(I find it extraordinary that the man is younger in this version and in no other that I have seen.  The 1860 makes a few minor changes and puts it on four lines.  Wehman, New Book of 200 Puzzles, 1908, p. 51 is an incomplete copy of this, comprising the first two lines of a four-line version, with the answer.)

Vyse.  Tutor's Guide.  1771? 

Prob. 3, pp. 135-136 & Key p. 177.  A = C + 4,  B = A + C + 9,  D = 45 = A + B + C.

Prob. 5, p. 137 & Key pp. 180-181.  B = 5A = 7(A-4).

Prob. 8, p. 136 & Key pp. 181-182.

          When first the Marriage-Knot was ty'd

                    Betwixt my Wife and me,

          My Age did her's as far exceed

                    As three Times three does three;

          But when ten Years, and Half ten Years,

                    We Man and Wife had been,

          Her Age came up as near to mine

                    As eight is to sixteen.

          Now, tell me, I pray,

                    What were our Ages on the Wedding Day?

John Mair.  Arithmetic, Rational and Practical....  Vol. III. Practical Arithmetic.  2nd ed., A. Kincaid & W. Creech, and J. Bell, Edinburgh, 1772.  P. 143, no. 11.  ??NX.  III‑(3, 15, 2).

                    When first the marriage-knot was tied

                              Betwixt my wife and me,

                    My age did hers as far exceed,

                              As three times three doth three;

                    But after ten and half ten years,

                              We man and wife had been,

                    Her age came up as near to mine,

                              As eight is to sixteen.

                    Now, Tyro, skill'd in numbers, say,

                              What were our ages on the wedding-day?

                                        Answer.

                    Sir, Forty-five years you had been,

                              Your Bride no more than just fifteen.

Dodson.  Math. Repository.  1775.

P. 6, Quest. XIV.  A + B + 25  =  2A;  A - B - 8  =  B.

P. 15, Quest. XL.  III-(4, 14, 2)  stated at the middle of the 14-year period.

Charles Hutton.  A Complete Treatise on Practical Arithmetic and Book-keeping.  Op. cit. in 7.G.2.  [c1780?]  1804: prob. 2, p. 132.  III-(5, 5, 4).

Bonnycastle.  Algebra.  1782.  Prob. 11, p. 84 (1815: prob. 10, p. 105).  A = 2B;  B = 3C;  A + B + C = 140.

Pike.  Arithmetic.  1788.  P. 335, no. 7.  B  =  3A/2;   C  =  21 (A + B)/10;   A + B + C  =  93.

Anon.  The American Tutor's Assistant.  (1st ed. is unknown;  2nd ed., Philadelphia, 1791);  1810 ed., Joseph Crukshank, Philadelphia.  ??NYS _ quoted in:  Lucas N. H. Bunt et al.; The Historical Roots of Elementary Mathematics; Prentice-Hall, 1976; p. 33.  III‑(3, 15, 2).

                    When first the marriage knot was ty'd

                              Between my wife and me,

                    My age was to that of my bride

                              As three times three to three

                    But now when ten and half ten years,

                              We man and wife have been,

                    Her age to mine exactly bears,

                              As eight is to sixteen;

                    Now tell, I pray, from what I've said,

                              What were our ages when we wed?

          Answer:

                    Thy age when marry'd must have been

                              Just forty-five; thy wife's fifteen.

Eadon.  Repository.  1794.  P. 297, no. 16.  III-(3, 15, 2) in verse.

                    When first the marriage knot was ty'd

                              Betwixt my wife and me,

                    My age did her's as far exceed,

                              As three times three doth three;

                    But after ten, and half ten years,

                              We man and wife had been,

                    Her age came up as near to mine,

                              As eight is to sixteen.

                    Now tell me (you who can) I pray,

                              What were our ages on the wedding day?

Hutton.  A Course of Mathematics.  1798?  Prob. 1,  1833 & 1857: 80.  III-(5, 5, 4).  On  1833: 224-231;  1857: 228-235,  he has an extensive discussion  Remarks upon Equations of the First Degree  concerning possible negative roots and considers  IV‑(42, 12, 4)  whose solution is  -2.

Bonnycastle.  Algebra.  10th ed., 1815.  P. 104, no. 7.  III-(3, 15, 2).

Manuel des Sorciers.  1825.  P. 81, art. 41.  ??NX  IV-(54, 18, 2).

Augustus De Morgan.  Arithmetic and Algebra.  (1831 ?).  Reprinted as the second, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  Art. 113, p. 29.  IV-(40, 8, 3)  _ "in how many years hence will the father's age be just three times the son's?"  Then gives general solution of  IV-(a, b, 3).  Observes that  a = 40,  b = 18  leads to  -7  and that  a = 3b  leads to  0  and discusses the meaning of both situations.  Then solves the general case  IV-(a, b, n).

Child.  Girl's Own Book.  Arithmetical Puzzles.  1832: No. 8, pp. 171 & 179;  1833: No. 8, pp. 185 & 193 (answer numbered 6);  1839: No. 8, pp. 165 & 173;  1842: No. 8, pp. 283 & 291;  1876: No. 6, pp. 231 & 244.  III‑(3, 15, 2)  given in verse which is the same in the 1832, 1839 and 1876 eds.  (Except 1839 & 1842 have  her's  in line three.)

 

                    When first the marriage knot was tied

                    Between my wife and me,

                    My age exceeded hers as much,

                    As three times three does three.

 

                    But when ten years and half ten years

                    We man and wife had been,

                    Her age approached as near to mine

                    As eight is to sixteen.

 

          The 1833 ed has the first lines of the second verse garbled as:  But when the man and wife had been, / For ten and half ten years.

Louis Pierre Marie Bourdon (1779-1854).  Élémens d'Algèbre.  7th ed., Bachelier, Paris, 1834.  (The 6th ed., was 1831, but I haven't yet looked for the early editions _ ??)  Art. 58, pp. 87-89.  Discusses the problem  IV-(a, b, 4)  and the significance of negative solutions, using  IV-(45, 15, 4)  as an example.

Walkingame.  Tutor's Assistant.  1835.  P. 180, prob. 59.  III-(3, 15, 2).

                    When first the marriage knot was ty'd

                              Between my wife and me,

                    My age did hers as far exceed,

                              As three times three does three;

                    But when ten years, and half ten years,

                              We man and wife had been,

                    Her age came then as near to mine,

                              As eight is to sixteen.

          = Depew; Cokesbury Game Book; 1939; Marriage problem, pp. 207-208.

Augustus De Morgan.  On The Study and Difficulties of Mathematics.  First, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  P. 53.  IV-(56, 29, 2)  _ "when will the father be twice as old as the son?"  Answer is  -2  and he discusses the meaning of this.

Unger.  Arithmetische Unterhaltungen.  1838. 

Pp. 108 & 255, no. 392.  III-(6, 3, 4).

Pp. 109 & 255, no. 402.  III-(7, 15, 2).

Pp. 110 & 255, no. 405.  III-(4, -2, 6),  phrased in terms of wealth and both give away  2.

Pp. 133 & 258, no. 508.  III-(14, 10, 5_).

Philip Kelland.  The Elements of Algebra.  A. & C. Black, Edinburgh, et al., 1839.  ??NX.

P. 15:  IV-(40, 20, 3).

P. 105:  "A's  money or debt is  a  times  B's;  if  A  lose  £10  to  B,  it will be  b  times  B's."  This is perhaps closer to the problems in 7.R.

Augustus De Morgan.  Negative and impossible numbers.  IN:  Penny Cyclopædia, vol. XVI, 1840, pp. 130-137.  ??NX  IV-(50, 40, 2)  _ "at what date is (was, or will be, as the case may be) the first twice as old as the second?" 

T. Tate.  Algebra Made Easy.  Op. cit. in 6.BF.3.  1848.

P. 37, no. 20.  III-(4, 4, 3).

P. 45, no. 11.  Couple are married for  _  of his life and  ¼  of hers.  Man is  8  years older than the wife, who survived him by  20  years.  How old were they at marriage?

P. 45, no. 12.  Man is  32  years older than  and  5  times as old as  his son.

P. 67, no. 6.  III-(4, 10, 5/2).

Phineas Taylor Barnum, c1848, is reputed to have given a friend a problem involving ages:  a man aged  30  with a child aged  1  is  30  times as old as his child, but in  30  years he is only twice as old, and in another  30  years he is only one-third older, ..., when does the child catch up with the father?  This is given in a cartoon biography of P. T. Barnum by Walt Kelly in the early 1930s _ reproduced in:  Outrageously Pogo; ed. by Mrs. Walt Kelly & Bill Crouch Jr; Simon & Schuster, NY, 1985, pp. 14-21.

Walter Taylor.  The Indian Juvenile Arithmetic ....  Op. cit. in 5.B.  1849.  P. 210, No. 4.  III‑(3, 15, 2).  No solution.

                    When first the marriage knot was tied, betwixt my wife and me;

                    My age did hers as far exceed, as three times three does three.

                    But when ten years and half ten years, we man and wife had been,

                    Her age came then as near to mine, as eight does to sixteen.

John H. Boardman.  Arithmetic: Rules and Reasons.  Macmillan, Cambridge, 1850.  P. 98.  "A.  is now twice as old as  B.;  eight years ago he was three times as old, and one year more;  find the age of each."

Anonymous.  A Treatise on Arithmetic in Theory and Practice: for the use of The Irish National Schools.  3rd ed., 1850.  Op. cit. in 7.H.

P. 353, no. 14.  III-(5, 5, 4).

P. 355, no. 1.  "Your age is now  1/5  of mine;  but  4  years ago it was only  1/7  of what mine is now".

John Radford Young (1799-1885).  Simple Arithmetic, Algebra, and the Elements of Euclid.  IN: The Mathematical Sciences, Orr's Circle of the Sciences series, Orr & Co., London, 1854.  [An apparently earlier version is described in 7.H.]

No. 3, p. 176.  III-(4, 5, 3)  phrased as five years ago.

No. 10, p. 177.  IV-(40, 9, 2).

The Family Friend (1856) 298 & 329.  A  is now one-fifth the age of  B,  but five years ago,  A  was one-seventh of  B's  present age.

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Arithmetical puzzles, no. 3, p. 173 (1868: 184).  "The square of my age  60  years ago is double my present age."

Magician's Own Book.  1857.  December and May, p. 246.  X + Y = 100,  Y = 4X/9.  = Boy's Own Conjuring Book, 1860, p. 216.  = Illustrated Boy's Own Treasury, 1860, prob. 19, pp. 428 & 432, but this has no title. 

Charades, Enigmas, and Riddles.  1859?: prob. 17, pp. 58 & 62;  1862?: prob. 560, pp. 105 & 152.  III‑(3, 15, 2).  (1865 differs very slightly in typography.)  This is essentially identical to Child.

                    When first the marriage knot was tied

                    Between my wife and me,

                    My age exceeded her's as much,

                    As three times three does Three:

                    But when Ten years and half ten years

                    We man and wife had been,

                    Her age approached as near to mine

                    As Eight is to sixteen.

Edward Brooks.  The Normal Mental Arithmetic  A Thorough and Complete Course by Analysis and Induction.  Sower, Potts & Co., Philadelphia, (1858), revised, 1863.  Further revised as:  The New Normal Mental Arithmetic: A Thorough and Complete Course by Analysis and Induction;  Sower, Potts & Co., Philadelphia, 1873.  Lots of examples.  I mention only those of some novelty to illustrate mid/late 19C texts.

1863 _ p. 129, no. 10;  1873 _ p. 158, no. 9.  A = 4B;  A + B = 25;  when will  A = 3B?

1863 _ p. 144, no. 13;  1873 _ p. 174, no. 4.  "Said  E  to  F,  my age is  5  years more than yours, but  4  years ago my age was  ½  of what yours will be  4  years hence;  what was the age of each?"

1873 _ p. 175, no. 14.  "Two years ago Mr. Smith was  5  times as old as his son John will be  2  years hence, and  3  years hence his age will equal  15  times John's age  3  years ago: required the age of each."

(Beeton's) Boy's Own Magazine 3:4 (Apr 1889) 175  &  3:6 (Jun 1889) 255.  (This is undoubtedly reprinted from Boy's Own Magazine 1 (1863).)  Mathematical question 33.  III‑(10/7, 2, 4/3)

Colenso.  Op. cit. in 7.H.  These are from the new material of 1864?

No. 23, pp. 190 & 215.  IV-(62, 30, 5).

No. 24, pp. 190 & 215.  I am  24  years older than my son.  When I am twice my present age, he will be  8  times his present age.

William J. Milne.  The Inductive Algebra ....  1881.  Op. cit. in 7.E.

No. 54, pp. 108 & 328.  III-(7/3, 6, 13/9).

No. 3, pp. 166 & 334.  Asks for general solution of  III-(a, b, c)  and for  III-(6, 3, 4).

No. 90, pp. 301 & 347.  III-(2, 12, 8/5).

Berkeley & Rowland.  Card Tricks and Puzzles.  1892.  Pp. 117-118.  III‑(7, 3, 4).

Hoffmann.  1893.  Chap. IV, nos. 20 and 29, pp. 149 & 194 and 151 & 196.  Simple age problems:  IV-(71, 34, 3),  IV-(71, 34, 2),  IV-(45, 12, 3)

Somerville Gibney.  So simple!  The Boy's Own Paper 20 (No. 992) (15 Jan 1898) 252  &  (No. 993) (22 Jan 1898) 267.  "When you are as old as I am, I shall be twice as old as you were when I was as old as you are."  This only gives the ages being in the ratio of  5  to  4.

P. Holland Lester.  Some "B.O.P." puzzles.  The Boy's Own Paper 20 (No. 1017) (9 Jul 1898) 655.  "I am twice as old as you were when I was what you are;  when you are what I am our united ages will be  63."  "You are twice as old as I was when you were what I am;  when you are twice what I shall be when you are twice what I am, our united ages will be  133." 

Clark.  Mental Nuts.

1904: no. 28: The candle question.  1916: no. 97: The candles.  "Suppose two candles, one of which will burn 4 hours and the other 5 hours, are lighted at once.  How soon will one be three times the length of the other?"  (1916 has a shortened version.)  I assumed that one candle starts out as  5/4  the length of the other candle.  Then this is like our age problem  III-(5/4, b, 3),  where the problem is to determine  b  in terms of  X  or  Y.  Or we might say that we have  X = 4,  Y = 5,  3(X-b) = (Y-b).  In any case, I get  b = 7/2 = 3 1/2.  However, Clark's answer is  3 7/11  and this arises by considering both candles to be the same length initially, but made of different materials, so they burn at different rates.  Taking the candles to be of unit length, the lengths after  b  hours are  (1 ‑ b/4)  and  (1 ‑ b/5).  Setting three time the first equal to the second gives  b = 40/11 = 3 7/11.  However, this formulation cannot be interpreted as an age problem as the candles age at different rates!

1904: no. 52;  1916: no. 64.  Man and wife.  "A man is twice as old as his wife was when he was as old as she is now.  When she reaches his present age, the sum of their ages will be 100 years.  What are their ages?"

Pearson.  1907.

   Part I, no. 39, pp. 123‑124 & 186‑187.  III-(3, 15, 2)  in verse with verse solution, very similar to previous examples.

   Part II.

No. 13: A brain twister, pp. 116 & 193.  Form I.

No. 57, pp. 124 & 202.  4  persons.

No. 70: How old was John?, pp. 128 & 205.  One person like Diophantos' age.

No. 105: A delicate question, pp. 136 & 212.  One person, in verse, using a square root.

No. 117: When was he born?, pp. 137 & 214.  One person.

No. 124: Ask any motorist, pp. 138‑139 & 215.  Car and tyres.

No. 156, pp. 144 & 222.  Two people.

No. 168: Very personal, pp. 146 & 224.  Two people, verse problem and solution.

Wehman.  New Book of 200 Puzzles.  1908.  The marriage knot, p. 51.  Gives only half of the problem!  Seems to be copied from a four-line version of Walkingame.

Loyd.  How Old was Mary?  Cyclopedia, 1914, pp. 53 & 346.  (= MPSL2, prob. 10, pp. 8 & 123.)  Form I _ he says this is a companion to his 'famous problem "How old was Ann"'.  He gives other age problems. 

                    Tell mother's age, pp. 84 & 349‑350.  (= MPSL1, prob. 85, pp. 82 & 151.) 

                    Pp. 216 & 367.  (= MPSL2, prob. 106: How old is Jimmy, pp. 75 & 155.)

G. G. Bain.  An Interview with Sam Loyd, 1907, op. cit. in 1, p. 777.  Refers to "How old was Mary?", and gives form I as in the Cyclopedia with slightly different wording but the same illustration.

A. C. White.  Sam Loyd and His Chess Problems.  1913.  Op. cit. in 1.  P. 101 calls it "How old was Mary?" and gives form I as in the Cyclopedia.

Dudeney.  AM.  1917.  Numerous problems, including the following.

Prob. 43: Mrs. Timpkins age, pp. 7 & 152.  III-(3, 18, 2).

Prob. 45: Mother and daughter, pp. 7 & 152.  IV-(45, 12, 3).

Prob. 47: Rover's age, pp. 7 & 152-153.  III-(5, 5, 3)  concealed by saying the sister was "four times older than the dog", meaning five times as old.

Prob. 51: How old was Mary?, pp. 8 & 153.  Form I, attributed to Loyd.

Adams.  Puzzles That Everyone Can Do.  1931.

Prob. 170, pp. 66 & 153.  A is older than B by as many years as B is older than C.  C is half as old as B was when B was half as old as A is now.  In a year's time, the combined ages of C and B will equal that of A.

Prob. 246, pp. 93 & 166: Molly & Polly.  M will be three times P in three year's time.  M is three times as old as P will be when P is three times as old as she is now.  (Since the second statement simply says  M = 9P,  this problem is actually  III‑(9, 3, 3).)

 

James Joyce.  Ulysses.  (Dijon, 1922);  Modern Library (Random House), NY, 1934, apparently printed 1946.  P. 663 (Gardner says the 1961 ed. has p. 679).  Humorous calculation assuming the ratio of ages could remain fixed at  17  to  1.  The numbers become wrong after a bit.

Ackermann.  1925. 

Pp. 93‑94: different problem, due to C. V. Boys.

Pp. 98‑100: form I, ending  "What are their present ages?"

Pp. 100‑102: complex version due to A. Honeysett, with four adults and an unspecified number of children, which turns out to be two, one of which has age zero.

Collins.  Book of Puzzles.  1927.  How old is Jane?, pp. 73-74.  Form I with Jane and Ann. 

Loyd Jr.  SLAHP.  1928.  How old is Ann?, pp. 4 & 87.  Gives the "original wording" as form I.  He gives numerous other age problems.

Perelman.  1937.  MCBF.  The equation does the thinking, p. 244.  IV-(32, 5, 10).

Evelyn August.  The Black-Out Book.  Op. cit. in 5.X.1.  1939. 

Number, please!, pp. 20 & 210.  One person, giving   x + 3  =  3 (x - 3).

Pp. 27 & 210.  A variation of type III, giving   X + 2  =  3Y;   X + 8   =   2 (Y + 8).

W. T. Williams & G. H. Savage.  The Penguin Problems Book.  Penguin, 1940.  No. 88: The monkey's mother, pp. 50 & 132-133.  Monkey on a rope over a pulley with a weight on the other side, but the weight of the rope is equal to the age of the monkey's mother, who was twice as old as the monkey was ....  What was the length of the rope?

Clark Kinnaird.  Loc. cit. in 1.  1946.  Pp. 265‑266, says Loyd Jr. is best known for this problem (form I).  On p. 267, he says it  "made Sam Loyd [Jr.] famous, although he did not originate it .... Loyd based his version upon a similar poser which went around like a chain‑letter fad in the early years of this century ...."  He also says  'References to "How Old is Ann" have been found as far back as 1789 ....', but he doesn't give any such references!

Owen Grant.  Popular Party Games.  Universal, London, nd [1940s?].  Prob. 17, pp. 39-40 & 52.  Form I with Mary and Ann replaced by Smith and Robinson.

Meyer.  Big Fun Book.  1940.  No. 10, pp. 167 * 753.  "The sum of our ages is  22.  I shall be  7  times as old as you are now when I become twice your age." 

Karl Menninger.  (Mathematik in deiner Welt.  Vandenhoek & Ruprecht, Göttingen, 1954;  revised, 1958.)  Translated as:  Mathematics in Your World; G. Bell, London, 1961.  The complicated problem of Anne and Mary, pp. 65‑66.  Gives form II with no history.

William R. Ransom.  Op. cit. in 6.M.  1955.  How old is Ann?, p. 91;  Mrs. M. & Miss A., p. 92.  He first gives form II and says  "This problem raged throughout the United States in the early 1900's.  It was concocted by Robert D. Towne, who died at the age of 86 in 1952."  He then gives form I and says  "This is a much older problem, of the same type as our "How old is Ann?" which has circulated mostly in England."

The Little Puzzle Book.  Op. cit. in 5.D.5.  1955.  P. 62: A problem of age.  "When I am as old as my father is now I shall be five times the age my son is now.  By then my son will be eight years older than I am now.  The combined ages of my father and myself total 100 years.  How old is my son?"

Doubleday - II.  1971.

Passing years, pp. 15-16.  Form I.

It's not so simple!, pp. 47-48.  Form II.

Angela Fox Dunn.  Mathematical Bafflers.  Dover, 1980.  Selected from Litton's Problematical Recreations, which appeared from 1971, ©1964.  An age problem, pp. 18 & 42.  "Lottie and Lucy Hill are both 90 years old.  Mary Jones, on the other hand, is half again as old as she was when she was half again as old as she was when she lacked 5 years of being half as old as she is now."  Te solution is 90, i.e. Mary is as old as the Hills!

David Singmaster.  Some diophantine recreations.  Op. cit. in 7.P.5.  1993.  Determines when integer data in  III-(a, b, c)  gives integer solutions.

 

          7.Y.    COMBINING AMOUNTS AND PRICES INCOHERENTLY

 

          Often called the applesellers' problem or the market women's problem.

          Two persons have the same number of items which they combine.  They average the number of items per unit cost rather than the cost per item.

          See Tropfke 652.

 

Alcuin.  9C.  Prob. 6: Propositio de duobus negotiatoribus  C  solidos communis habentibus.  Buy  125 & 125  at  5 for 2,  costing  100,  sell at  2 for 1  and  3 for 1, leaving  5 & 5  left over, so they received  100  and the left overs are worth  4 1/6.

Ibn Ezra, Abraham (= ibn Ezra, Abraham ben Meir  = ibn Ezra, Abu Ishaq Ibrahim al‑Majid).  Sefer ha‑Mispar.  c1163.  Translated by Moritz Silberberg as:  Das Buch der Zahl  ein hebräisch-arithmetisches Werk; J. Kauffmann, Frankfurt a. M., 1895.  P. 42.  Buys  100 for 100,  sells  50 & 50  at  5 for 4  and  3 for 4,  makes  6_.  Silberberg's note 95, p. 107, says that the the same problem occurs in Elia Misrachi, c1500.

Fibonacci.  1202.  P. 281.  Buys  100 for 100  and sells  50 & 50  at  4/5  and  4/3.  What profit?

Provençale Arithmétique.  c1430.  Op. cit. in 7.E.  F. 113r, pp. 59-60.  Buy  60 for 24,  sell as  x & y  at  a & b  to make profit of  1.  How?  This is actually an indeterminate question, complicated by prices which are non-integral.  The author seems to give just one solution:  x = y = 30,  a = 1/2,  b = 1/3.

AR.  c1450.  Prob. 354, pp. 154, 183, 229‑230.  Buy  100  at  5 for 2,  sell  50 & 50  at  3 for 1  and  2 for 1,  making  1_.

Pacioli.  De Viribus.  c1500.  Prob. 66: De uno ch' compra  60  perle et revendele a ponto per quelli ch' gli stanno et guadagna.  Buy  60  at  5 for 2,  sell  30 & 30  at  2 for 1  and  3 for 1.

Tartaglia.  General Trattato, 1556, art. 160‑162, p. 259v.

160:  buy  60 & 60  at  5 for 1,  sell at  2 for ½  and  3 for ½,  make 1.

161:  buy  42 & 42  at  7 for 12,  sell at  3 for 6  and  4 for 6,  make 3.

163:  buy  270 & 270  at  9 for 12,  sell at  4 for 6  and  5 for 6,  make 9.

Buteo.  Logistica.  1559.  Prob. 17, pp. 215-217.  Buy  60 for 24,  i.e. at  5 for 2.  How were they sold 'at the same price' to make  1?  Sell at  2 for 1  and  3 for 1  and gain  1.  Says this was proposed by Stephanus and that the apples aren't sold at the same price as they were bought.  (H&S 53 gives Latin.)

Ozanam.  1725.  Prob. 51, art. 1, 1725: 258.  Buy  20  at  5 for 2;  sell  10 & 9  at  2 for 1  and  3 for 1  to recover cost and have  1  left over.

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790), prob. XVII, pp. 87-88 (1790: prob. XXV, p. 87).  Buy  x & x  at  2 for 1  and  3 for 1,  sell at  5 for 2,  lose 4 _ what was  x?

Vyse.  Tutor's Guide.  1771?  Prob. 28, p. 62 & Key p. 68.  Buy  120 & 120  at  2 for 1  and  3 for 1,  sell at  5 for 2,  lose 4  _ "Pray how comes that about?"

Dodson.  Math. Repository.  1775.  P. 24, Quest LXII.  Same as Simpson.

Bonnycastle.  Algebra.  1782.  P. 82, no. 9 (1815: pp. 101-102, no. 8).  Same as Simpson.

Pike.  Arithmetic.  1788.  P. 492, no. 9.  Same as Simpson.

John King, ed.  John King  1795  Arithmetical Book.  Published by the editor, who is the great-great-grandson of the 1795 writer, Twickenham, 1995.  P. 55.  Buy  120 & 120  at  3 for 1d  and  2 for 1d.  Do I gain or lose by selling at  5 for 2d?  He finds a loss of  4d.  On pp. 152-153, the Editor discusses Alcuin, where pigs are sold at three different prices, but he knows no example with four or more different prices.  On p. 163, he mentions almost the same problem, with prices multiplied by  12.  [I think he has misinterpreted Alcuin _ the left over pigs are sold at the same prices as the others.]

Robert Goodacre.  Arithmetic.  2nd ed., T. Ostell & C. Law, London, 1804.  Miscellaneous Questions, no. 128, p. 205.  Buy  30 & 30  at  3 for 1  and  2 for 1,  sell at  5 for 2.  Does he gain or lose?

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions.  No. 21, pp. 19 & 76‑77.  Buy  120 & 120  at  2 for 1  and  3 for 1,  sell at  5 for 2  and discover a loss of  4.  c= Magician's Own Book (UK version), 1871, The costermonger's puzzle, pp. 38-39.

Boy's Own Book.  Profit and loss.  1828: 414;  1828-2: 418-419;  1829 (US): 212;  1855: 566;  1868: 669.  Buy  96 & 96  at  3 for 1  and  2 for 1,  sell at  5 for 2  giving  2  left over and a loss of  4.

Augustus De Morgan.  Examples of the Processes of Arithmetic and Algebra.  Third, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  Prob. 11, p. 28.  Buy  150  at  3 for 1  and  100  at  2 for 1,  sell at  5 for 2  leads to no loss.  But buying  150 & 150  at  3 for 1  and  2 for 1,  selling at  5 for 2  leads to a loss.  "What was the reason of this?"

Hutton-Rutherford.  A Course of Mathematics.  1841?  Prob. 13,  1857: 81.  Buy  180 & 180  oranges at  3 for 1d  and  2 for 1d.  Does he gain or lose by selling at  5 for 2d?  He finds a loss of  6d. 

Boy's Treasury.  1844.  Amusements in arithmetic: Profit and loss, p. 306.  Same as Boy's Own Book, but says the loss is a fraction above  3½.

Magician's Own Book.  1857.  The astonished farmer, p. 244.  Compare selling  30 & 30  at  3 for 1  and  2 for 1  versus selling  60  at  5 for 2,  which makes  1  less.  = Boy's Own Conjuring Book, 1860, p. 214.  = Illustrated Boy's Own Treasury, 1860, prob. 16, pp. 428 & 432. 

The Sociable.  1858.  Prob. 35: The market woman's puzzle, pp. 297 & 315.  Same as Jackson.  = Book of 500 Puzzles, 1859, prob. 35, pp. 15 & 33.  = Wehman, New Book of 200 Puzzles, 1908, p. 29.

Book of 500 Puzzles.  1859. 

Prob. 35: The market woman's puzzle, pp. 15 & 33.  Same as Jackson and The Sociable.

The astonished farmer, p. 58.  Identical to Magician's Own Book.

[Chambers].  Arithmetic.  Op. cit. in 7.H.  1866?  P. 250, quest. 6.  Buy  x & x  at  2 for 1  and  3 for 1.  Sell at  5 for 2  and lose  8d.  How many were bought?

William J. Milne.  The Inductive Algebra ....  1881.  Op. cit. in 7.E.  Buy  x  apples at  5 for 2;  sell half at  2 for 1,  the other half at  3 for 1;  make  1.

Cassell's.  1881.  Pp. 100‑101: The costermonger's puzzle.  Buy  120 & 120  at  2 for 1  and  3 for 1,  sell at  5 for 2  and lose  4.

Hoffmann.  1893.  Chap. IV, no. 6: A little miscalculation, pp. 146 & 182.  Buy  120 & 120  at  4 for 1  and  6 for 1,  then sell at  10 for 2.

Loyd.  The lost cent.  Cyclopedia, 1914, pp. 39 & 344 (plus mention on p. 153).  = MPSL1, prob. 60 _ The missing pennies, pp. 58 & 142  = SLAHP: The apple mystery, pp. 44 & 100.  Calls it also the Covent Garden Problem.  Two applesellers with  x & x  being sold at  2 for 1  and  3 for 1.  They combine and sell at  5 for 2,  losing  7.  Further the proceeds are divided equally _ how much did the  2 for 1  lady lose?

Ahrens.  A&N, 1918, pp. 85‑87, discusses this problem, gives Alcuin's problem and the following.  Sell  30 & 30  at  1 for 5  (= 2 for 10)  and  3 for 10,  combine and sell  60  at  5 for 20,  losing  10  thereby.

Dudeney.  PCP.  1932.  Prob. 17: The missing penny, pp. 18 & 129.  = 536, prob. 23, pp. 8 & 229.

McKay.  Party Night.  1940.  No. 26, p. 182.  "A man bought equal quantities of apples at  2  a penny and at  3  a penny.  He sold them all at  5  for twopence.  Did he gain or lose?"  Does with  30 & 30.

Sullivan.  Unusual.  1947.  Prob. 33: Another missing dollar.  Compare selling  30 & 30  at  3 for 1  and  2 for 1  versus selling  60  at  5 for 2,  which makes  1  less.

 

          7.Y.1. REVERSAL OF AVERAGES PARADOX

 

          Example.  Player  A  gets  1 for 1  and  1 for 2  while player  B  gets  8 for 9  and  1 for 3.  A  has averaged better than  B  in each part, but overall  A  has  2  for  3  while  B  has  9  for  12  and has averaged better overall.  I have now computed small examples and the examples using the smallest integer values are:  2 for 1  and  2 for 4  versus  4 for 3  and  1 for 3  and  3 for 1  and  3 for 4  versus  4 for 2  and  1 for 2.  Allowing numerators of  0  gives a simpler? example:  2 for 1  and  1 for 3  versus  3 for 2  and  0 for 1.  By going up to values of  15,  one can have  A  being twice as good as  B  in each part but  B  is twice as good overall!.  E.g.  2 for 1  and  2 for 15  versus  13 for 13  and  1 for 15.

          This is sometimes called Simpson's Paradox.

          New section.

 

Morris R. Cohen & Ernest Nagel.  An Introduction to Logic and Scientific Method.  Harcourt, 1934.  ??NYS _ cited by Newson below.  I have an abridged student's edition which doesn't seem to have the example described by Newson.

Rupert T. Gould.  The Stargazer Talks.  Geoffrey Bles, London, 1944.  A Few Puzzles _ write up of a BBC talk on 10 Jan 1939, pp. 106-113.  Cricket version.  A  takes  5  wickets for  30  runs,  B  takes  5 for 31.  Then  A  takes  3 for 12  and  B  takes  7 for 29.  But overall  B  is better as the totals are  8 for 42  and  12 for 60.  On p. 113, his Postscript gives another version due to a correspondent:  28 for 60  and  28 for 60  combined with  4 for 36  and  1 for 27  give the same total averages of  32 for 96  and  29 for 87.

E. H. Simpson.  ??  J. Royal Statistical Society B, 13:2 (1951) ??NYS _ cited by Newson below.

R. L. Bolt.  Class Room Note 19:  Cricket averages.  MG 42 (No. 340) (May 1958) 119‑120.  In cricket one says, e.g. that a bowler takes  28  wickets for  60  runs, but one considers the runs per wicket.  A  and  B  both take  28  wickets for  60  runs, then continue with  4 for 36  and  1 for 27.  A  seems to be better than  B,  but both have averaged  3  runs per wicket.  He gives a clear graphic explanation of such perplexities and constructs  10 for 26  and  5 for 44  versus  10 for 22  and  10 for 78  as an example where the second is better at each stage but worse overall.

Colin R. Blyth.  On Simpson's paradox and the sure-thing principle.  J. American Statistical Association 67 (Jun 1972) 364-381.  ??NYS _ cited by Gardner below.

Gardner.  SA (Apr 1976)  c= Time Travel, chap. 19.  Time Travel gives a number of more recent references up to 1985.

Birtwistle.  Calculator Puzzle Book.  1978.  Prob. 18: Cricket commentary, pp. 16 & 75-76.  Two bowlers have each taken  10  wickets for  70  runs, then get  1 for 15  and  2 for 26.  The latter has done better in this last match, but is worse overall.

Clifford Wagner.  Simpson's paradox in real life.  American Statistician 36:1 (Feb 1982) 46‑48.  ??NYS _ cited by Gardner in Time Travel and by Newson below.

Donald Watson.  Note 72.23:  Combination of ratios.  MG 72 (No. 460) (Jun 1988) 126‑127.  Graphical and other analyses.

Nick Lord.  Note 74.11:  From vectors to reversal paradoxes.  MG 74 (No. 467) (Mar 1990) 55‑58.  He says the paradox is called "Simpson's reversal paradox", but gives no reference.  [Another source cites E. H. Simpson without reference.]  He discusses various interpretations of the phenomenon.

Graham Newson.  Simpson's paradox revisited.  MG 75 (No. 473) (Oct 1991) 290-293.  Cites the Simpson paper and some other recent papers, but with a lamentable lack of details.  He gives an example from "old SMP Puzzle Corners" which deals with cricket averages in 1906-07.  He quotes three examples from Wagner, one of which is taken from Cohen & Nagel referring to incidence of tuberculosis in 1910.

 

          7.Y.2. UNFAIR DIVISION

 

New section.  This might be considered as having some resemblence to 7.E and 7.J.

 

Magician's Own Book.  1857.  The unfair division, p. 255.  Farmer is to give  2/5  of his yield to the landlord, but the farmer uses  45  bushels of the harvest before they can divide it.  He then proposes to give  18  bushels to the landlord and then divide up the rest.  Is this correct?  = Book of 500 Puzzles, 1859, p. 64.  = Boy's Own Conjuring Book, 1860, p. 225, which has some mathematical misprints.

 

          7.Z.    MISSING DOLLAR AND OTHER ERRONEOUS ACCOUNTING

 

          The withdrawal version is the confusion caused by adding the amounts remaining, which has no meaning. 

          The missing dollar version mixes payments and refunds.  E.g. three people pay  $10  each for a triple room.  The landlord decides they were overcharged and sends  $5  back with the bellhop.  Perplexed by dividing  $5  by 3, he appropriates  $2  and refunds each person  $1.  Now they have paid  $9  each, making  $27,  and the bellhop has  $2,  making  $29 in all.  But there was  $30  originally.  What happened to the other dollar?

 

Walkingame.  Tutor's Assistant.  1751.  1777: p. 177, prob. 118;  1835, p. 180, prob. 57;  1860: p. 185, prob. 116.  "If  48  taken from  120  leaves  72,  and  72  taken from  91  leaves  19,  and  7  taken from thence leaves  12,  what number is that, out of which, when you have taken  48, 72, 19, and 7,  leaves  12?"  Though this is not the same as the withdrawal problems below, the mixing of amounts subtracted and remainders makes me think that this kind of problem may have been the basis of the later kind.

Cecil B. Read.  Mathematical fallacies.  SSM 33 (1933) 575‑589.  Gives a version with  $50  in the bank being withdrawn.  Withdraw  $20  leaving  $30;  withdraw  $15  leaving  $15;  withdraw  $9  leaving  $6;  withdraw  $6  leaving  $0.  But  $30 + $15 + $6 = $51.

Evelyn August.  The Black-Out Book.  Op. cit. in 5.X.1.  1939.  What happened to the shilling?, pp. 82 & 213.  Three girls paying  5s  each to share a room, landlord refunds  5s,  lift boy appropriates  2s.

Meyer.  Big Fun Book.  1940.  Where did the dollar go?, pp. 111 & 735.  3  men paying  $30.  = Jerome S. Meyer; Fun-to-do; op. cit. in 5.C; 1948; prob. 71: Where did the dollar go?, pp. 55-56 & 195.

Harriet Ventress Heald.  Mathematical Puzzles.  Booklet 171,  Educational Research Bureau, Washington, 1941.  Prob. 9, pp. 6‑7.  $50  being withdrawn from a bank.

Northrop.  Riddles in Mathematics.  1944.  1944: 8-9;  1945: 8;  1961: 18.  3  men paying a bill of  $30  or  30s.

W. A. Bagley.  Paradox Pie.  Op. cit in 6.BN.  1944.  No. 12: The financier, p. 15.  Withdrawing  £100.

Leeming.  1946.  Chap. 3, prob. 5: What happened to the dollar?, pp. 19‑20 & 152.  3  men paying  $30.

Sullivan.  Unusual. 

1943.  Prob. 8: Balancing the checkbook.  Withdrawing  $50.

1947.  Prob. 32: Where is the dollar?  3  men paying  $30.

"Willane".  Willane's Wizardry.  Academy of Recorded Crafts, Arts and Sciences, Croydon, 1947.  Easy money, p. 47.  Withdrawing  £100.

G. A. Briggs.  Puzzle and Humour Book.  Published by the author, Ilkley, 1966.  Prob. 4/26 (b): The dishonest waiter, pp. 44 & 85.

 

          7.AA. NEGATIVE DIGITS

 

          This is the use of  ‑5, ‑4, ..., 0, 1, ..., 5  instead of  0, 1, ..., 9.  See Knuth in 7.AA.1 for a general discussion of positional number systems, including negative digits, e.g. balanced ternary.

 

J. Colson.  A short account of negativo‑affirmative arithmetick.  Philos. Trans. Roy. Soy. 34 (1726) 161‑173.  He describes the use of negative digits quite clearly.  All work is done in the decimal system.  In concluding, he mentions "the several other species, as duodecimal, sexagesimal, centesimal, etc."

John Leslie.  The Philosophy of Arithmetic.  Constable, Edinburgh & Longman, London, 1817.  Pp. 33-34, 54, 64-65, 117, 150.  ??NYS.

A. L. Cauchy.  Comptes Rendus Acad. Sci. 11 (1840) 789-798.  ??NYS.

Léon Lalanne.  Comptes Rendus Acad. Sci. 11 (1840) 903-905.  ??NYS _ described in Knuth, op. cit. in 7.AA.1.  Introduces balanced ternary.  Knuth cites some mid 20C discussions of the system as a possible system for computers.

J. Halcro Johnston.  The Reverse Notation, Introducing Negative Digits with  12  as the Base.  Blackie, London, 1937.  ??NYS.

J. Halcro Johnston.  Two‑way arithmetic.  MiS 1:6 (Sep 1972) 10‑12.

C. A. B. Smith.  Looking glass numbers.  JRM 7 (1974) 299‑305.

E. Hillman, A. Paul & C. A. B. Smith.  History of two‑way numbers  &  Bibliography of two‑way numbers.  Colson News 1:4 (Dec 1984) 45‑46  &  47.  The bibliography lists 14 items which are all that are known to the authors.  Additional references in 2:1 (Mar 1985) 1.  [Smith produced about 12 issues of this newsletter devoted to unusual numbe systems.]

 

          7.AA.1.        NEGATIVE BASES, ETC.

 

          For binary, see 7.M.

          The Duodecimal Bulletin has regular articles discussing various bases.

 

Georg Cantor.  Zeitschrift für Math. und Physik 14 (1869) 121-128.  ??NYS _ Knuth, below, says this is the first general treatment of mixed base systems.

Vittorio Grünwald.  Intorno all'arithmetica dei sisteme numerici a base negativa.  Giornale di Matematiche di Battaglini 23 (1885) 203‑221 & 367.  ??NYS _ cited by Glaser, op. cit. in 7.M, pp. 94 & 109..

N. G. de Bruijn.  On bases for the set of integers.  Publ. Math. Debrecen 1 (1950) 232‑242.  ??NYS _ cited by Knuth & Gardner, below.  Knuth says it has representations with negative bases, but doesn't do arithmetic.

Donald E. Knuth.  Paper submitted to a Science Talent Search for high-school seniors in 1955.  ??NYS _ described in Knuth, below.  Discussed negative bases and complex bases. 

G. F. Songster.  Master's thesis, Univ. of Pennsylvania, 1956.  ??NYS _ Knuth, below, says it studies base  -2.

Z. Pawlak & A. Wakulicz.  Bull. de l'Acad. Polonaise des Sciences, Classe III, 5 (1957) 233‑236; Série des sciences techniques 7 (1959) 713-721.  ??NYS _ Knuth, below, cites this and the next item as the first mentions of negative base arithmetic in print.

Louis R. Wadel.  Letter.  IRE Transactions on Electronic Computers EC‑6 (1957) 123.  ??NYS _ cited by Knuth & Gardner, below.

W. Parry.  Acta Math. (Hungar.) 11 (1960) 401-416.  ??NYS _ Knuth says he treats irrational bases.

Donald E. Knuth.  The Art of Computer Programming:  Vol. 2: Seminumerical Algorithms.  Addison-Wesley, Reading, Massachusetts, 1969.  Section 4.1: Positional arithmetic, pp. 161-180 is an exposition of various bases.  The fact that any positive number can be used as base seems to first appear in Pascal's De numeris multiplicibus of c1658.  He suggested base  12.  Erhard Wiegel proposed base  4  from 1673.  Joshua Jordaine's Duodecimal Arithmetick, London, 1687, obviously expounded base  12.  Juan Caramuel Lobkowitz's Mathesis biceps 1 (Campaniae, 1670) 45-48 discussed bases  2, 3, 4, 5, 6, 7, 8, 9, 10, 12 and 60.  Charles XII of Sweden seems to have invented base  8  c1717 and also considered base  64.  John W. Nystrom developed base  16  in:  J. Franklin Inst. 46 (1863) 263-275, 337-348 & 402-407.  Knuth then discusses negative and complex bases _ see above items _ and describes bases  2i  and  i-1  and the use of balanced ternary and negative digits in general.

Gardner.  SA (Apr 1973)  c= Knotted, chap. 8.  Exposits negative bases, which seem to have been invented c1955 by Donald Knuth.  But the Addendum in Knotted cites Grünwald via Glaser.  Gardner cites several other articles.

Daniel Goffinet.  Number systems with a complex base: a fractal tool for teaching topology.  AMM 98 (1991) 249-255.  Explores the set of all sums of distinct powers of the base, using base  1/2,  i,  etc.

Fred Newhall.  History of the duo-decimal, base  12,  dozenal idea, chronologically.  The Duodecimal Bulletin 37;:2; (No. 73; (11*2) 4‑6 [i.e. 43:2 (No. 87) (1994) 4-6].  Outline chronology with 42 entries.

 

          7.AB. PERFECT NUMBERS, ETC.

 

          This is too lengthy a subject to cover in detail here.  Below are a few landmarks.  See Dickson I, ch. I for an extended history.  Heath's notes summarise the history.  I have a separate file on the history of Mersenne and perfect numbers. 

          Let  M_n  =  2ⁿ - 1  and  P_n  =  2^n‑1(2ⁿ-1).  M_n  is known to be prime, and hence  P_n  is perfect for  n =  2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433.

          Paul Poulet (1918) found two amicable chains, one starting at 12496, and Henri Cohen found seven more, one of which starts at 14316 and has 28 links.

 

Euclid.  (The Thirteen Books of Euclid's Elements, edited by Sir Thomas L. Heath;  2nd ed., (CUP, 1925??);  Dover, vol. 2, pp. 278, 293-294 & 421-426.)

VII, def. 22.  "A perfect number is that which is equal to its own parts." 

IX, prop. 36.  "If as many numbers as we please beginning from an unit be set out continuously in double proportion, until the sum of all becomes prime, and if the sum multiplied into the last make some number, the product will be perfect."

Nicomachus (c100) and Theon of Smyrna (c125) define abundant and deficient numbers.

Nicomachus seems to be the first to claim the perfect numbers alternately end in  6  and  8.

Iamblichus.  On Nicomachus's Introduction to Arithmetic.  c325.  ??NYS _ cited in Dickson I 38.  Gives first four perfect numbers.  Earliest known reference to amicable numbers, giving the first pair:  220, 284. 

There was Arabic interest in these numbers.  Thabit ibn Qurra (Th_bit ibn Qurra) (836-901) gave a complex rule to produce amicable numbers, but apparently could not find any.   Ibn al-Banna (Ibn al-Bann_’) (1256-1321) discovered the second known pair of amicable numbers:  17296 and 18416,  apparently using Thabit ibn Qurra's rule.  [Q. Mushtaq & A. L. Tan; Mathematics: The Islamic Legacy; Noor Publishing House, Farashkhan, Delhi, 1993, pp. 93-94.]

Jordanus de Nemore.  De Arithmetica.  c1225.  An edition by Lefèvre d'Etaples was printed in Paris in 1496.  Book VII, prop. 53-60.  ??NYS _ described in: Nobuo Miura; Charles de Bovelles and perfect numbers; Historia Scientarum 34 (1988) 1-10.  He knows of the existence of odd abundant numbers, despite Dickson's assertion to the contrary.

Chuquet.  1484.  Triparty, part 1.  FHM 57-59 & 360 gives English and discussion.  Believes the pattern is regular, giving  P_n  for  n = 2, 3, 5, 7, 9, 11, 13  _ but the  5th  &  6th  of these are not perfect.

Pacioli.  Summa.  1494.  Ff. 6v-8v.  Discussion of perfect numbers.  Cites Euclid.  Says  3, 7, 31, 127  are prime and so  6, 28, 496, 8128  are perfect.  Says endings alternate.

Charles de Bovelles (Carolus Bovillus).  Liber de perfectis numeris.  This is a small treatise included in an untitled collection of his works, Paris, 1510 (reprinted Stuttgart/Bad Cannstatt, 1970), ff. 172r-180r.  ??NYS _ described in: Nobuo Miura; Charles de Bovelles and perfect numbers; Historia Scientarum 34 (1988) 1-10.  This is probably the first printed work on number theory, but it claims that  P_n  is prime when  n  is odd!!  Gives values of the 'perfect' number for  n = 1, 2, 3, 5, 7, 9, ..., 39.

Tonstall.  De Arte Supputandi.  1522.  Pp. 222-223.  Brief discussion of abundant, deficient and perfect numbers, only mentioning  6  and  28.

Cardan.  De Numerorum Proprietatibus.  ??, ??NYS.  = Opera Omnia, vol. IV, pp 2-4, sections 4 & 5.  (It is possible that this is the first publication of this item??)  Brief discussion, mentioning  6, 28, 496, 8128  and Euclid.

Cardan.  Practica Arithmetice.  1539.  Chap. 42, sections 2 & 3, ff. H.i.r - H.i.v (p. 52).  Mentions  220 & 284,  then similar to above.

Robert Record.  The Whetstone of Witte.  John Kyngstone, London, 1557.  Facsimile by Da Capo Press, NY  &  Theatrum Orbis Terrarum, Amsterdam, 1969.  Nombers perfecte, ff. A.iv.r - A.iv.v.  Discusses perfect numbers briefly and asserts  P_n  is perfect for   n  = 2, 3, 5, 7, 9, 11, 13, 15.

van Etten.  1624.  Prob. 70 (63), parts IIII & VII, pp. 66-67 (92-93).  Mentions perfect numbers and says they occur for  n = 2, 3, 5, 7, 9, 11, 13,  with  486  for  496.  He asserts the endings alternate between  6  and  28.  Henrion's 1630 Notte, pp. 22‑23, refers to Euclid, corrects  486  to  496,  and says someone has recently claimed  120  is a perfect number.  Deblaye, op. cit. in 1, copies  486  as  286!  The 1653 English ed copies  486,  gives  120816  for  130816,  extends the list with  n = 15, 17, 19  and asserts  n = 39  gives the  20th  perfect number.  Describes  220  and  284. 

M. Mersenne.  Letter to Descartes, 1631.  (??NYS _ cited in:  Ore, Number Theory and Its History, 95  and  Dickson I 33.)  Raises question of multiply perfect numbers and states(?)  120  is  3‑perfect.

Thomas Stanley.  Pythagoras.  The Ninth Part of The History of Philosophy, (1655-1662), collected ed., 1687, pp. 491-576.  Reprinted by The Philosophical Research Society, Los Angeles, 1970.  P. 552 discusses the Pythagorean attitude to perfect numbers.  "... nor without reason is the number  6  the foundation of generation, for the Greeks call it τελείov, we perfect; because its three parts,  1/6  and  1/3  and  1/2  (that is  1,  2,  and  3.) perfect it".

Leybourn.  Pleasure with Profit.  1694.  Observation 2, p. 3.  Essentially taken from the English ed. of van Etten, with the same numerical mistakes and a further mistake in copying the 20th case.  Mentions  220, 284.

Ozanam.  1694.  Prob. 5, quest. 17-19, 1696: 14-18;  1708: 13-15.  Prob. 8, quest. 17-19, 1725: 29-41.  Chap. 3, art. 11-12 & 15, 1778: 32-38;  1803: 35-40;  1814: 32-36;  1840: 19-21.  1696 says  2^p‑1(2^p-1)  is perfect if  2^p - 1  is prime, but asserts this is true for  p = 11.  1696 also notes that  120  is  3-perfect and gives several amicable numbers.  1725 extends the remarks on amicable numbers.  1778 notes that  p = 11  fails, stating that Ozanam forgot that  2^p - 1  had to be a prime, and gives the first  8  perfect numbers.  1778 also gives some amicable pairs. 

Manuel des Sorciers.  1825.  P. 86.  ??NX  Gives the "perfect numbers" corresponding to  p = 2, 3, 5, 7, 9, 11, 13,  giving  486  for  496  and saying the endings alternate btween 6 and 8.  Probably copied from van Etten.  I've included this because it is surprisingly late to be so erroneous!

B. N. I. Paganini.  Atti della Reale Accademica delle Scienze di Torino 2 (1866‑1887) 362.  ??NYS.  Discovery of second smallest amicable  pair:  1184, 1210.

Pearson.  1907.  Part II: Amicable numbers, pp. 35‑36.  Asserts that  220, 284;  17296, 18416;  93 63584, 94 37056  are the only amicable pairs below  10 millions.

Alan Turing and/or colleagues was the first to use a computer to search for new Mersenne primes on the Manchester Baby in 1949, but it could not easily deal with numbers greater than  M₃₅₃.

R. M. Robinson wrote a program to search for Mersenne primes using the Lucas-Lehmer test on the SWAC in late 1951/early 1952.  It was his first program.  On 30 Jan 1952. it was loaded and ran!  It discovered the 13th and 14th Mersenne primes:  M₅₂₁  (at about 10:00 pm, taking about a minute) and  M₆₀₇  (just before midnight).  M₁₂₇₉,  M₂₂₀₃  and  M₂₂₈₁  were found in the next months.  The program comprised  184  machine instructions on  24  feet of paper tape and would handle cases up through  2297.  It ran successfully on its first trial!  Lehmer was present when the program was tested on  M₂₅₇,  which Lehmer spent some  700  hours in testing  c1932,  and the program confirmed this in a fraction of a second.  c1982,  Robinson ran his program on an early PC which only ran about twice as fast as the SWAC.

Alan L. Brown.  Multiperfect numbers _ cousins of the perfect numbers.  RMM 14 (Jan‑Feb 1964) 31‑39.  Lists all known  3‑,  4‑,  5‑perfects and the first  100  6‑perfects.

Elvin J. Lee & Joseph S. Madachy.  The history and discovery of amicable numbers _ Parts 1, 2, 3.  JRM 5 (1972) 77‑93, 153‑173, 231‑249.  Part 1 is the main history.  Parts 2 and 3 give all  1107  amicable pairs known at the time, with notes explaining the listings.

B. L. van der Waerden.  A History of Algebra.  Springer, Berlin, 1985.  Pp. 21‑23 describes the work of Tabit ibn Qurra (824?‑901) on amicable numbers and its development by Fermat, Descartes, Euler and Legendre.

Jan P. Hogendijk.  Th_bit ibn Qurra and the pair of amicable numbers  17296, 18416.  HM 12 (1985) 269-273.  This pair is often named for Fermat, who first mentions it in Europe.  Th_bit gives a general rule which would yield this pair as the second example, though he doesn't give the values.  Hogendijk analyses Th_bit's work and concludes that he must have known these values.  In the same issue, a review by Hogendijk (pp. 295-296) mentions that the pair in question was known in 14C Persia and that the pair  9363584, 9437056,  usually ascribed to Descartes, was known c1600 in Persia.

Ettore Picutti.  Pour l'histoire des septs premiers nombres parfaits.  HM 16 (1989) 123-126.

 

          7.AC. CRYPTARITHMS, ALPHAMETICS AND SKELETON

                                        ARITHMETIC

          A skeleton problem shows all the working with most digits indicated by the same symbol, e.g.  *,  and only a few digits are left in place.

          A cryptarithm or alphametic usually shows just the data and the result with digits replaced by letters as in a substitution cipher.

          The opening section includes some miscellaneous numerical-alphabetical recreations which I haven't yet classified in subsections.

 

C. Dudley Langford.  Some missing figure problems and coded sums.  MG 24 (No. 261) (Oct 1940) 247‑253.  Lots of examples of various forms.

[J. S. Madachy?]  Alphametics.  RMM 6 (Dec 1961) 27,  7 (Feb 1962) 13  &  10 (Aug 1962) 11.  Historical comments.  Cites Berwick, Schuh, Dudeney, Minos, Hunter.  Says Strand Mag. (1921) is first division with letters instead of uniform  *.

"Fomalhaut".  Cryptophile cryptofile:  Cryptarithms.  World Game Review 8 (Jul 1988) 5‑12.  Survey of various forms of these puzzles and related books and magazines.

Graham Hawes.  Wordplay.  M500 116 (Nov 1989) 6‑7.  Using the numerological mapping  A = 1, B = 2, ..., Z = 26,  he finds two numbers, in British usage, whose numerological value is itself.  One is "two hundred and fifty one".  (The use of 'and' is British, but not American.)  He has found none in American usage.  Again in British usage, the sum  73 + 89 = 162   gives a correct sum for its numerological values:  166 + 116 = 282.

 

          7.AC.1.        CRYPTARITHMS:  SEND + MORE  =  MONEY,  ETC.

 

American Agriculturist (Dec 1864).  ??NYR _ copy sent by Shortz.   Multiplication problem where the letters for  1 ‑ 0  spell Palmerston.

Loyd.  Cyclopedia.  1914. 

Alphabetical addition, pp. 233 & 370. 

   BOW + APPLE + CHOPS + HASHES + CHEESE + APPLES + EHW  =  PALEALE;  B + LAY + TEN + DOZ  =  DNLL. 

Pp. 238 & 371.  = SLAHP: Masquerading digits, pp. 86 & 119.  JGDCH * IFABE  =  BIBDEB.

Smith.  Number Stories.  1919.  See 7.AC.2 for examples with full layout.

Dudeney.  Perplexities: Verbal arithmetic.  Strand Mag. (Jul 1924).  ??NYS.  SEND + MORE  =  MONEY;   EIGHT ‑ FIVE  =  FOUR;   TWO * TWO  =  THREE;   SEVEN/TWO  =  TWO  (with full division layout).

Dudeney.  Problem ?: The Arab's puzzle.  Strand Mag. (Early 1926?).  ??NX.  ABCD * EFGHI  =  ACGEFHIBD.

Loyd Jr.  SLAHP.  1928.  Kindergarten algebra, pp. 48 & 102.  AB * AB  =  CDDD.

MINOS [Simon Vatriquant].  Sphinx 1 (May 1931) 50.  Introduces word "cryptarithmie".  "A charming cryptarithm should  (1)  make sense in the given letters as well as the solved digits,  (2)  involve all the digits,  (3)  have a unique solution, and  (4)  be such that it can be broken by logic, without recourse to trial and error."  (Translation by C. W. Trigg in CM 4 (1978) 68.)

C. O. Oakley, proposer;  W. E. Buker, solver.  Problem E7.  AMM 39 (1932) 548 ??NYS  &  40 (1933) 176.  SEND + MORE  =  MONEY.  Editorial comment in solution cites L'Echiquier (June 1928) and Sphinx.

H. Phillips.  The Playtime Omnibus.  Op. cit. in 6.AF.  1933.  Section XVI, prob. 7: The money code, pp. 49‑50 & 234.  SEND + MORE  =  MONEY.

Rudin.  1936.  No. 84, pp. 28-29 & 92.  SEND + MORE  =  MONEY.

Dr. Th. Wolff.  Die lächelnde Sphinx.  Academia Verlagsbuchhandlung, Prague, 1937.  Prob. 7, pp. 188 & 197-198.  SEND + MORE = MONEY.

Adams.  Puzzle Book.  1939.  Several straightforward examples and the following.  Prob. C.38: Economy, pp. 133 & 176.  SAVE + MORE  =  MONEY.  Four solutions given.

Alan Wayne.  The Cryptogram (c1945) (a US puzzle magazine), ??NYS.   Introduces 'doubly true additions', e.g.  SEVEN + SEVEN + SIX  =  TWENTY.  (See Trigg cited above at MINOS.)

Alan Wayne, proposer;  A. Chulick, solver;  editorial note by Howard Eves.  Problem E751 _ A cryptarithm.  AMM 54 (1947) 38  &  412‑414.  FORTY + TEN + TEN  =  SIXTY.  Editor cites Wayne in The Cryptogram for several others:  SEVEN + SEVEN + SIX  = TWENTY;   SEVEN + THREE + TWO  =  TWELVE;   TWENTY + FIFTY + NINE + ONE  =  EIGHTY.

Morley Adams.  Puzzle Parade.  Faber, London, 1948.

Chap. 3, no. 25: A wordy sum, pp. 46 & 52.  ONE + TWO + FIVE  =  EIGHT.  Answer starts "Here is one solution".

Chap. 9, no. 32: Simple as ABC, pp. 149 & 151.  ABC + ABC/5  =  CBA.  Answer is  495.

Jerome S. Meyer.  Fun-to-do.  Op. cit. in 5.C. 1948.  Prob. 27: The Cleveland butcher, pp. 34 & 186.  PORK/CHOP  =  C  >  2.  Unique answer is  9867/3289 = 3.  [Can there be answers with  C = 2??]

Problems Drive, 1953‑54.  Eureka 17 (1954) 8 & 16.  ONE + TWO + FOUR  =  SEVEN.  One solution (not unique) given.

J. A. H. Hunter.  Fun with figures.  Globe & Mail (Toronto) (27 Oct 1955) 27  &  (28 Oct 1955) 29.  "It's just an easy alphametic today."   ABLE/RE  =  SIR  given in full diagram, determine the value of  MAIL  (=  8940).  Brooke (below), p. 45, reproduces Hunter's column.  Brooke says:  'Hunter received a letter from a reader referring to a "alphametical problem in which letters take the the place of figures".'

Maxey Brooke.  150 Puzzles in Crypt‑Arithmetic.  Dover, (1963), 2nd ed. 1969.  On p. 4, he asserts that "Arithmetical Restorations" ... "were probably invented in India during the Middle Ages", but he gives no evidence on this point.

J. A. H. Hunter.  Note 3104:  CROSS + ROADS  =  DANGER.  MG 48 (No. 366) (Dec 1964) 433‑434.  This was posed by E. A. Maxwell in MG (Feb 1964) 114.  Hunter finds the unique solution.

Roy Childs.  Letter of 4 Aug 1999.  He has used a computer to study 'doubly-true additions' with sums up to  TWENTY.  In 14 cases, there are unique solutions, of which all but the first have sum  TWENTY. 

          THREE + THREE + TWO + TWO + ONE; 

          FIVE + THREE + THREE + THREE + THREE + THREE;  

          SEVEN + THREE + THREE + ONE + ONE + ONE + ONE + ONE + ONE + ONE;  

          SEVEN + THREE + THREE + TWO + TWO + ONE + ONE + ONE;  

          SEVEN + FIVE + TWO + TWO + ONE + ONE + ONE + ONE;  

          SEVEN + FIVE + TWO + TWO + TWO + ONE + ONE;  

          SEVEN + FIVE + FIVE + TWO + ONE;  

          SEVEN + SEVEN + TWO + ONE + ONE + ONE + ONE;  

          SEVEN + SEVEN + TWO + TWO + TWO;  

          SEVEN + SEVEN + SIX;  

          EIGHT + EIGHT + TWO + ONE + ONE;  

          ELEVEN + THREE + THREE + ONE + ONE + ONE;  

          ELEVEN + THREE + THREE + TWO + ONE; 

          ELEVEN + THREE + THREE + THREE

 

          7.AC.2.        SKELETON ARITHMETIC:  SOLITARY SEVEN, ETC.

 

W. P. Workman.  The Tutorial Arithmetic, op. cit. in 7.H.1, 1902.  See also comments in Ackermann, under Berwick, below.  Chap. VI _ Examples XIX, probs. 30‑41, pp. 48‑49 & 503 (= 50‑51 & 529 in c1928 ed.).  Simple problems, e.g. prob. 30:  2982**  divided by  456  leaves remainder  1.  Probs. 31‑34 are skeleton multiplications;  35‑37 are skeleton divisions.

W. E. H. Berwick.  School World 8 (Jul & Aug 1906) 280 & 320.  ??NYS.  Division with seven sevens given:  7375428413 / 125473  =  58781.  Actually, there are  13  7s  in the layout, so not all the  7s  are shown.  Ackermann, below, says Berwick composed this, at age 18, after seeing some examples in Workman.

Pearson.  1907. 

Part II, no. 5: Fill in the gaps, pp. 114 & 191.  Division layout with some numbers given.

Part II, no. 11: Find the multiplier, pp. 115‑116 & 193.  Multiplication layout with some numbers given.

???  J. Indian Math Club, 1910.  ??NYS _ cited by Archibald who says it has skeleton divisions with  4  digits given.

Smith.  Number Stories.  1919.  Pp. 111‑112 & 139‑140.  Two cryptarithmic multiplications and two cryptarithmic divisions, but with full layouts.

W. E. H. Berwick.  MG (Mar 1920) 43.  ??NYS _ cited by Archibald.  Four fours.

F. Schuh.  Een tweetal rekenkundige ardigheden.  Nieuw Tijdschrift voor Wiskunde 8 (1920‑1921) 64.  Skeleton division with no digits given, but the quotient has a repeating decimal and the divisor and dividend are relatively prime.  The problem is reproduced as Note 16, AMM 28 (1921) 278, signed ARC [= R. C. Archibald], with solution by D. R. Curtiss and comment by A. A. Bennett in AMM 29 (1922) 210‑213.  Bennett shows that relatively primality is not essential.  The problem and solution are given as Section 258: Repeating division puzzle, pp. 320‑322, in Schuh's The Master Book of Mathematical Recreations, Dover, 1968.  (Originally Wonderlijke Problemen; Leerzaam Tijdverdrijf Door Puzzle en Spel, Thieme, Zutphen, 1943.)  7752341 / 667334  =  11.6168830001168830001....

R. C. Archibald.  AMM 28 (1921) 37 _ sketches the history:  J. Indian Math Club;  Berwick's 'seven sevens' and 'four fours' cited to MG (Mar 1920) 43.

W. E. H. Berwick.  Problem 555: The four fours (in Dudeney's column).  Strand Mag. (1921 or 1922?)  ??NYS.  Four fours given in the skeleton of  1200474 / 846  =  1419,  but there are other fours present and there are three other solutions.

Egbert F. Odling.  Problem 627: Solitary seven (in Dudeney's column).  Strand Mag. (Nov  &  Dec 1922), ??NYS.  Skeleton of  12128316 / 124  =  97809,  with only the  7  given.  Unique solution and the  7  only occurs once.

Anon.  Note 671.  MG 11 (1922‑23) 338.  Gives Odling's problem and solution and some comments.

Ackermann.  1925.  Pp. 109‑115.  Discusses Berwick's problems, referring to MG of Mar 1920, Dec 1921 and Jan 1922 (??NYS) for four fours, five fives, three threes and six sixes.  Gives one of Smith's problems, Berwick's seven sevens and five fives, Schuh's problem (attributed to Ball) and Odling's problem.

Dudeney.  MP.  1926.  Prob. 70: The solitary seven, pp. 26‑27 & 118.  (= 536, prob. 144, pp. 43 & 259.)  Cites EFO [= Odling].  "It is the first example I have seen ... in which only one figure is given."

A. A. Bennett, proposer;  H. Langman, solver.  Problem 3212.  AMM 33 (1926) 429  &  34 (1927) 538-540.  Skeleton division   xxxxxxxxxxxxxccfx / xxxxabxxxx  =  xcxxxxx,  with numerous further positions given by definite letters.  Solution notes that some of the information is not needed, e.g.  f  can be replaced by  x.  Answer is   70900515872010075 / 68253968253  =  1038775.

Collins.  Fun with Figures.  1928.  Through a knot-hole, p. 189.  *7*9* / 215  =  1**   with some further figures given.

H. E. Slaught, proposer;  C. A. Rupp, solver.  Problem E1.  AMM 39 (1932) 489  &  40 (1933) 111-112.  Odling's 'Solitary Seven' problem.  Editor's note says it has appeared in Le Sphinx.

Perelman.  FFF.  1934.  Mysterious division & Another division.  1957: probs. 104 & 105, pp. 138 & 145;  1979: probs. 107 & 108, pp. 167 & 176.  = MCBF, probs. 107 & 108, pp. 168 & 178-179.  Berwick's 'four fours' and 'seven sevens', with all four solutions of the latter, rather poorly attributed to "the American publications School World (1906) and Mathematical Magazine (1920)".

Perelman.  FMP.  c1935?  Mysterious division, pp. 256 & 268‑269.  Skeleton of  11268996 / 124  =  90879,  with only the  7  given.  The quotient is unique, but there are  11  possible divisors:  114, 115, ..., 124,  of which  115, 116 and 120  give no other  7s  in the layout.

A. G. Sillito.  Note 1424:  Division without figures.  MG 23 (No. 257) (Dec 1939) 467‑468.  Two divisions like Schuh's:  16 / 41  and  81 / 91.

W. T. Williams & G. H. Savage.  The Penguin Problems Book.  Penguin, 1940. 

No. 67: A skeleton square root, pp. 40 & 125.  A dotted diagram with two letters marked in, one  8  times, the other  4  times.

No. 87: A skeleton division, pp. 49 & 132.  = Odling, 1922.

Sullivan.  Unusual.  1947.  Prob. 37: Lost and found.  Skeleton division of  1089708 / 12  =  90809   with only the  8  of the quotient given.

P. L. Chessin, proposer;  ???, solver.  Problem E1111.  AMM 61 (Apr 1954) 712  &  ???.  ??NYS _ given in the Otto Dunkel Memorial Problem Book, ed. by Howard Eves and E. P. Starke, AMM 64:7 part II (Aug-Sep 1957) 6, where it is described as the most popular problem ever published in the AMM, with 70 solvers.  Also given by Gardner, SA (May 1959) = 2nd Book, chap. 14, prob. 5: The lonesome  8,  pp. 154-155 & 160‑161.  Skeleton division of   10020316 / 124  =  80809   with only the middle  8  of the quotient given.  Answer is unique.

William R. Ransom.  Op. cit. in 6.M.  1955.  Only one digit known, p. 134.  Skeleton of   11260316 / 124  =  90809  with only the  8  given.  Answer is unique and  8  only appears once.

G. A. Guillotte.  Note 2865:  Missing digits.  MG 43 (No. 345) (Oct 1959) 200.  Long division with  17  0s  specified.

Anonymous postcard to The Science Correspondent, "The Glasgow Herald", 8 May 1963, found in Prof. Lenihan's copy of Gardner's More Mathematical Puzzles and Diversions and given in Jay Books sale catalogue 129 (Feb? 1992) and 130 (Jun 1992).  Full skeleton of  1062 / 16 = 66.375   with no digits specified.  The solution is unique.

Birtwistle.  Math. Puzzles & Perplexities.  1971.

Seven and eight, p. 149.  **7** / **  = 8***,  full skeleton shown, but no other values given.  Unique solution is  99708 / 12  =  8309.

Long division, pp. 150, 185 & 197.  Skeleton of  123195 / 215  =  573  with four values given.

Letter division, pp. 151, 186 & 198.  Complete layout of a division with letters denoting the digits, as in a cryptarithm.  Two possible answers:  6420 / 20  =  321  or  6930 / 30  =  231.

On all fours, pp. 151 & 198.  Skeleton of  31666 / 142  =  223  with all fours given.

 

          7.AC.3.        PAN‑DIGITAL SUMS

 

          These are generally of the form   ABC + DEF  =  GHI.   The digits may be positive or  9  of the  10  digits.  One can also have a  10‑digital form, e.g.   ABC + DEF  =  GHIJ.

          I also include here problems like inserting  +  and  -  (and perhaps  ´  and  ¸)  signs into  12...9  to yield  100,  which I will call  'insertion to make  100'.  This has two quite different sets of answers depending on whether the operations are carried out sequentially (as on an old calculator) or in algebraic order of precedence (as on a computer or modern calculator).  See also 7.AC.6 for similar problems.

 

Leske.  Illustriertes Spielbuch für Mädchen.  1864?  Prob. 564-16, pp. 253 & 395.  Make  100 from  1, 2, ..., 9, 0.  Answer:  9x8 + 7 + 6 + 5 + 4 + 3 + 2 + 1.  (0 isn't used.)

Dudeney.  Problem 64: The lockers puzzle.  Tit‑Bits 33 (18 Dec 1897  &  5 Feb 1898) 220  &  355.  = AM, prob. 79, pp. 14 & 156.  Find   ABC + DEF  =  GHI   using  9  of the  10  digits which have the least result, the greatest result and a result whose digits are distinct from the first two.  Answers:  107 + 249  =  356; 

          235 + 746  or  324 + 657   =  981; 

          134 + 586  =  720   or   134 + 568  =  702   or   138 + 269  =  407.

Anon. & Dudeney.  A chat with the Puzzle King.  The Captain 2 (Dec? 1899) 314-320;  2:6 (Mar 1900) 598-599  &  3:1 (Apr 1900) 89.  Insert as few signs as possible in  12...9  to make  100.  Usual answer is  1 + 2 + 3 + 4 + 5 + 6 + 7 + (8x9)  (cf Leske),  but he gives  123 ‑ 45 ‑ 67 + 89.  Cf. AM, 1917.

Dudeney.  AM.  1917. 

Prob. 77: Digits and squares, pp. 14 & 155.  For  ABC + DEF  =  GHI,  he wants  DEF  =  2 * ABC,  so   GHI  =  3 * ABC,  using the  9  positive digits.  Says there are four solutions, the tops being   192,  219,  273,  327.

Prob. 94: The digital century, pp. 16-17 & 159-160.  Insert signs into  12...9  to make  100,  using:  (1) as few signs as possible;  (2) as few strokes as possible, with  -  counting as  1  stroke;  +,  ()  &  ´  counting as  2;   ¸  counting as  3.  He finds his 1899 result is best under both criteria.  Cf. Anon & Dudeney, 1899.

Adams.  Puzzles That Everyone Can Do.  1931.

Prob. 89, pp. 38 & 142: Doubles and trebles.  Same as Dudeney, AM, prob. 77.

Prob. 262, pp. 98 & 169: A number puzzle.  Insert 'mathematical signs' into  4 3 2 1  to make 100.  Answer:  Ö4 * [(3 + 2) / .1] .

Perelman.  1934.  See in 7.AC.6 for pandigital sum yielding  1  and  100.

McKay.  At Home Tonight.  1940.  Prob. 19: Centuries, pp. 66 & 80.  Insertion to make  100.  -1x2 - 3 - 4 - 5 + 6x7 + 8x9.   -1x2 -3 - 4 + 5x6 +7 + 8x9.   1x2x3x4 + 5 + 6 + 7x8 + 9.

Morley Adams.  Puzzle Parade.  Op. cit. in 7.AC.1.  1948.  No. 12: Figure square, pp. 146 & 150.  As in Dudeney, AM, prob. 77.  Says there are four solutions, but wants the one with minimal  E.  Solution:   219 + 438  =  657.

Philip E. Bath.  Fun with Figures.  Op. cit. in 5.C.  1959.

No. 15: Skeleton addition, pp. 10 & 41.  Complete   ABC + DE7  =  GH8   using all nine positive digits.  Gets four forms by reversing  A  and  D  and/or  B  and  E.

No. 48: A tricky problem, pp. 20 & 48.  "Can you replace the asterisks by the digits of the number  216345879  in this order so the resulting total is  100?   * + * + * + * + * + *"   Answer:  2/1 + 6/3 + 4 + 5 + 8 + 79.  His fraction bars are horizontal, but this problem seems a bit unreasonable to me.

Richard E. Bellman.  On some mathematical recreations.  AMM 69:7 (Aug/Sep 1962) 640‑643.  Develops a general theory for the number of ways  n  can be obtained by inserting  +  or  x  into   a₁a₂...a_N   and for determining the minimum number of  +  signs occurring.  He computes the example of inserting into  12...9  to make  100  by use of recursion, finding that   1x2x3x4 + 5 + 6 + 7x8 + 9  =  100   has the minimal number of  +  signs.

Gardner.  SA (Oct 1962) c= Unexpected, chap. 15.  Insertion to make  100.  Cites Dudeney.  Asks for minimum number of insertions into  98...1  to  make  100.  Answer with four signs.

Gardner.  SA (Jan 1965) c= Magic Numbers, chap. 6.  Considers inserting  +  and  -  signs in  12...9  or  98..1  to yield  100.  Says he posed this in SA (Oct 1962) c= Unexpected, chap. 15 and many solutions for both the ascending and descending series were printed in Letters in SA (Jan 1963).  He gives a table of all the answers for both cases:  11  solutions for the ascending series and  15  solutions for the descending series.  He extends the problem slightly by allowing a  -  in front of the first term and finds  1  and  3  new solutions.

Jonathan Always.  Puzzles to Puzzle You.  Op. cit. in 5.K.2.  1965.  No. 135: All the numbers again, pp. 42 & 90.  As in Dudeney, AM, prob. 77.  Gives one solution:   192 + 384  =  576.

Wickelgren.  How to Solve Problems.  Op. cit. in 5.O.  1974.  Integer‑path‑addition

          problem, pp. 130‑132.  Wants the  9  positive digits in a pan-digital sum, so the                   1 2 9

          resulting  3 x 3  array has each digit  i  horizontally or vertically adjacent                         +  4 3 8

          to the digit  i+1.  Says there is one solution, shown at the right.                                     =  5 6 7

Shakuntala Devi.  Puzzles to Puzzle You.  Op. cit. in 5.D.1.  1976.  Prob. 41: The

          number and the square, pp. 31 and 107.  As in Dudeney, AM, prob. 77.  Gives all solutions.

Birtwistle.  Calculator Puzzle Book.  1978. 

Prob. 65: Key to the problem, pp. 47-48 & 104.  Using a calculator, insert operations in  012...9  and  98...10  to produce  100.  Gives one example of each:  0 + 1/2 + 3x4x5 + 6 + 7 + 8 + 9;   9x8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0.

Prob. 87: Three by three _ part two, pp. 61 & 120-121.  Which distributions of the nine positive digits as  ABC, DEF, GHI  have the lowest sum and product?  147 + 258 + 369  =  774   gives the lowest sum and the digits in each position can be permuted _ e.g.  348 + 257 + 169  gives the same sum.  The lowest product is uniquely given by   147 x 258 x 369  =  13994694.

Johannes Lehmann.  Kurzweil durch Mathe.  Urania Verlag, Leipzig, 1980.  No. 14, pp. 39 & 139-140.   A + B + C  =  D + E + F  =  G + H + I   has just two solutions using the positive digits.  [Interestingly, one gets no more solutions using the ten digits.]

Steven Kahane.  Sign in, please!  JRM 23:1 (1991) 19-25.  Considers inserting  +  and  -  signs in  12...n  and in  n...21  to produce various results, e.g.  0,  n+1,  n,  among others.

Ken Russell & Philip Carter.  Intelligent Puzzles.  Foulsham, Slough, 1992. 

Prob. 102, p. 79 & Answer 30, p. 140: One hundred puzzles.  Three problems to insert the nine positive digits into formulae to make 100.  E.g.  + + + - - -  =  100  is solved as:  3² + 91 + 7 + 8 - 6 - 5 - 4.

Prob. 140, p. 106 & Answer 104, p. 181: Plus and Minus.  Find all ways to insert  +  and  -  signs into  1 2 ... 9  to yield 100.  Finds 12 ways, one with a leading minus, as given by Gardner, 1962.

David Singmaster.  Determination of all pan-digital sums with two summands.  JRM 27:3 (1995) 183-190.  AB + CDE = FGHI  has no solutions with the nine positive digits and ten basic solutions using nine of the ten digits.  AB + CDEF = GHIJ  has nine basic solutions.  (Each basic solution gives four or eight equivalent solutions.)  ABC + DEF = 1GHI  has 12 basic solutions, which can be paired.  ABC + DEF = GHI  has 216 basic solutions, but 80 have  A = 0.  42 cases use the positive nine digits.  The 216 can be grouped into 72 triples and a canonical example is given for each triple.  The cases where the three terms form a simple proportion are listed.

 

          7.AC.4.        PAN‑DIGITAL PRODUCTS

 

Loyd.  Problem 36: Juggling with figures.  Tit‑Bits 32 (5 Jun  &  3 Jul 1897) 173  &  258.  10‑digital product with smallest result is   3907 * 4  =  15628.   Solution also gives the largest case:   9403 * 7  =  65821.   [I have verified that these are correct.]

Dudeney.  The miller's puzzle.  The Canterbury puzzles.  London Mag. 8 (No. 46) (May 1902) 367-371  &  8 (No. 47) (Jun 1902) 480-482.  = CP, prob. 3, pp. 26 & 164-165.  Find a solution of   A*BC  =  D*EF  =  GHI  using the  9  positive digits and which is closest to a given pattern.  Answer says there are four solutions, the closest is   2 * 78  =  4 * 39  =  156.

Dudeney.  CP.  1907. 

Prob. 93: The number blocks, pp. 139-140 & 238.  Find a solution of   AB*CDE  =  FG*HIJ.   Answer:  64 * 915  =  80 * 732.

Prob. 101: The three motor-cars, pp. 147-149 & 242-243.  Wants a solution of   AB*CDE  =  FGHIJ   such that  AB  divides  CDE.  Answer is   27 * 594  =  16038   and he says it is hard to show this is unique.  He says there are many solutions for   A*BCDE  =  FGHIJ,   e.g.   3 * 5694  =  17082.  = Wood, 1927, prob. 62.

M. Thié.  ??  Nouv. Ann. Math. (4) 11 (1911) 46.  ??NYS _ cited by Dickson I 463, item 62a.  Found examples with  9  positive digits like   12 * 483  =  5796.

T. C. Lewis.  ??  L'Intermédiaire des Math. 19 (1912) 26-27  &  187.  ??NYS _ cited by Dickson I 463, item 66.  Examples with the  10  digits, like   7 * 9403  =  65821   and   3 * 1458  =  6 * 0729.

Dudeney.  AM.  1917. 

Prob. 80: The three groups, pp. 14 & 156.  Cites Thié and his example as being an extension of CP, prob. 101.  Asks for solutions to Thié's form and to   A*BCDE  =  FGHI,   with the  9  positive digits, e.g.   4 * 1738  =  6952.  Answer gives  7  solutions in the first case and  2  in the second case.

Prob. 81: The nine counters, pp. 14 & 156.  Find solution of   AB*CDE  =  FG*HI,   with the  9  positive digits, such that the product is maximal.  Answer:  32 * 174  =  58 * 96  =  5568. 

Prob. 82: The ten counters, pp. 15 & 156.  Divide the  10  digits into two equal products giving maximum and minimum products.  Answers:   2 * 3485  =  1 * 6970  =  6970   &   64 * 915  =  80 * 732  =  58560.  = Wood, 1927, probs. 59 & 60.

Prob. 85: The cab numbers, pp. 15 & 157.  Find two numbers, using all  9  positive digits, whose product contains all  9  positive digits and is maximal.  He believes the maximum is   96 * 8745231  =  839542176.

Prob. 86: Queer multiplication, pp. 15-16 & 157.  Examples of  A*BCDEFGHI  =  abcdefghi,  where both sides use the  9  positive digits:   3 * 51249876   and   9 * 16583742.   Asks for a solution with   A = 6.  Answer:  6 * 32547891.

Peano.  Giochi.  1924.  Prob. 34, p. 9.  Notes   2 * 78  =  39 * 4  =  156.  (Cf Dudeney, 1902.)

Wood.  Oddities.  1927.

Prob. 14: A problem in multiplication, p. 16.  A * BC  =  D * EF  =  GHI.  Gives four solutions:  2 * 78  =  4 * 39  =  156;   3 * 58  =  6 * 29  =  174  and their reversals (i.e.  4 * 39  =  2 * 78  =  156;   6 * 29  =  3 * 58  =  174) and implies there are no more.

Probs. 59 & 60: Number blocks  &  More number blocks, p. 47.  Same as Dudeney's AM prob. 82.

Prob. 62, pp. 47-48.  Same as Dudeney's CP prob. 101.

W. F. Cheney Jr, proposer;  Victor Thébault, solver.  Problem E13.  AMM 39 (1932) 606  &  41 (1934) 265‑266.  Two factor products using all the digits just once.  Gives all solutions without  0:  2 of form   A*BCED  =  FGHI  (confirming results of Buker in AMM 40 (1933) 559 ??NYS);  7  of form   AB*CDE  =  FGHI.  Gives some solutions with  0:  4  of form   A*BCDE  =  FGHIJ;  3  of form   AB*CDE  =  FGHIJ.

Perelman.  FFF.  1934.  Tricky multiplication.  1957: prob. 45, pp. 56 & 61;  1979: prob. 48, pp. 71 & 77.  = MCBF: prob. 48, pp. 69 & 74.  Gives all  9  solutions without  0  to Dudeney's AM prob. 80.

Victor Thébault, proposer;  G. H. Biucliu & L. Tits, solvers.  Mathesis 44 (1935) 205‑207.  ??NYS _ described in CM 9 (1983) 89.  All  94  solutions of   n*ABCDE  =  FGHIJ.

Jerome S. Meyer.  Arithmetricks.  Scholastic Book Services, NY, 1965.  No repeating digits, pp. 16-17.  Says   A*BCDEFGHI  =  abcdefghi   with  A = 9  has four solutions such that we also have   2*abcdefghi  =  abcdefghij.   9 * 81274365;   9 * 72645831;   9 * 58132764;   9 * 76125483  (where the last  3  is misprinted as  4).  However this doesn't include Dudeney's example in AM 86.

Charles L. Baker.  (Presumably in RMM.  ??NYS)  Reported in Madachy's Mathematical Recreations, op. cit. in 6.D, 1966, pp. 183‑185.  Confirms Thébault's (1934) and Perelman's results without  0  and presents all two‑factor products with  0:  13  of form   A*BCED  =  FGHIJ;  9  of form   AB*CDE  =  FGHIJ.

Charles W. Trigg, proposer;  Edward Moylan, solver;  David Daykin, commenter.  Problem 691 _ A product of integers.  MM 41:3 (1968) 158;  42:1 (Jan 1969) 44-45  &  42:2 (Mar 1969) 102-103.  Solve   a = 8b,   where  a  and  b  together use the  9  positive digits once each.  Must have the form   ABCDE  =  8*FGHI   and there are  46  solutions, all listed.  D. Sumner assumed that both  a  and  b  contained all nine positive digits and found a unique solution with  b = 123456789.  Daykin gives the number of solutions of the first problem in base  β,  with multiplier  m,  for  2 £ m < β £ 15,  and also considering the use of  0  as a digit and the use of just odd or just even digits.  The tables show surprising irregularity.  [Is this really surprising??]

Stewart Metchette.  A note on digital products.  JRM 10 (1977‑78) 270‑271.  Extends Thébault and Baker to three factor products and gives all of the following forms:  12  of form   A*BC*DE  =  FGHI;  10  of form   A*BC*DE  =  FGHIJ;  2  of form   A*BC*DEF  =  GHIJ.

Birtwistle.  Calculator Puzzle Book.  1978. 

Prob. 25, pp. 20 & 79.  Same as Dudeney's AM 82.

Prob. 87: Three by three _ part two, pp. 61 & 120-121.  See 7.AC.4 for the distribution of the nine positive digits as   ABC, DEF, GHI   with the lowest product, which is uniquely given by   147 x 258 x 369  =  13994694.

David Singmaster.  Work of 30 Jul 1998 in response to a letter of J. I. Collings.  There are 99 solutions of  AB*CDE = FG*HIJ,  with  A < F.  Of these,  35  have a leading zero.  Trailing zeroes lead to  13  pairs of related solutions, e.g.  23 * 760  =  76 * 230  =  95 * 184.  The largest value of the common product is  58560  =  64 * 915  =  80 * 732,  as given by Dudeney, AM, prob. 82.  The smallest common product is  3588  =  04 * 897  =  23 * 156,  while the smallest without a leading zero is  8262  =  18 * 459  =  27 * 306.  There are two cases with the same common product and further one of the two products is the same:  18 * 465  =  30 * 279  =  45 * 186.  Collings notes that there is only one solution with  E  or  J  being three and no leading zeroes

 

          7.AC.5.        PAN‑DIGITAL FRACTIONS

 

Pearson.  1907.  Part II: Juggling with the digits, pp. 40‑41.  Examples of   ABCD/EFGHI =  1/n   for  n = 3, 4, ..., 9.

Dudeney.  AM.  1917. 

Prob. 88: Digital division, pp. 16 & 158.  Gives  13458/6729 = 2.  Find solutions of   ABCDE/FGHI  =  n   for  n = 3, 4, ..., 9.  Also find the smallest solutions in each case _ e.g.  14658/7329  =  2  is a larger solution than the first example.

Prob. 90: The century puzzle, pp. 16 & 158-159.  Write a mixed number, using the  9  positive digits, equal to  100,  e.g.  91  5742/638.  Says Lucas found  7  ways, but he has shown that there are just  11  ways.  One of these has a single digit integer part _ find it.  Answer gives all  11  solutions. 

Prob. 91: More mixed fractions, pp. 16 & 159.  Says he has tried the same question with  100  replaced by other values and gives  12  values to try.  However, two of these are impossible.  He has found solutions for all values from  1  to  100,  except that  1, 2, 3, 4, 15, 18  are impossible, though  15  and  18  can easily be expressed if the integer part is permitted to be zero or if compound fractions are permitted, e.g.  3 (8952/746)/1.

Prob. 92: Digital square numbers, pp. 16 & 159.  Find largest and smallest squares using all  9  positive digits.  Answers:  923,187,456  =  30384²;  139,854,276  =  11826².

Peano.  Giochi.  1924.  Prob. 35, p. 10.  Gives  6  solutions of   9  =  ABCDE/FGHIJ,   three of which have  F = 0.

Adams.  Puzzle Book.  1939.  Prob. B.83: Figure juggling, part 2, pp. 78 & 107.  Asks for an example (of what??) for  n = 2  and remarks that solutions exist for  n = 3, ..., 9.

George S. Terry.  The Dozen System.  Longmans, Green & Co., NY, 1941.  ??NYS _ quoted in Underwood Dudley; Mathematical Cranks; MAA, 1992, p. 25.  Express unity as a sum of two fractions which contain all the digits once only, duodecimally.  E.g.   136/270 + 48χ/95ε  =  1   (χ = 10, ε = 11).  Says about five dozen.  Terry (or Dudley) says the decimal answer is about one dozen.

Ripley's Believe It or Not, 24th series.  Pocket Books, NY, 1975.  P. 76.  3/6  =  7/14  =  29/58  uses all nine positive digits.  [Are there other examples or other forms??]

Michael Holt.  Math Puzzles and Games.  Walker Publishing Co., NY, (1977), PB ed., 1983.  9  in ten digits, pp. 26 & 98.   ABCDE/FGHIJ  =  9   has six solutions, all given.

James W. Carroll.  Letter:  Computerizing Sam Loyd.  Games 7:5 (May 1983) 6.  ABCD/EFGHI  =  1/n   for  n = 2, 3, ..., 9  has  12, 2, 4, 12, 3, 7, 46, 3  solutions.

Nob Yoshigahara.  Puzzle problem used on his TV(?) program in Japan and communicated to me at 13th International Puzzle Party, August, 1993.  Use the nine positive digits to make   A/BC + D/EF + G/HI  =  1.  There is a unique solution:   5/34 + 7/68 + 9/12.

 

          7.AC.6.        OTHER PAN‑DIGITAL EXPRESSIONS

 

          See also 7.I and 7.I.1 for related problems.

 

The Family Friend (1856) 149 & 180.  Enigmas, Charades, &c.  87  Mathematical Puzzle.  "Take all the figures, (i.e., 1 2 3 4 5 6 7 8 9 0,) and place them in such a mode, that, when they are added up, they may be equal to 100."  Signed  S. W. S.  Answer is   76 + 3 10/5 + 8 + 9 4/2.

Magician's Own Book.  1857.  The united digits, p. 246.  "Arrange the figures  1  to  9  in such order that, by adding them together, they amount to  100."   15 + 36 + 47  =  98 + 2  =  100.   = Book of 500 Puzzles, 1859, p. 60.  = Boy's Own Conjuring Book, 1860, p. 216.

Leske.  Illustriertes Spielbuch für Mädchen.  1864? 

Prob. 564-16, pp. 253 & 395.  Combine  1, 2, ..., 9  to make  100.  Answer:  1 + 3/6 + 27/54 + 98.

Prob. 564-19, pp. 253 & 395.  Add the nine digits to make  100.  Answer: 75 9/18  +  24 3/6.

Boy's Own Book.  The united digits.  1868: 429.  "The figures  1  to  9  may be placed in such order that the whole added together make exactly  100.  Thus _  15 + 36 + 47  =  98 + 2  =  100."

Hanky Panky.  1872.  The century of cards, p. 294.   15 + 36 + 47  =  98 + 2  =  100   given with cards.

Berkeley & Rowland.  Card Tricks and Puzzles.  1892.  Card Puzzles No. 4: Century addition, p. 4.  Use  1, ..., 9  to add up to  100.  Solutions are:   74 + 12 + 3  =  89 + 6 + 5  =  100;   same as Book of 500 Puzzles;   19 + 28 + 6  =  53 + 47  =  100.

Anonymous problem proposal with solutions by  K. K.;  R. Ichikawa;  S. Tamano  and two papers by T. Hayashi.  J. of the Physics School in Tokyo 5 (1896) 82, 99-103, 153-156 & 266-267, ??NYS  Abstracted in:  Yoshio Mikami, ed.; Mathematical Papers from the Far East; AGM 28 (1910) 16-20; as: A queer number.  In base  b,  we have  [12...b * (b‑2)] + b-1  =  b...21.  [I wonder about other solutions of   a * x + b  =  y,   where  a, b  are digits and  x, y  are pandigital expressions (either with the  9  positive digits or all  10  digits, either separately or together) or  y  is the reversal of  x,  etc.]

T. Hayashi.  On the examination of perfect squares among numbers formed by the arrangements of the nine effective figures.  J. of the Physics School in Tokyo 5 (1896) 203-206, ??NYS.  Abstracted in:  Yoshio Mikami, ed.; Mathematical Papers from the Far East; AGM 28 (1910) 23-25.  Says Artemas Martin asked which squares contain all nine positive digits once each and that Biddle found  29  of these.  (Cites same J. 5 (1896) 171, ??NYS, for the solutions.)  [What about with all 10 digits?].

Ball.  MRE, 4th ed., 1905.  P. 14.  Use the  10  digits to total  1  _ a solution is  35/70 + 148/296  _ or to total  100  _ a solution is  50 + 49 + 1/2 + 38/76.  Use the  9  digits to make four numbers which total  100  _ a solution is  78 + 15 + ²Ö9 + ³Ö64.

Ball.  MRE, 5th ed., 1911.  Pp. 13-14.  Briefly restates the material in the 4th ed. as "questions which have been propounded in recent years.  ...   To the making of such questions of this kind there is no limit, but their solution involves little or no mathematical skill."

Adams.  Indoor Games.  1912.  A clever arrangement, p. 353.  Same as Boy's Own Book.

Ball.  MRE, 6th ed., 1914.  Pp. 13-14.  Restates the material in the 5th ed. as "... numerous empirical problems, ....  To the making of such questions there is no limit, but their solution involves little or no mathematical skill."

                    He then introduces the "Four Digits Problem".  "I suggest the following problem as being more interesting."  Using the digits  1, 2, ..., n,  express the integers from  1  up using four different digits and the operations of sum, product, positive integral power and base-10  (or also allowing iterated square roots and factorials).  With  n = 4,  he can get to  88  or to  264.  With  n = 5,  he can get to  231  or  790.  Using  0, 1, 2, 3,  he can get to  36  (or  40). 

Dudeney.  AM.  1917.  Prob. 13: A new money puzzle, pp. 2-3 & 148-149.  States largest amount of old English money expressible with all nine positive digits is  £98765  4s  3½d.  Asks for the smallest amount.  Answer:  £2567  18s  9¾d.

Ball.  MRE, 9th ed., 1920.  Pp. 13-14.  In the "Four Digits Problem", he considers  n = 4,  i.e. using  1, 2, 3, 4,  and discusses the operations in more detail.  Using sum, product, positive integral power and base-10  notation, he can get to  88.  Allowing also finitely iterated square roots and factorials, he can get to  264.  Allowing also negative integral indices, he can get to  276.  Allowing also fractional indices, he can get to  312.  He then mentions using  0, 1, 2, 3  or four of the five digits  1, ..., 5.

Ball.  MRE, 10th ed., 1922.  Pp. 13-14.  In the "Four Digits Problem", he repeats the material of the 9th ed., but at the end he adds that using all of the five digits,  1, ..., 5,  he has gotten to  3832  or  4282,  depending on whether negative and fractional indices are excluded or allowed.

Wood.  Oddities.  1927.

Prob. 53: Can you do this?, p. 44.  Consider  1, 2, 3, 4  and  5, 7, 8, 9.  Rearrange the sets so both have the same sum!  He phrases it in terms of numbers on jerseys of football players.  Cf Morris, 1991.

Prob. 55: A matter of multiplication, p. 45.  AB*CDE  consists of the same five digits and  A = 1.  Answer is:  14 * 926 = 12964.  He says the only solutions when the condition  A = 1  is dropped are:  24 * 651 = 15624;  42 * 678 = 28476;  51 * 246 = 12546;  57 * 834 = 47538;  65 * 281 = 18265;  65 * 983 = 63895;  72 * 936 = 67392;  75 * 231 = 17325;  78 * 624 = 48672;  86 * 251 = 21586;  87 * 435 = 37845.  Cf Ripley below.

A. B. Nordmann.  One Hundred More Parlour Tricks and Problems.  Wells, Gardner, Darton & Co., London, nd [1927 _ BMC].  No. 94: The "100" problem, pp. 88 & 114.  Use  1, 2, ..., 9  to make  100.  Answer:  15 + 37 + 46  =  98  + 2  =  100.

Perelman.  FFF.  1934.  1957: prob. 99 & 101, pp. 137 & 144;  1979: probs. 102 & 104, pp. 166-167 & 174-175.  = MCBF, probs. 102 & 104, pp. 167 & 177-178.

102: One.  "Write one by using all the ten digits."   148/296 + 35/70.   Also  (123456789)⁰,  etc.

104: Ten Digits.  "Write 100 using all the ten digits.  How many ways are there of doing it?  We know at least four."   70 + 24 9/18 + 5 3/6   and three other similar answers.

See: Meyers in 7.I.1 for the largest integer constructible with various sets of numbers.

J. R. Evans.  The Junior Week‑End Book.  Op. cit. in 6.AF.  1939.  Prob. 32, pp. 264 & 270.  Find largest and smallest amounts in pounds, shillings, pence and farthings using the nine positive digits.  Solutions as in Dudeney.

The Little Puzzle Book.  Op. cit. in 5.D.5.  1955.  P. 56: One hundred per cent.  Form  100  using all  10  digits, but not in order.  Gives just one solution.

Philip E. Bath.  Fun with Figures.  Op. cit. in 5.C.  1959.  No. 27: Another sum of money, pp. 14 & 48.  Least amount of English money using all  10  digits.  Answer:  £20567  18s  9¾d.

Robert Harbin.  Party Lines.  Op. cit. in 5.B.1.  1963.  1, 2, 3 ... 100,  p. 77.  Use each of the  10  figures to make a total of  100.  Answer:  57 + 23  =  80   + 1 + 4 + 6 + 9  =  100.

Jonathan Always.  Puzzles to Puzzle You.  Op. cit. in 5.K.2.  1965.  No. 91: Monetary matter, pp. 30 & 82.  Find smallest amount, as in Dudeney.

Ripley's Believe it or Not, 14th series.  Pocket Books, NY, 1968.  Unpaged _ about 85% of the way through.   AB * CDE   gives a product of five digits which is a permutation of ABCDE.  E.g.   14 * 926 = 12964.   Asserts there are just  12  "sets of figures in which the multiplicand and the multiplier reappear in the product."  Gives six examples.  Cf Wood, 1927.

Ripley's Believe it or Not, 15th series.  Pocket Books, NY, 1968.  Unpaged _ about 30% of the way through.  Gives  7  examples of the above situation, one of which was given above, saying  "The original figures reappear in the results ...."  Cf Wood, 1927.  They do not seem to have considered other forms, e.g. I find   3 * 51  =  153;   6 * 21  =  126;   8 * 86  =  688.

Shakuntala Devi.  Puzzles to Puzzle You.  Op. cit. in 5.D.1.  1976.  Prob. 20: Make a century, pp. 20 & 99-100.  Express  100  as a mixed fraction using the nine positive digits, e.g.  81  5643/297.   There are  10  of this form and one other:   3  69258/714.

Liz Allen.  Brain Sharpeners.  Op. cit. in 5.B.  1991.  Fodder for number crunchers, pp. 69 & 128.  Find the  10-digit numbers, using all  10  digits, which are divisible by  1, 2, ..., 18  [i.e. are multiples of  12252240].  There are four such and none is divisible by  19.  [I checked and found no smaller multiples of  12252240  with distinct digits.  I found that there are  94  multiples of  9!  (=  362880)  with distinct digits.  Of these, there are  2, 3, 16, 73  with  7, 8, 9, 10  digits.  (Any multiple of  10!  (=  3628800)  has its last two digits equal.)]

Scot Morris.  The Next Book of Omni Games.  Op. cit. in 7.E.  1991.  Pp. 55 & 192.  Form the digits into two sums:  2 + 6 + 7 + 9  and  1 + 3 + 4 + 5 + 8.  Make the sums equal by moving one number!  Cf Wood, 1927.

Ed Barbeau.  After Math.  Wall & Emerson, Toronto, 1995.  Four problems with whole numbers, pp. 127-130. 

                    1.  Find all integers such that the number and its square contain all nine positive digits just once.  Answers:  567  and  854.

                    3.  Find all integers such that its cube and its fourth power contain all ten digits just once.  Answer:  18.

 

          7.AC.7.        SELF-DESCRIPTIVE NUMBERS, PANGRAMS, ETC.

 

          New section.  Are there older examples?

 

Solomon W. Golomb.  Shift Register Sequences.  Holden-Day, 1967.  ??NYS _ cited by a 1996 article, but I cannot locate the material in the revised ed., Aegean Park Press, Laguna Hills, California, 1982; perhaps it is in some other work of Golomb ??check.  Find a non-decreasing sequence of positive integers,  (a_i),  i = 1, 2, ...,  such that  i  appears  a_i  times.  Unique answer is:  1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, ....

Douglas Hofstadter.  SA (Jan 1982)  c= Metamagical Themas, Basic Books, NY, 1985, chap. 2, pp. 25-48.  In Jan 1981, he had discussed self-referential sentences, and readers sent in a number of numerical ones. 

          Jonathan Post:

                    This sentence contains ten words, eighteen syllables and sixty four letters. 

          John Atkins:

                    'Has eighteen letters' does. 

          Howard Bergerson:

                    In this sentence, the word and occurs twice, the word eight occurs twice, the word four occurs twice, the word fourteen occurs four times, the word in occurs twice, the word seven occurs twice, the word the occurs fourteen times, the word this occurs twice, the word times occurs seven times, the word twice occurs eight times and the word word occurs fourteen times.

          Lee Sallows:

                    Only the fool would take trouble to verify that his sentence was composed of ten a's, three b's, four c's, four d's, forty-six e's, sixteen f's, four g's, thirteen h's, fifteen i's, two k's, nine l's, four m's, twenty-five n's, twenty-four o's, five p's, sixteen r's, forty-one s's, thirty seven t's, ten u's, eight v's, eight w's, four x's, eleven y's, twenty-seven commas, twenty-three apostrophes, seven hyphens, and last, but not least, a single !

          Raphael Robinson asks to fill in the blanks in the following and says there are two

                              solutions:

                    In this sentence, the number of occurrences of  0  is  __,  of  1  is  __,  of  2  is  __,  of  3  is  __,  of  4  is  __,  of  5  is  __,  of  6  is  __,  of  7  is  __,  of  8  is  __,  and of  9  is  __.

          The supplemental material in the book includes

          J. K. Aronson:

                    'T' is the first, fourth, eleventh, sixteenth, twenty-fourth, twenty-ninth, thirty-third, ....

          See below for further material.

Douglas Hofstadter.  Metamagical Themas, Basic Books, NY, 1985, chap. 3, pp. 49-69.  In the Post Scriptum, pp. 68-69, he reports on Sallows' search for a 'pangram'.  Apparently the first of the type he wants is in Dutch, by Rudy Kousbroek and Sarah Hart:  Dit pangram bevat vijf a's, twee b's, ....  After some months search, Sallows' computer found: 

                    This pangram tallies five a's, one b, one c, two d's, twenty-eight e's, eight f's, six g's, eight h's, thirteen i's, one j, one k, three l's, two m's, eighteen n's, fifteen o's, two p's, one q, seven r's, twenty-five s's, twenty-two t's, four u's, four v's, nine w's, two x's, four y's, and one z. 

          He challenges people to compute a version starting: 

                    This computer-generated pangram contains ....

Douglas Hofstadter.  Metamagical Themas, Basic Books, NY, 1985, chap. 16 pp. 364-395.  In the Post Scriptum, pp. 387-395, he continues his discussion of the above material.  He notes that Robinson's problem is convergent in the sense that if one inserts a random sequence of numbers, then counts the occurrences of the numbers and uses the counts as a new number, etc., then this iterative process usually converges to a solution.  There are two solutions, but there is also a two term cycle and Hofstadter conjectures all initial values converge to one of these three situations.  Sallows' challenge was given in A. K. Dewdeney's Computer Recreations column (SA, Oct 1984) and Larry Tesler used an iterative program on it.  Tesler soon found a loop and modified the program a bit to obtain a solution: 

                    This computer-generated pangram contains six a's, one b, three c, three d's, thirty-seven e's, six f's, three g's, nine h's, twelve i's, one j, one k, two l's, three m's, twenty-two n's, thirteen o's, three p's, one q, forteen r's, twenty-nine s's, twenty-four t's, five u's, six v's, seven w's, four x's, five y's, and one z. 

Lee Sallows.  In Quest of a Pangram.  Published by the author, Holland, nd [1985?].  Describes his search for a pangram.

Lee Sallows.  Reflexicons.  Word Ways 25 (1992) 131-141.  A 'reflexicon' is a list of numbers and letters which specifies the number of times the letter occurs in the list.  There are two in English. 

                    fifteen e's,  seven f's,  four g's,  six h's,  eight i's,  four n's,  five o's,  six r's,  eighteen s's,  eight t's,  four u's,  three v's,  two w's,  three x's.

                    sixteen e's,  five f's,  three g's,  six h's,  nine i's,  five n's,  four o's,  six r's,  eighteen s's,  eight t's,  three u's,  three v's,  two w's,  four x's.

                    He also discusses 'pangrams', which are sentences containing the above kind of information _ e.g.  This sentence contains one hundred and ninety-seven letters:  four a's, ....  The search for these is described in his booklet cited above.  He then discusses crosswords using the number names.

Tony Gardiner.  Challenge! What is the title of this article?  Mathematics Review 4:4 (Apr 1994) 28-29.  Following on a previous article in 4:1, he discusses self-describing sequences, where the description arises by reading the sequence.  E.g.  22  is read as  'two twos';  31 12 33 15  is  'three ones, one two, three threes, one five'.  He also mentions self-describing lists, e.g.  1210   contains  'one 0, two 1s, one 2, zero 3s'.

Ed Barbeau.  After Math.  Wall & Emerson, Toronto, 1995.  Pp. 117-122.  Considers self-describing lists of length  n  and shows there are only the following:  1210,  2020,  21200  and, for  n > 6,  (n-4)2100...001000

Lee Sallows.  Problem proposal to Puzzle Panel, 18 Jun 1998.  "How many letters would this question contain, if the answer wasn't already seventy three?"

 

          7.AD. SELLING, BUYING AND SELLING SAME ITEM

 

          The problem is to determine the profit in a series of transactions involving the same item, but there is usually insufficient information.  But see Clark and Sullivan for an unusual answer.

 

Clark.  Mental Nuts.  1904: no. 2;  1916, no. 24.  The horse question.  Sell a horse for $90,  buy back at $80,  resell at $100.  "What did he make on the transaction?"  Answer is $20,  but this assumes the horse had no initial cost.  If the item has no initial cost and the prices are  a, b, c,  then the gain is  a - b + c.  But if the question is asking for the profit, then the data are insufficient as the base cost is not given.

Loyd.  The trader's profit.  Cyclopedia, 1914, pp. 291 (no solution).  (= MPSL1, prob. 13 _ What was the profit?, pp. 12 & 125.)  Sells a bicycle at  50,  buys back at  40,  sells again at  45.  Lengthy discussion of various 'solutions' of  15, 5 and 10.  He says "the President of the New York Stock Exchange was bold enough to maintain over his own signature that the profit should be  $10."  Gardner points out that there isn't enough information.

H. E. Licks.  Op. cit. in 5.A.  1917.  Art. 25, p. 18.  Sells at  50,  buys back at  45,  sells at  60  _ what is the profit?  Author labels it impossible.

Smith.  Number Stories.  1919.  Pp. 126‑127 & 146.  Buys at  5000,  sells at  5000,  buys back at  4500,  sells again at 5500.

Loyd Jr.  SLAHP.  1928.  The used‑car puzzle, pp. 9‑10 & 88.  Sell a used car at  100,  buy back at  80,  resell at  90.  "This popped into my head one morning ...."  Gives arguments for profits of  30, 10 and 20.  Solution says the information is insufficient.

Collins.  Fun with Figures.  1928.  This sticks 'em up, p. 69.  Buys at  $55,  sells at  $55,  buys back at  $50,  sells again at  $60.

Rudin.  1936.  No. 160, pp. 57 & 113.  Buy for  $70,  sell for  $80,  buy back for  $90,  sell for  $100.

Sullivan.  Unusual.  1943.  Prob. 2: A business transaction.  Sell for  $4000,  buy back for  $3500,  sell again for  $4500.  Says the gain is  $5000,  composed of the initial  $4000  plus the gain on the buying and selling.  This is like Clark and markedly different than most approaches, which refer to profit.

Hubert Phillips.  Something to Think About.  Ptarmigan (Penguin), 1945.  Problem 12: Alf's bike, pp. 15 & 89.  More complex version with four persons and each person's percentage profit or loss given.

The Little Puzzle Book.  Op. cit. in 5.D.5.  1955.  P. 40: Multiple choice (B).  Sell a cow for  $100,  buy back at  $90,  resell at  $120.  Says the profit is  $30.

 

          7.AD.1.        PAWNING MONEY

 

Viscount Simon.  [Memory of Lewis Carroll.]  Loc. cit. in 7.S.2.  He says Carroll gave the following.  Man pawns  12d  for  9d  and sells the ticket to a friend for  9d.  Who loses and how much?  Simon says that when he said that the friend lost  6d,  Carroll pointed out that pawnbrokers charge interest.

Don Lemon.  Everybody's Pocket Cyclopedia.  Revised 8th ed., 1890.  Op. cit. in 5.A.  P. 155 gives current pawnbrokers' charges.  These would be  ½d  for the ticket and  ½d  per  2s  or part thereof per month or part thereof.  So the above will cost at least  ¾d.

Smith.  Number Stories.  1919.  Pp. 131 & 148.  Pawn  $1  for  $.75  and sell ticket for  $.50.  Doesn't consider interest.

R. Ripley.  Believe It Or Not!  Book 2.  Op. cit. in 7.J.  1931.  P. 143.  "A man owed  $3.00.  He had a  $2.00  bill, which he pawned for  $1.50,  and then sold the pawn ticket to another man for  $1.50,  who redeemed the  $2.00  bill.  Who lost?"  No answer given.

H. Phillips.  The Playtime Omnibus.  Op. cit. in 6.AF.  1933.  Section XVII, prob. 7, pp. 54 & 236.  Jones pawns  6d  for  5d,  then sells the ticket to Brown for  4d.  Who lost?  Doesn't consider interest.

John Paul Adams.  We Dare You to Solve This!.  Op cit. in 5.C.  1957?  Prob. 52: Losers weepers, pp. 32 & 49.  Pawn  $5  for  $3,  sell ticket to a friend for  $3  and he redeems it.  Who lost and how much?  Doesn't consider interest.

 

          7.AE.  USE OF COUNTERFEIT BILL OR FORGED CHEQUE

 

          These problems give one or several transactions involving a bill or cheque which is then found to be counterfeit or forged.  Who loses and how much?  In the classic version of the hatter, the straightforward answer is that he loses the value of the counterfeit bill.  However, there are two values for the hat _ the sale price and the cost price, whose difference is the profit that would be made in a normal sale.  One can argue that one has only lost the cost price of the hat, so the loss is the value of the couterfeit bill less the expected profit.

 

Magician's Own Book.  1857.  The unlucky hatter, p. 245.  Man buys  $8  hat with counterfeit  $50  bill.  "... and in almost every case the first impression is, that the hatter lost  $50  besides the hat, though it is evident he was paid for the hat...."  = Book of 500 Puzzles, 1859, p. 59.  = Boy's Own Conjuring Book, 1860, pp. 215‑216, but this spells out  $  as dollars.. 

Lemon.  1890.  The unlucky hatter, no. 225, pp. 34 & 106.  Man pays for  $8  hat with counterfeit  $50  bill.  "In almost every case the first impression ... is that the hatter lost  $50  beside [sic] the hat..." 

Hoffmann.  1893.  Chap. IV, no. 43: What did he lose?, pp. 153‑154 & 205‑206.  Man pays for hat with a counterfeit bill.  How much does the seller lose?  Answer says that "The reply of most people is, almost invariably, that the hatter lost [the change] and the value of the hat, but a little consideration will show that this is incorrect."  He then says that the seller loses the amount given in change less his profit on the goods sold;  "the nett value of the hat, plus such trade profit, being balanced by the difference ... which he retained out of the proceeds of the note."  This is wrong, as the sale price is part of the the refund that he has to make to the person who changed the note.

Clark.  Mental Nuts.  1904: no. 21;  1916: no. 7.  The shoe question.  Boy pays for  $4  pair of shoes with counterfeit  $10  bill.  Answer says he lost  $6  and the pair of shoes.

Dudeney.  Some much‑discussed puzzles.  Op. cit. in 2.  1908.  Dud cheque used to buy goods and get cash.  "Perpetually cropping up in various guises."

H. E. Licks.  Op. cit. in 5.A.  1917.  Art. 21, p. 17.  Use of a counterfeit bill.

Lynn Rohrbough, ed.  Mental Games.  Handy Series, Kit E, Cooperative Recreation Service, Delaware, Ohio, 1927.  Counterfeit Bill, p. 10.  Man buys $6 pair of shoes with a phoney $20 bill.  How much did seller lose?  No solution given.

Ahrens.  A&N, 1918, pp. 95‑96 gives such a problem.

Dudeney.  PCP.  1932.  Prob. 34: The banker and the note, pp. 21 & 131.  = 536; prob. 31: The banker and the counterfeit bill, pp. 10‑11 & 230.  Counterfeit bill goes in a circle so no one loses.

Phillips.  Week‑End.  1932.  Time tests of intelligence, no. 15, pp. 14‑15 & 188.  Forged cheque used to settle account and receive cash.

Robert A. Streeter & Robert G. Hoehn.  Are You a Genius?  Vol. 1, 1932; vol. 2, 1933, Frederick A. Stokes Co., NY.  Combined ed., Blue Ribbon Books, NY, 1936.  Vol. 1, p. 46, no. 10: "Brain twister".  Man owes me  40¢,  he gives me a knife worth at least  60¢  and I give him  20¢.  I then find the knife was stolen and I pay the owner  75¢,  which is its value.  How much have I lost?  Answer is  60¢.  This is based on my payment of  75¢  being a fair purchase, so my loses are the  40¢  debt and the  20¢  change, which are now irrecoverable.  However, if the first transaction is considered fair, then I've lost  75¢.  Further, I might consider the original debt as a past loss, which would reduce the present loss to  20¢  or  35¢.

Phillips.  The Playtime Omnibus.  Op. cit. in 6.AF.  1933.  Section XVII, prob. 5: Pinchem, pp. 53 & 236.  Identical to Week‑End.

Dr. Th. Wolff.  Die lächelnde Sphinx.  Academia Verlagsbuchhandlung, Prague, 1937.  Prob. 3, pp. 187-188 & 196-197.  Sell bracelet worth  60  for phoney  100  bill, which is changed by the neighbouring shopkeeper.  Author says he gets many answers, including  140,  200  and even  340,  but the right answer is  100.

Depew.  Cokesbury Game Book.  1939.  Shoe dealer, p. 211.  Sell shoes worth  $8  for phoney  $20  bill.  Loss is  $12  plus value of shoes.

McKay.  Party Night.  1940.  No. 7, p. 177.  Man buys boots worth  15s  with a bad  £1  note.  Says he gets answers up to "35s  and a  15s  pair of boots".  Notes that the neighbouring grocer who changed the bill is irrelevant and the bootseller is simply  £1  out of pocket.  This ignores his profit on the boots. 

Meyer.  Big Fun Book.  1940.  No. 10, pp. 171 & 754.  Same as Streeter & Hoehn.

Shakuntala Devi.  Puzzles to Puzzle You.  Op. cit. in 5.D.1.  1976.  Prob. 13: The counterfeit note, pp. 17 & 98.  Counterfeit note goes in a circle.  She claims all the transactions are invalid _ but the bill has simply allowed a circuit of debts to be cancelled and hence no one has lost and there is no reason to cancel anything.

 

          7.AF.  ARITHMETIC PROGRESSIONS

 

          See Tropfke 625.

          There are many problems which are based on this.  Some occur in 7.H.7 and 10.A.  Here I only include the most interesting.

 

Bakhshali MS.  c7C.  In:  G. R. Kaye, The Bakhsh_li manuscript;  J. Asiatic Soc. Bengal (2) 8:9 (Sep 1912) 349‑361; p. 358  and in  Kaye I 43 & III 176-177, ff. 4r-5r, sutra 18  and in  Gupta.  Consider two APs  a, a + b, ...,  and  c, c + d, ...,  and suppose the sums after  n  terms are equal to  S.  In the notation of 10.A, this is O-(a, b; c, d).  Then  n = 2(a ‑ c)/(d - b) + 1.  Does examples with  (a, b; c, d)  =  (4, 3; 6, 1);  (2, 3; 3, 2)  and  (5, 6; 10, 3).

                    Kaye III 174, f. 4v  &  Gupta.  This is a problem of the same type, but most of it is lost and the scribe seems confused.  Gupta attempts to explain the confusion as due to using the data   a, b; c, d  =  3, 4; 1, 2,   with the rule   n  =  2(c‑a)/(b-d) + 1,   where the scribe takes the absolute values of the differences rather than their signed values.  In this way he gets  n = 3  rather than  n = -1.

Pacioli.  Summa.  1494.  F. 44v, prob. 32.  1 + 2 + ... + 10½.  He gets  10½ x 11½ / 2.

Unger.  Arithmetische Unterhaltungen.  1838.  Pp. 182 & 263, no. 693.  Man digging a well  49  feet deep.  First foot costs  15,  but each successive foot costs  6  more than the previous.  Find cost of last foot and total cost.  So this is really an arithmetic progression problem, but I haven't seen others of those using this context.

M. Ph. André.  Éléments d'Arithmétique (N^o 3) a l'usage de toutes les institutions ....  3rd ed., Librairie Classique de F.-E. André-Guédon, Paris, 1876.  Prob. 550, p. 239.  How many edges and diagonals does a convex octagon have? 

(Beeton's) Boy's Own Magazine 3:6 (Jun 1889) 255  &  3:8 (Aug 1889) 351.  (This is undoubtedly reprinted from Boy's Own magazine 1 (1863).)  Mathematical question 59.  Seller of  12  acres asks  1  farthing for the first acre,  4  for the second acre,  16 for the third acre, ....  Buyer offers  £100  for the first acre,  £150 for the second acre,  £200  for the third acre, ....  What is the difference in the prices asked and offered?  Also entered in 7.L.

Perelman.  1937.  MCBF, A team of diggers, prob. 195, pp. 372-373.  A team can dig a ditch in  24  hours, but just one digger begins and then the others join in at equal intervals, with the work finished in one interval after the last man joined.  The first man works  11  times as long as the last man.  How long did the last man work?  Perelman finds this noteworthy (and I agree) because the number of men in the team cannot be determined!

 

          7.AF.1.         COLLECTING STONES

 

Alcuin.  9C.  Prob. 42: Propositio de scala habente gradus centum.  Computes  1 + 2 + ... + 100   as  100 + (1+99) + (2+98) + ... + 50.

Pacioli.  Summa.  1494.  F. 44v, prob. 31.  Collect  100  oranges.

Pacioli.  De Viribus.  c1500.  Prob. 73: Levare  100  saxa a filo.  Wager on the number of steps to pick up  100  sacks (or apples or nuts).

Cardan.  Practica Arithmetice.  1539.  Chap. 66, section 55, f. EE.iiii.r (pp. 150-151).  (The 55 is not printed in the Opera Omnia.)  Picking up  100  stones in a line.  (H&S 56-57 gives Latin with English summary.)

Buteo.  Logistica.  1559.  Prob. 87, pp. 299-300.  Ant collecting  100  grains.  (H&S 56.)

H&S 56 says this occurs in Trenchant (1566), ??NYS.

Baker.  Well Spring of Sciences.  1562?  ??check if this is in the Graves copy of the 1562/1568 ed.  Prob. 2,  1580?: f. 36r;  1646: p. 56;  1670: pp. 72-73.  100  stones.

van Etten.  1624.  Prob. 87 (84), part IV (8), p. 114 (184).  100  apples, eggs or stones.  Henrion's Notte, p. 38, observes that there are many arithmetical errors in prob. 87 which the reader can easily correct.

Ozanam.  1694.  Prob. 7, question 6, 1696: 53;  1708: 29.  Prob. 10, question 6, 1725: 65‑66.  Prob. 1, 1778: 64-65;  1803: 66-67;  1814: 59-60;  1840: 32.  100  apples, becoming stones in 1778 et seq.  1778 describes a bet based on this process versus a straight run of the same distance.

Dilworth.  Schoolmaster's Assistant.  1743.  P. 94, no. 6.  100  stones  2  yards apart.  Answer in miles, furlongs and yards.

Walkingame.  Tutor's Assistant.  1751.  Arithmetical Progression, prob. 3, 1777: p. 90;  1835: p. 98;  1860: p. 118.  100  eggs a yard apart.  Answer in miles and yards.

Edmund Wingate (1596-1656).  A Plain and Familiar Method for Attaining the Knowledge and Practice of Common Arithmetic.  ....  19th ed., previous ed. by John Kersey (1616-1677) and George Shell(e)y, now by James Dodson.  C. Hitch and L. Hawes, et al., 1760.  Quest. 44, p. 366.  100  stones a yard apart.

Euler.  Algebra.  1770.  I.III.IV: Questions for practice, no. 4, p. 139.  100  stones a yard apart.  Answer:  5 miles, 1300 yards.

Vyse.  Tutor's Guide.  1771? 

Prob. 3, p. 141 & Key p. 186.  94 eggs.

Prob. 4, p. 141 & Key p. 186.  100 stones.

Prob. 21, p. 146 & Key pp. 188-190.  1000 stones,  2 yards apart, gathered by ten men, each man to collect ten and then the next man to collect the next ten, etc.  Can they do it in 24 hours?  How far did each man run?

Charles Hutton.  A Complete Treatise on Practical Arithmetic and Book-keeping.  Op. cit. in 7.G.2.  [c1780?] 

1804: prob. 4, p. 125.  100  stones a yard apart.  Converts answer to miles.

1804: prob. 74, p. 140.  100  eggs a yard apart.  Again converts to miles.

Bonnycastle.  Algebra.  1782.  P. 60, no. 7 (1815: p. 76, no. 8).  100  stones, a yard apart.

Pike.  Arithmetic.  1788.  P. 221, no. 3.  Stones laid a yard apart over a mile, starting a yard from the basket.  Finds the travel is  1761  miles.

Bullen.  Op. cit. in 7.G.1.  1789.  Chap. 30, prob. 6, pp. 213‑214.  100  stones, at  2  yard intervals.  Converts to miles, furlongs and yards.

Eadon.  Repository.  1794. 

P. 235, ex. 4.  Collect  500  stones a yard apart.  This takes  142  miles and some, so he could sooner run from Sheffield to York and back, since they are  50  miles apart.

P. 374, no. 19.  1760  stones a yard apart.  In six days a man only manages to collect  769  of them.  How far has he gone and how much farther has he to go?

John King, ed.  John King  1795  Arithmetical Book.  Published by the editor, who is the great-great-grandson of the 1795 writer, Twickenham, 1995.  P. 100.  100  eggs a yard apart.  He also gives a variation:  100  sheep are priced in arithmetic progression, with the first costing  1s  and the last costing  £9 19s (= 199s);  what do the sheep cost all together?

Hutton.  A Course of Mathematics.  1798?  Prob. 7,  1833: 277;  1857: 313.  100  stones  2  yards apart.  Gives answer in miles and yards.

Manuel des Sorciers.  1825.  Pp. 83-84.  ??NX  120 stones 6 feet apart.

Endless Amusement II.  1826?  Pp. 115-116:  "If a hundred Stones ...."

Boy's Own Book.  The basket and stones.  1828: 176;  1828-2: 239;  1829 (US): 107;  1855: 394;  1868: 432.  100  stones a yard apart.

Louis Pierre Marie Bourdon (1779-1854).  Élémens d'Algèbre.  7th ed., Bachelier, Paris, 1834.  (The 6th ed., was 1831, but I haven't yet looked for the early editions _ ??)  Art. 190, question 5, p. 319.  Loads of sand (or grit) have to be delivered, one load at a time, to  100  vehicles in a line,  6 meters apart, from a pile  40 meters from the end of the line.

Hutton-Rutherford.  A Course of Mathematics.  1841?  Prob. 24,  1857: 82.  100 eggs a yard apart.

Nuts to Crack XIV (1845), no. 76.  The basket and stones.  Almost identical to Boy's Own Book.

Boy's Treasury.  1844.  Amusements in arithmetic: The basket and stones, pp. 299-300.

Walter Taylor.  The Indian Juvenile Arithmetic ....  Op. cit. in 5.B.  1849.  P. 197: "The following is a favorite old problem."  100  stones, a yard apart.

John Radford Young (1799-1885).  Simple Arithmetic, Algebra, and the Elements of Euclid.  IN: The Mathematical Sciences, Orr's Circle of the Sciences series, Orr & Co., London, 1854.  [An apparently earlier version is described in 7.H.]  No. 2, p. 228.  200  stones.

Magician's Own Book.  1857.  The basket and stones, pp. 246-247.  100  stones, one yard apart.  = Book of 500 Puzzles, 1859, pp. 60-61.  = Boy's Own Conjuring Book, 1860, p. 218.  = Indoor & Outdoor, c1859, part II, prob. 20, pp. 136-137.

Illustrated Boy's Own Treasury.  1860.  Prob. 30, pp. 429-430 & 434.  100  trees to be watered, five steps apart.

Leske.  Illustriertes Spielbuch für Mädchen.  1864?  Prob. 564-32, pp. 255 & 396: Am Feste der bemalten Eier.  100  eggs, a fathom apart.  Says that the 'egg gathering' is a traditional race.

M. Ph. André.  Éléments d'Arithmétique (N^o 3) a l'usage de toutes les institutions ....  3rd ed., Librairie Classique de F.-E. André-Guédon, Paris, 1876.  Probs. 551 & 552, p. 239, are similar to Bourdon.

W. W. Rouse Ball.  Elementary Algebra.  CUP, (1890), 2nd ed., 1897.  Prob. 38, pp. 347 & 480.  10  balls equally spaced in a row, starting  12 ft  from the basket.  A boy picks them up in the usual way and find he has travelled  ¼ mile.  What was the spacing?

Lucas.  L'Arithmétique Amusante.  1895.  Prob. XXX: La course des œufs, pp. 110-111.  100  apples.  Says it is played on the beaches with eggs in good weather.

A. Sonnenschein & H. A. Nesbit.  The New Science and Art of Arithmetic For the Use of Schools.  A. & C. Black, London, 1903.  Pp. 342 & 489.  Potato race with  50  potatoes a yard apart to be retrieved.

Pearson.  1907.  Part II, no. 143: The stone carrier, pp. 142 & 219.  52  stones, with spacing  1, 3, 5, ..., 103,  to be brought to the first stone, yielding   2 (1 + 3 + 5 + ... + 103)  =  2 * 52².

Wehman.  New Book of 200 Puzzles.  1908.  An egg problem, p. 53.  100 eggs, a yard apart.

Joseph & Lenore Scott.  Master Mind Brain Teasers.  Op. cit. in 5.E.  1973.  More mileage possible, pp. 187-188.  Consider  10  points equally spaced on a line.  Starting at one of them and visiting each of them once, what is the maximum mileage possible?  Though not the same as the above problems, this uses the same set-up, but one does not return to base after each visit.

 

          7.AF.2.         CLOCK STRIKING

 

Tagliente.  Libro de Abaco.  (1515).  1541.  Prob. 113, f. 56v.

Rudolff.  Künstliche rechnung, 1526, op. cit. in 7.L.2.b.  1540 ed., f. N.vii.v.  ??NYS.  (H&S 57 gives the German.)

Apianus.  Kauffmanss Rechnung.  1527.  How many times does a clock strike during  1  to  12?

H&S 57 says this is in Buteo (1559), ??NYS.

Wingate.  Arithmetic.  1629?  Cf Wingate/Kersey.  This item is in a section which Kersey did not revise.  P. 296 in the 1678? ed.  "How many strokes the Clock strikes betwixt midnight and noon."

Baker.  Well Spring of Sciences.  Prob. 1, 1670: p. 71, ??NX.  1 + ... + 12  strokes.

A Manual of Curious and Useful Questions.  MS of 30 Jan 1743(OS) owned by Susan Cunnington (??NYS) and described in her:  The Story of Arithmetic; Swan Sonnenschein, London, 1904, pp. 155-157.  "How many strokes do y^e clocks of Venice (which go on to  24  o' th' clock) strike in the compass of a natural day?"  [Where is this MS??]

Dilworth.  Schoolmaster's Assistant.  1743.  P. 93, no. 1.  "How many strokes does the hammer of a clock strike in  12  hours?"

Walkingame.  Tutor's Assistant.  1751.  Arithmetical Progression, prob. 1, 1777: p. 90;  1835: p. 98;  1860: p. 118.   How many strokes in  12  hours?

Euler.  Algebra.  1770.  I.III.IV: Questions for practice, no. 3, p. 139.  "The clocks of Italy go on to  24  hours:  how many strokes do they strike in a complete revolution of the index?"

Vyse.  Tutor's Guide.  1771? 

Prob. 1, p. 141 & Key p. 186.  "How many strokes do the Clocks at Venice (which go on to  24  o'Clock) strike in the Compass of a natural Day?"

Page 2, p. 141 & Key p. 186.  "How many Strokes does the Hammer of a Clock strike in  12  Hours?"

Pike.  Arithmetic.  1788.  P. 221, no. 2.  "It is required to find out how many strokes the hammer of a clock would strike in a week, or  168  hours, provided it increased at each hour?"

Bonnycastle.  Algebra.  1782.  P. 60, no. 5 (1815: p. 75, no. 5).  "How many strokes do the clocks in Venice, which go on to  24  o'clock, strike in the compass of a day?"  (1815 omits "the compass of".)

Hutton.  A Course of Mathematics.  1798? 

Prob. I-2,  1833 & 1857: 66.  "It is required to find the number of all the strokes a clock strikes in one whole revolution of the index, or in 12 hours?"

Prob. I-3,  1833 & 1857: 67.  "How many strokes do the clocks of Venice strike in the compass of the day, which go right on from 1 to 24 o'clock?"

Prob. 5,  1833: 277;  1857: 313.  "How many strokes do the clocks of Venice, which go on to  24  o'clock, strike in the compass of a day?"  See Bonnycastle, 1782.

Hutton-Rutherford.  A Course of Mathematics.  1841?

          Prob. 25,  1857: 82.  "The clocks of Italy go on to 24 hours; then how many strokes do they strike in one complete revolution of the index?"

Walter Taylor.  The Indian Juvenile Arithmetic ....  Op. cit. in 5.B.  1849.  P. 196.  "How many strokes does a common clock strike in the compas of 12 hours?"

[Chambers].  Arithmetic.  Op. cit. in 7.H.  1866?  P. 224, quest. 5.  "How many times does a common clock strike in a day?"  Answer:  156.

James Cornwell & Joshua G. Fitch.  The Science of Arithmetic: ....  11th ed., Simpkin, Marshall, & Co., London, et al., 1867.  (The 1888 ed. is almost identical to this, so I suspect they are close to identical to the 2nd ed. of 1856.)  Exercises CXXXVIII, no. 9, pp. 291 & 370.  "How many times does the hammer of a clock strike in a week?"

Lucas.  L'Arithmétique Amusante.  1895.  Prob. XXXI: Les quatre cents coups, p. 111.  A clock that goes to  12,  strikes  78 + 78 = 156  hours in a day.  If it also strikes quarters, it makes  240  of those, totalling  396,  nearly  400  blows per day.

 

          7.AG. 2592

 

Dudeney.  AM.  1917.  Prob. 115: A printer's error, pp. 20 & 162.   ABCA  =  A^BC^A   has the unique solution  2592.

Hubert Phillips.  Question Time.  Op. cit. in 5.U.  1937.  Prob. 133: Phoney 'phone, pp. 87 & 219.   ABCD  =  A^BC^D.   Answer asserts that  2592  is the unique solution.

Donald L. Vanderpool.  Printer's "errors".  RMM 10 (Aug 1962) 38.  Extends Dudeney's examples, e.g.   2⁵ * 25/31  =  25  25/31;   3⁴ * 425  =  34425   (which can be multiplied by any power of  10).

M. H. Greenblatt.  Mathematical Entertainments, op. cit. in 6.U.2, 1968.  Expression for which  2,592  is a solution, pp. 15‑16.  He asserts that Dudeney's expression was discovered by his officemate, Larry, in response to a colleague, B. Rothlein, complaining that he could not remember his telephone number,  EVergreen 2592.  Rothlein had this telephone on 41st St., Philadelphia, during 1943‑47.  [A likely story??]

 

          7.AH. MULTIPLYING BY REVERSING

 

          For two-digit numbers, reversing and shifting are the same, but I will consider them here as the transformation  ab  to  ba  seems more like a reversal than a shift.

          I have just noted the items by Langford in 7.AR and the item by Meyer below which make me realise that this section is connected to 7.AR.  Meyer doubles the result of  1089  which comes up in 7.AR and gets  2178  which I remembered occurs in this section.  Upon investigating, we see Langford notes that   1089  =  1100 - 11   so that  k * 1089  =  kk00 ‑ kk  =  k,k‑1,9‑k,10-k   in base  10.  From this we see that  k*1089  is the reverse of  (10-k)*1089.  Now  10-k  is a multiple of  k  for  k = 1, 2,  but we get some new types of solution for  k = 3, 4,  namely:   7 * 3267  =  3 * 7623;   6 * 4356  =  4 * 6534.

          This pattern leads to all solutions in 7.AR and also leads to further solutions here, using   11000 ‑ 11 = 10989,  etc., but this does not seem to give a complete solution here.

 

Charles Babbage.  The Philosophy of Analysis _ unpublished collection of MSS in the BM as Add. MS 37202, c1820.  ??NX.  See 4.B.1 for more details.  F. 4 is "Analysis of the Essay of Games".  F. 4.r has entry 12:  "Given product of a number consisting of  n  figures mult. by same [some??] figures in an inverted order  query num"  Though a bit cryptic, this seems to refer to the problem of this section.

Unger.  Arithmetische Unterhaltungen.  1838.  Pp. 136 & 258, no. 525.  ab  =  2_ * ba.  He adds a further condition, but this is not needed.

Sphinx.  1895.  Arithmetical, no. 217, pp. 33 & 104.   x²  =  4*(x reversed).

T. C. Lewis.  L'Intermédiaire des Mathematiciens 18 (1911) 128.  ??NYS. Dickson, Vol. 1, p. 463, item 70 says he discusses "number divisible by the same number reversed".

Ball.  MRE, 6th ed., 1914, p. 12.  Mentions the general problem and gives   8712  =  4*2178  and   9801  =  9*1089   as examples.  Gives four citations to L'Intermédiaire des Maths., none of them the same as the above!

R. Burg.  Sitzungsber. Berlin Math. Gesell. 15 (1915) 8‑18.  ??NYS. Dickson, vol. 1, p. 464, item 83, says he found those  N, base 10,  whose reversal is  kN, in particular for  k = 9, 4.

Wood.  Oddities.  1927.  Prob. 56: Wizard stunts, pp. 45-46.  Notes that  83 * 41096 = 3410968  and asks for more examples of  AB * n = BnA.  Finds  86 * 8 = 688  and then says the next example is 'infinitely harder to find':  71 * 16 39344 26229 50819 67213 11475 40983 60655 73770 49180 32787. 

Ball.  MRE, 11th ed., 1939, p. 13.  Adds some different types of examples.   312*221  =  68952;   213*122  =  25986.   See also 7.AJ.

G. H. Hardy.  A Mathematician's Apology.  CUP, 1940.  Pp. 44-45.  "8712  and  9801  are the only four-figure numbers which are integral multiples of their 'reversals': ....  These are odd facts, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in them which appeals to a mathematician."  He cites Ball for this information.

Morley Adams.  1948.  See item in 7.AC.1 for an example of   6 * ABC  =  5 * CBA.

K. Subba Rao.  An interesting property of numbers.  The Mathematics Student 27 (1959) 57‑58.  Easily shows the multiplier must be  1, 4 or 9  and describes all solutions.

Jonathan Always.  Puzzling You Again.  Tandem, London, 1969.  Prob. 32: Four different answers, pp. 23 & 80.   AB * 7/4  =  BA   has four solutions.

Liz Allen.  Brain Sharpeners.  Op. cit. in 5.B.  1991.  A tram called Alec, pp. 89 & 137-138.   4 * TRAMS  =  SMART.

Jerome S. Meyer.  Arithmetricks.  Scholastic Book Services, NY, 1965. 

The  2178  trick, pp. 3-4.  Essentially the trick which gives  1089,  but he doubles the result at the end to get  2178.

Juggling numbers no. 2, pp. 83 & 88.  4 * ABCDE  =  EDCBA  with solution  21978.

 

          7.AH.1.        OTHER REVERSAL PROBLEMS

 

          New section.  There are many forms of this that I have not recorded before.

 

Augustus De Morgan.  Arithmetic and Algebra.  (1831 ?).  Reprinted as the second, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  Art. 114, pp. 29-30.  ab + 18 = ba  with  a + b = 6.

Birtwistle.  Calculator Puzzle Book.  1978.  Prob. 37: In reverse, pp. 28 & 84.  Which two digit numbers squared are the reversals of the squares of their reversals?  These are given by  12  and  13.

 

          7.AI.  IMPOSSIBLE EXCHANGE RATES

 

F. & V. Meynell.  The Week‑End Book.  Op. cit. in 7.E.  1924.  2nd ed., prob. three, p. 274;  5th? ed., prob. six, pp. 407‑408.  US & Mexico value each other's dollar at  58/60  (i.e. at  4s 10d : 5s).

Phillips.  Week‑End.  1932.  Time tests of intelligence, no. 36, pp. 20‑21 & 192.  Two countries each value the other's money at  90%  of its own.  The solution says this appeared in the New Statesman and Nation in late 1931, ??NYS.

Northrop.  Riddles in Mathematics.  1944.  1944: 9;  1945: 8‑9;  1961: 18‑19.  As in Phillips.

W. A. Bagley.  Paradox Pie.  Op. cit in 6.BN.  1944.  No. 3: South of the border, pp. 8-9.  US & Mexico each value other's dollar at  $1.05.

John Fisher.  John Fisher's Magic Book.  Muller, London, 1968.  Magical shopping, pp. 120‑121.  North and South Fantasi value each other's  £  as  19s  (i.e.  95%)  of its own.

 

          7.AJ.  MULTIPLYING BY SHIFTING

 

          That is, we want solutions of   k * a₁ ... a_m b₁ ... b_n  =  b₁ ... b_n a₁ ... a_m.   E.g., for  m = 1,   3 * 142857  =  428571;  for  n = 1,   4 * 025641  =  102564,   4 * 102564  =  410256,  5 * 142857  =  714285.   If  1/n  has a repeating decimal of period  n-1,  then all multiples of it by  1, 2, ..., n-1  are shifts of it _ e.g.   1/7  =  .142857142857...   gives   2 * 142857  =  285714,   etc.  Such an  n  must have  10  as a primitive root.

          Early versions of the idea are simply observations of the properties of  1/7  etc.

          Note that two-digit problems, i.e.  m = n = 1,  look more like reversals and are considered in 7.AH.

 

Charles Babbage.  The Philosophy of Analysis _ unpublished collection of MSS in the BM as Add. MS 37202, c1820.  ??NX.  See 4.B.1 for more details.  F. 4 is "Analysis of the Essay of Games".  F. 4.r has entry 13:  "What number is that whose  6  first multiples are the same digits differently arranged".

Ainsworth & Yeats.  Op. cit. in 7.H.4.  1854.  Exercise XXXVI, pp. 73 & 176.

No. 9:  k = 3,  m = 1,  a₁ = 1,  n = 5.

No. 10:  k = 3,  m = 1,  a₁ = 2,  n = 5.

Birger Hausted.  ??  Tidsskrift for Math. 2 (1878) 28.  ??NYS _ cited by Dickson, vol. 1, pp. 170‑171, item 81.  Studies problem with all shifts of the same number, but starting with the case  n = 1.  He finds that   k * a₁...a_mB  =  Ba₁...a_m   gives   B/(10k‑1)  =  (a₁...a_mB) / (10^m+1 ‑ 1).   He allows  k  to be rational.

Dickson, vol. 1, pp. 174‑179, items 101, 102, 106, 114, 120, 137, 150.

Hoffmann.  1893.  Chap. IV.

No. 15: A peculiar number, pp. 148 & 192‑193.   2 * 142857  =  285714,   which has  m = 2.

No. 49: A peculiar number, pp. 155 & 208.   5 * 142857  =  714285;   3 * 142857 =  428571;   6 * 142857  =  857142.

L. E. Dickson.  ??  Quarterly J. Math.  27 (1895) 366‑377.  [Item 106 above.]  Shows that all  ks  are integers and  a₁ ¹ 0  only for  142857  (in base  10).

Anonymous note.  J. of the Physics School in Tokyo 6? (1897?) ??NYS  Abstracted in:  Yoshio Mikami, ed.; Mathematical Papers from the Far East; AGM 28 (1910) 21; as: Another queer number.   3 * 526 31578 94736 84210  =  1578 94736 84210 52630,   which is not quite a pure shift of the first number, but would be if the final zeroes were dropped.  Also multiplication by  4  or  8  or division by  5  give similar results.

T. Hayashi.  On a number that changes its figures only cyclically when multiplied or divided by any number.  J. of the Physics School in Tokyo 6 (1897) 148-149.  Abstracted ibid., p. 21.  Notes that the above number is based on the form   10 Σ (10r)ⁱ,   which explains and generalizes its properties.

U. Fujimaki.  Another queer number.  J. of the Physics School in Tokyo 6 (1897) 148-149.  Abstracted ibid., pp. 22-23.  Notes that the above number has the form   (10^m - 1) / n,  which also explains and generalizes its properties.

Pearson.  1907.  Part II, no. 31: A large order, pp. 120 & 197‑198.  n = 1,  m = 21,  k  =  b₁  =  7.

Schubert.  Op. cit. in 7.H.4.  1913.  Section 16, no. 186, pp. 51 & 137.  k = 3,  m = 1,  a₁ = 1,  n = 5.

Peano.  Giochi.  1924.

Prob. 39, p. 10.  Notes cyclic property of  142857.

Prob. 40, pp. 10-11.  Notes cyclic property of   (10¹⁸ - 1) / 19,  says an English book calls them phoenix numbers and says similar properties hold for  (10^n-1 - 1) / n   for   n = 7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 130, 149, ....

Ackermann.  1925.  Pp. 107‑108.  k = 7,  n = 1,  b₁ = 7.

W. B. Chadwick.  On placing the last digit first.  AMM 48 (1941) 251-255.  n = 1.  Shows   b₁ ³ k   and   N  =  a₁ ... a_m b₁  =  b₁ (10^m+1 - 1) / (10k-1).  Finds all solutions for   k = 2, 3, ..., 9.  Cites Guttman, AMM 41 (1934) 159 (??NYS) for properties of these 'cyclic numbers'.  An editorial note adds that the expression for  N  works for any base.

Jerome S. Meyer.  Fun-to-do.  Op. cit. in 5.C. 1948.  No. 76: Mathematical whiz, pp. 59-60.  Discusses cyclic properties of the period of  1/17,  though he doesn't identify it as arising from  1/17.

J. Bronowski.  Christmas Teasers.  New Statesman and Nation (24 Dec 1949).  ??NYS.  Case  m = 1,  k = 3/2.

J. H. Clarke.  Note 2298:  A digital puzzle.  MG 36 (No. 318) (Dec 1952) 276.  Solves Bronowski's problem, leading to   3 * 10ⁿ  º  2 (mod 17)   for an  n+1  digit answer.

D. E. Littlewood.  Note 2494:  On Note 2298:  A digital puzzle.  MG 39 (No. 327) (Feb 1955) 58.  Easy solution of Bronowski's problem, via  20/17.

E. J. F. Primrose.  Note 2495:  A digital puzzle.  Ibid, 58‑59.  Case   m = 1,  k = p/q.

R. Sibson.  Note 2496:  On Note 2298.  Ibid, 59.  Simple but fortuitous solution via  20/17.

R. L. Goodstein.  Note 2600:  Digit transfers.  MG 40 (No. 332) (May 1956) 131‑132.  Shows Littlewood's method gives an easier solution to Primrose's version.

Gardner.  SA (Jan 1961) c= Magic Numbers, chap. 2.  Mentions   4*102564  =  410256   and the general problem with  n = 1  and  k = b₁.  Gives some references to the most general problem in various standard works and some journal references up to 1968.

D. St. P. Barnard.  A Book of Mathematical and Reasoning Problems.  Op. cit. in 7.C.  1962.  Prob. 23: A safe number, pp. 30, 62 & 85‑86.  Wants the smallest number with  n = 1  and  k = b₁,  i.e. such that   b * a₁a₂...a_mb  =  ba₁a₂...a_m.   Finds it is  102564.

Charles W. Trigg.  Division of integers by transposition.  JRM 1:3 (1968) 180-182.  Considers the case  n = 1,  k = b₁,  i.e. the same as Barnard.  He gives the smallest answers for bases  3, 4, ..., 12  and  k = 2, ..., b-1.  Cites some earlier versions not given above.

Steven Kahan.  k-transposable integers.  MM 49:1 (1976) 27-28.  Studies the case  m = 1  and shows that then only  k = 3  can work.  The basic solutions are  142857  and  285714  and all other solutions are obtained by repeating these, e.g.  142857142857.

Steven Kahan.  k-reverse-transposable integers.  JRM 9:1 (1976-77) 15-20.  Studies the case  n = 1  and finds all solutions.

Birtwistle.  Calculator Puzzle Book.  1978. 

Prob. 67: Moving around, pp. 49 & 105-106.   4 * abcde4  =  4abcde,  i.e.  k  =  b₁  =  4,  m = 5,  n = 1.  Notes properties of  142857.

Prob. 78: Change around, pp. 55 & 113.  k = 2,  n = 1,  b₁ = 5,  formulated as a division ignoring remainder.  Finds   52631/2  =  26315,   which is a bit of a cheat, but he notes that this is part of the decimal expansion of  1/19  =  .0526315789473684210526...   which would give a proper answer.

Warren Page.  A general approach to p.q r‑cycles.  In:  Warren Page, ed.; Two‑Year College Mathematics Readings; MAA, 1981, pp. 263‑274.  He considers rational  k  (=  his  q/p)  and given  n  (=  his  r).  He also studies the case where multiples of   a₁...a_mb₁...b_n   have the same properties.  He gives tables of all solutions for  n = 1  and  n = 2.  11 references to similar work.

Anne L. Ludington.  Transposable integers in arbitrary bases.  Fibonacci Quarterly 25:3 (1987) 263-267.  Considers case  m = 1.  Cites Kahan's results.  Considers base  g  and shows that there is some  k  if and only if  g = 5  or  g ³ 7.  Shows that there are only a finite number of basic solutions for any  k  and hence for any  g  and describes how to find them.

Anne L. Ludington.  Generalized transposable integers.  Fibonacci Quarterly 26:1 (1988) 58‑63.  Considers general case with shift of  j  (=  m  in my notation) and arbitrary base  g.  Shows there is such a  k  for all  j ³ 0  if and only if  g = 5  or  g ³ 7.  For  g = 3, 4, 6,  there is such a  k  for all  j ³ 2.  For fixed  j,  there are only a finite number of solutions.

Keith Devlin.  Better by degrees _ Micromaths column in The Guardian (17 Nov 1988) 31.  Let  F, C  be the same temperature in Fahrenheit and Centigrade (= Celsius).  When can  F = a₁b₁...b_n,   C = b₁...b_na₁?   First answer is  527.

 

          7.AJ.1.         MULTIPLYING BY APPENDING DIGITS

 

          New section.

 

Tony Gardiner.  Challenge! What is the title of this article?  Mathematics Review 4:4 (Apr 1994) 28-29.  Find an integer  A  such that adding digits of  1  at each end multiplies it by  99.  If  A  has  n  digits, then this gives us   10ⁿ⁺¹ + 10A + 1  =  99A,   so  A  =  (10ⁿ⁺¹ + 1) / 89.  He leaves the rest to the reader.

 

          7.AK. LAZY WORKER

 

          A worker earns  a  for each day he works and forfeits  b  for each day he doesn't work.  After  c  days, he has gained  d.  This gives   x + y = c,  ax ‑ by = d   or  (a+b)x - bc = d.   This is a straightforward problem, giving   x = (bc + d)/(a + b),   so I only give some early examples.  There are also many different forms of problem which give the same equations.  PROBLEM _ for which integral  a, b, c, d,  is it true that  x, y  are integral?

          NOTATION:  we denote this by  (a, b, c, d). 

          For general solutions, see:  Hutton, 1798?;  De Morgan.

          See Tropfke 603.

 

         2½            1          30          54              Hutton, c1780?;  Hutton, 1798?; 

         4½         1½          24         4½              Hutton-Rutherford

         4½         1½          24          78              Hutton-Rutherford

            5            3          28            0              Riese, 1524; 

            5            6          30            0              Chuquet; 

            5            6          30          12              Chuquet; 

            5            9          30          10              Columbia Alg.; 

            5            9          30          15              Gherardi; 

            5          12          30          99              Robinson

            6            5          30            0              Muscarello; 

         6½         5¾          50          18              Unger; 

            7            3         365            0              Dodson

            7            4          30            1              Fibonacci;  Columbia Alg; 

            7            4          30          30              Fibonacci; 

            7            5          30            0              Riese, 1522; 

            9          11          30            0              Bartoli

         9¼          6_          70         180              Unger; 

          10            4          30         132              BR; 

     10/30       6/30          30           -2              al-Karkhi; 

     10/30       6/30          30            0              al-Karkhi; 

     10/30       6/30          30            4              al-Karkhi; 

          10          12          30            0              P. M. Calandri;  Calandri, 1491; 

          10          12          40            0              AR;  Wagner; 

          10          14          20          15              Tartaglia; 

          11            8          36          12              Riese, 1524

          12            8         390            0              Vyse

          16            8          12         126              Eadon; 

          16          20          30            0              Eadon; 

          16          24          36            0              Tartaglia; 

          16          24          36          60              Tartaglia; 

          18          16          30            0              Pacioli; 

          20            8          40         372              Hutton, 1798?

          20            8          40         380              Simpson;  Bonnycastle

          20          10          40         500              Vyse

          20          28          40            0              Borghi; 

          20          28          40          30              Borghi; 

          24          12          48         504              Bourdon

          30          15          40         660              Ozanam-Montucla

          40            5          60         980              Les Amusemens; 

 

Muhammad ibn Muammad ibn Yahy_ al-Buza__n_ Ab_ al-Waf_’  = Ab_ al-Waf_’ al‑B_zaj_n_.  Arithmetic.  c980.  Arabic text edited by  Ahmad Salim Saidan;  Arabic Arithmetic; Amman, 1971.  P. 353.  ??NYS _ mentioned by Hermelink, op. cit. in 3.A.

al‑Karkhi.  c1010.  Sect. I, no. 12‑14, p. 83.  (10/30, 6/30, 30, d)  with  d = 0, 4, ‑2.

Tabari.  Mift_h al-mu‘_mal_t.  c1075.  P. 113, no. 23.  ??NYS _ Hermelink, op. cit. in 3.A, gives al‑Karkhi's problem and then says  "This problem occurs also in Tabar_ ...."  Tropfke 603 gives the same reference.

Fibonacci.  1202.

Pp. 160‑161: De laboratore laborante in quodam opere.  (7, 4, 30, 30).

Pp. 323‑324: De laboratore, question notabilis.  (7, 4, 30, 1).

BR.  c1305.  No. 51, pp. 68‑69.  (10, 4, 30, 132).

Gherardi.  Libro di ragioni.  1327.  Pp. 48‑49.  (5, 9, 30, 15)

Columbia Algorism.  c1370.

Prob. 64, pp. 85‑86.  (5, 9, 30, 0).

Prob. 65, pp. 86‑87.  (7, 4, 30, 1).

Dresden C80.  ??NYS _ asserted in BR, p. 158.

Bartoli.  Memoriale.  c1420.  Prob. 7, ff. 75r - 75v (= Sesiano, pp. 137 & 147).  (9, 11, 30, 0).

AR.  c1450.  Prob. 183, pp. 86, 176, 222‑223.  (10, 12, 40, 0).

Muscarello.  1478.  Ff. 73v-74r, p. 188.  Worker building a house, (6, 5, 30, 0).

P. M. Calandri.  c1480.  P. 88.  (10, 12, 30, 0).

Wagner.  Das Bamberger Rechenbuch, op. cit. in 7.G.1.  1483.  Von Tagelohn oder Arbeit, pp. 98 & 216.  (10, 12, 40, 0).  (= AR.)

Chuquet.  1484.

Prob. 51, English in FHM 209.  (5, 6, 30, 0).

Prob. 52.  (5, 6, 30, 12).

Borghi.  Arithmetica.  1484. 

Ff. 111v-112r (1509: f. 94r).  (20, 28, 40, 0).

Ff. 112r-113r (1509: ff. 94v-95r).  (20, 28, 40, 30).

Calandri.  Arimethrica.  1491.  F. 69v.  (10, 12, 30, 0).

Pacioli.  Summa.  1494.  F. 99r, prob. 11.  (18, 16, 30, 0).

Riese.  Rechnung.  1522.  1544 ed. _ pp. 89‑90;  1574 ed. _ pp. 60v‑61r.  (7, 5, 30, 0).

Riese.  Die Coss.  1524.

No. 37, p. 45.  (5, 3, 28, 0).

No. 110, pp. 54‑55.  (11, 8, 36, 12).

Tartaglia.  General Trattato.  1556.  Book 17, art. 38, 39, 42, pp. 275r-275v & 277r.

Art. 38.  (16, 24, 36, 0).

Art. 39.  (16, 24, 36, 60).

Art. 42.  (10, 14, 20, 15).

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790), prob. VII, pp. 80-81 (1790: prob. XIX, p. 84).  (20, 8, 40, 380).

Les Amusemens.  1749.  Prob. 112, p. 254.  (40, 5, 60, 960).

Vyse.  Tutor's Guide.  1771? 

Prob. 21, pp. 85-86 & Key p. 111.  (12, 8, 390, 0).

Prob. 23, p. 86 & Key pp. 111-112.  (20, 10, 40, 500).

Prob. 15, p. 139 & Key p. 184.  (12, 8, 390, 0)  solved a different way.

Dodson.  Math. Repository.  1775.  P. 9, Quest. XXI.  (7, 3, 365, 0).

Ozanam‑Montucla.  1778.  Prob. 8, 1778: 194-195;  1803: 192;  1814: 166‑167;  1840: 86.  (30, 15, 40, 660).  1790 has  620  for  660,  apparently a misprint (check if this is in 1778 _ ??).

Charles Hutton.  A Complete Treatise on Practical Arithmetic and Book-keeping.  Op. cit. in 7.G.2.  [c1780?]  1804: prob. 3, p. 135.  (2½, 1, 30, 54).

Bonnycastle.  Algebra.  1782.  Prob. 6, p. 80.  Same as Simpson.

Eadon.  Repository.  1794. 

P. 297, no. 14.  (16, 8, 12, 126).

Pp. 374-375, no. 20.  (16, 20, 30, 0).  The answer has days worked and days away reversed.

Hutton.  A Course of Mathematics.  1798? 

Prob. 2,  1833 & 1857: 80.  (2½, 1, 30, 54).

Prob. 7,  1833: 212;  1857:: 216.  (20, 8, 40, 372),  then done in general.

Prob. 46,  1833: 224;  1857: 228.  Man to pay his friend  a  for each shot he misses and to receive  b  for each hit.  After  n  shots,  he owes  c,  where  c  may be positive, zero or negative.

Louis Pierre Marie Bourdon (1779-1854).  Élémens d'Algèbre.  7th ed., Bachelier, Paris, 1834.  (The 6th ed., was 1831, but I haven't yet looked for the early editions _ ??)  Art. 47, prob. 2, pp. 64-65.  (24, 12, 48, 504)  and the general problem.  On pp. 102-104, he discusses the problem algebraically and the meaning of the signs involved.

Augustus De Morgan.  On The Study and Difficulties of Mathematics.  First, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  Workman earns  a  each day he works, but has expenses of  b  every day.  After  m  days, he has earned  c.  This is equivalent to  (a-b, b, m, c),  giving  ax - bm = c.  Gives general solution and discusses problems with negative values.

Unger.  Arithmetische Unterhaltungen.  1838.  Pp. 41-45 & 248, nos. 176-190.  These cover a variety of problems leading to the same equations.

Pp. 41-42, no. 176.  (9¼, 6_, 70, 180).

P. 45, no. 190.  Purely arithmetic formulation of  (6½, 5¾, 50, 18).

Hutton-Rutherford.  A Course of Mathematics.  1841?.  Prob. 43,  1857: 43.  4½, 1½, 24, 78  and  4½, 1½, 24, 4½.

Horatio N. Robinson.  New Elementary Algebra: Containing the Rudiments of the Science for Schools and Academies.  Ivison, Blakeman, Taylor & Co., New York, 1875.  This has several versions of the problem, but on p. 77, probs. 77-78 use the following novel fomulation: a boy is to deliver some glass vessels and will be paid  a  for each success and will forfeit  b  for each breakage; after  c  vessels, he has earned  d.  Does  (5, 12, 30, 99)  and the general case.

 

          7.AL.  IF  A  IS  B,  WHAT IS  C?

 

          Generally, this is done by proportion of some sort.  In the simplest case _ if  a  is  b,  what is  c?  _ the answer is  cb/a.  In the more complex case _ if  ab = c,  what is  de = f?  _ the answer is usually  f = cde/ab.  Many problems have  e  and  f  given and ask for  d,  which is then  d = abf/ce.  However, some authors say that all answers should be multiplied by  c/ab  and hence give  d = cf/abe,  which I will call inverted reasoning or the inverted answer _ see:  P. M. Calandri, p. 65;  The Sociable (& Book of 500 Puzzles);  Lemon;  Hoffmann;  Pearson;  Loyd.

 

Fibonacci.  1202.  P. 170.  He discusses this and says that it is just a proportion which is commonly stated in this way.  E.g. if  5  is  9,  what is  11?  He says  99/5.  Also, if  7  is  half  12,  what is half of  10?  He says  35/6.

Gherardi.  Libro di ragioni.  1327.  P. 17: Questi sono numeri.  "If  9  is the half of  16,  what part is  12  of  25?"  Answer is  8/9  of  12/25.

Bartoli.  Memoriale.  c1420.

Prob. 5, f. 75r (= Sesiano, pp. 137 & 147).  If  7  is  1/8  of  49,  what is  1/3  of  57?  He seems to say: "If  7  becomes  6 1/8,  what becomes  19?"  but he computes the inverse result.

Prob. 6, f. 75r (= Sesiano, pp. 137 & 147).  If  3  times  6  makes  17,  what will  7  times  8  be?  Obtains the normal answer.

AR.  c1450.  Prob. 303, 305, pp. 135‑136, 178, 225.

303:  If  4  is  ½  of  10,  what is  _  of  24?   Answer:  6 2/5  =  8 * 4/5.

305:  If  3 * 3  is  10,  what is  4 * 4?   Answer:  17 7/9  =  16 * 10/9.

Vogel says these problems also occur in Widman, 1489, ??NYS, and versions are in al‑Khowarizmi.  However al‑Khowarizmi's examples are straightforward rules of three, e.g  "If you are told 'ten for six, how much for four?'...."

The Treviso Arithmetic = Larte de labbacho.  Op. cit. in 7.H.  1478.  Ff. 31r‑33v (= Swetz, pp. 103‑109).  Three examples in abstract rule of three, e.g.  "If  8  should become  11,  what would  12  become?"   Answer:  12 * 11/8.

P. M. Calandri.  c1480.

P. 64.  If  3 * 3  =  10,  what is  10 * 10  by the same rule?  Answer:  100 * 10/9.

P. 65.  If  3  is half of  7,  what is the half of  7?   Answer:  7/6 * 7/2, by inverted reasoning.

Chuquet.  1484.  Triparty, part 1.  FHM 73-74 gives a number of problems which are treated as proportions.

"If  3  times  4  will lead to  9,  what will  4  times  5  lead to?  ...  If  12  is worth  9,  what is  20  worth?" 

"If  7  is the  ½  of  12,  what is  _  of  9?"   Answer:  3½.

"If  2/3  will be  3/4  of  4/5,  what will be the  5/6  of  6/7?"   Answer:  50/63.

"If  7  were the  ½  of  12  one asks what part  3½  would be of  9."   Answer:  3½.

"If  2/3  were the  3/4  of  4/5,  one asks what part would  50/63  of  6/7  be."  Answer:  5/6.

Pacioli.  Summa.  1494.

Ff. 99r-99v, prob. 13.  If  ½  of  5  is  3,  when is  ¼  of  x  equal to  5.  Gets  x = 16_.  Then does:  If  4  is  6,  what is  10?  He gets  (6/4)10  =  15  and then says that if one rephrases it, then one gets  (4/6)10  =  6_.  Then mentions:  If  ½  of  7  is  3,  what is  _  of  9?  _ no answer given.

F. 99v, prob. 14.  If  1/3  is  1/2  of  1/5,  when is  1/5  of  1/x  equal to  1/4?  The first statement says  1/5 = 2/3,  so  2/3x = 1/4,  giving  x = 8/3,  which is the direct answer.

F. 99v, prob. 15.  If  3½  is  ½  of that number of which  5  is  _,  of what number is  3  the  ½?  Takes  3/2  of  5  =  7½  and then half of that, i.e.  3½  and says  "If  3½  is  3¾,  what is  3?"  Gets  45/14  and doubles it.

F. 99v, prob. 18.  If  3  is the  ½  of  7,  what part of  11  is  4?  Says "if  3  is  3½,  what is  4?"  and gets  14/3  which is  14/33  of  11.  Says this may not be the right way to do it.

Tonstall.  De Arte Supputandi.  1522.  Pp. 223-224.

If  4  is  6,  what makes  10?   (4/6)10  =  6_.

If  ½  of  5  is  3,  when is  ¼  of  x  equal to  5.  Gets  16_.

If  ½  of  7  is  3,  what part of  11  is  4?  His method would give  33/14,  but he has  33/13  due to an error.

Dilworth.  Schoolmaster's Assistant.  1743.  P. 157, no. 3.  "If the  _  of  6  be  3,  what will  ¼  of  20  be?"  Answer:  7½,  which is the direct answer.

Les Amusemens.  1749.  P. xxv.  See 7.AN for a problem that looks like it belongs here.

Walkingame.  Tutor's Assistant.  1751.  1777: p. 172, prob. 48;  1835: p. 178, prob 27;  1860: p. 180, prob. 47.  Identical to Dilworth.

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions.  No. 2, pp. 15 & 71.  "If the half of five be seven, What part of nine will be eleven?"  Answer:  55/126  =  (5/2)(11/7)/9.

The Sociable.  1858.  Prob. 46: A dozen quibbles: part. 7, pp. 300 & 319.  "If  5  times  4  are thirty-three, what will the fourth of twenty be?"  Answer is  8¼  with no explanation, which is the inverted answer.  = Book of 500 Puzzles, 1859, prob. 46: part 7, pp. 18 & 37.  = Magician's Own Book (UK version), 1871, Paradoxes [no. 2], p. 37.

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Arithmetical puzzles, no. 7, p. 174 (1868: 185).  "If six the third of twenty be, what is the fourth of thirty‑three?"  Gets   (6 * 33/4) / (20/3)  =  7 17/40.

Lewis Carroll.  Alice in Wonderland.  1865.  Chap. II.  In:  M. Gardner; The Annotated Alice; revised ed., Penguin, 1970, p. 38.  "Let me see:  four times five is twelve, and four times six is thirteen, and four times seven is _ oh dear!  I shall never get to twenty at that rate!"  Gardner's note 3 gives various explanations, the simplest of which is that   4 * 12  =  19   and Alice only knows tables up to 12 times.

Lemon.  1890.  Quibbles, no. 254(b), pp. 37 & 107 (= Sphinx, no. 453(b), pp. 63 & 113.)  "If five times four are thirty‑three, what will the fourth of twenty be?"  Inverted answer of  8¼  with no explanation _ see The Sociable.

Hoffmann.  1893.  Chap. IX, no. 34: A new valuation, pp. 320 & 327.  Identical to Lemon.  Answer is  8¼,  but he gives no real reason.

Pearson.  1907.  Part II, no. 95, pp. 134 & 210.  Same as Lemon.

Loyd.  Cyclopedia.  1914.  Sam Loyd's perplexed professor, pp. 332 & 383.  = SLAHP: If things were different, pp. 56 & 106.  "If five times six were  33,  what would the half of  20  be?"  Answer:  "If five times six is  33  _ ten would naturally be  1‑3  of what  30  would be, viz:  11."  This is the inverted answer.  Loyd Jr. gives a bit more explanation.

Loyd.  Cyclopedia.  1914.  p. 317 (no solution).  As in Lemon.

Perelman.  MCBF.  1937.  Imaginary nonsense, prob. 143, pp. 243-244.  "What is the number  84  if  8 x 8  is  54?"  Finds that the base is  12,  so the answer is   84₁₂  =  100.

Evelyn August.  The Black-Out Book.  Op. cit. in 5.X.1.  1939.

Number, please! _ no. 3, pp. 95 & 214.  "If five times  8  made  60,  what would a quarter of  40  be?"  Answer is  15,  which is the inverted answer.

Is your brain working? _ no. 3, pp. 148 & 215.  "If five times four made thirty three,  What would a fifth of fifth be?"  Gives the answer  15½,  which is described as half of  33!  I.e. the answer is intended to be the inverted answer  16½.

The Little Puzzle Book.  Op. cit. in 5.D.5.  1955.  P. 8: Mathematical "if".  "If a fourth of forty is six;  what is a third of twenty?"  Says four, basically by saying the result is  6/10  of the real result.

G. A. Briggs.  Puzzle and Humour Book.  Published by the author, Ilkley, 1966.  Prob. 1/9, pp. 12 & 71.  If a third of six were three, what would the half of twenty be?"  Answer is  15,  which is the normal answer, but he gives no reason.

Liz Allen.  Brain Sharpeners.  Op. cit. in 5.B.  1991.  If, pp. 28 & 109.  "If a third of six was three, what would a quarter of twenty be?"  Takes  3/2  of the correct answer, as in the usual method.

 

          7.AM.          CROSSNUMBER PUZZLES

 

          New section _ there may be older examples, but see Barnard.

 

Dudeney.  Strand Magazine (1926).  ??NYS _ cited by Angela Newing.

Richard Hoadley Tingley.  Mathematical Cross Number Puzzle[s].  In:  S. Loyd Jr., ed.; Tricks and Puzzles; op. cit. in 5.D.1, 1927.  Pp. 103‑105 & Answers pp. 11-12.  Three examples with  13 x 13  frames.  Tingley's name is only on the first example.  "This puzzle is radically different from the usual type ....  We have named this new brain teaser "Cross Number Puzzle" ...."  I found several errors in the second example.

Dudeney.  PCP.  1932.  Prob. 173: Cross‑figure puzzle, pp. 48 & 148.  11 x 11  frame.  Erroneous set of clues and solution.  Corrected in the revised ed. of 1935?, pp. 48‑49 & 148.

Michael H. Dorey.  "Little Pigley" [or "Little Pigsby"].  1936.  This is also called "Dog's Mead" _ original ??NYS.  A 1939 version with this attribution and date are given in:  Tim Sole; The Ticket to Heaven and Other Superior Puzzles; Penguin, 1988, pp. 92 & 108.  A 1935 version is given in:  Williams & Savage, 1940, below.  A 1936 version is given in:  Philip Carter & Ken Russell; Classic Puzzles; Sphere, London, 1990, pp. 62-63 & 128.  A 1939 version is given as The Little Pigsby Farm Puzzle in:  David Ahl & Burchenal Green, eds.; The Best of Creative Computing, vol. 3; Creative Computing Press, Morristown, NJ, 1980, p. 177; no solution.

Phillips.  Brush.  1936.  4 x 4  numerical crosswords.

Prob. H.5, pp. 26 & 91.

Prob. J.5, pp. 35 & 96.

Prob. T.3, pp. 68‑69 & 115.

Jerome S. Meyer.  Fun for the Family.  (Greenberg Publishers, 1937);  Permabooks, NY, 1959.  No. 28: Family skeleton, pp. 41 & 240.  4 x 4  based on ages in a family.

W. T. Williams & G. H. Savage.  The Penguin Problems Book.  Penguin, 1940.

No. 18: Mr Turtle, pp. 17 & 108.  4 x 4  square with  11  clues.

No. 51: Little Pigley Farm, 1935, pp. 32-33 & 119.  7 x 7  square with  21  clues and some more in the lead-in.

M. A. Porter.  Note 1982:  The missing clue.  MG 31 (No. 296) (Oct 1947) 237.  4 x 4  puzzle.

Anonymous examples in Eureka.

12 (1949) ? (??NYS)  &  13 (1950) 21.  ??confusion _ check

13 (1950) 15  &  14 (1951) 23.  This uses mathematical symbols as well as numbers.

14 (1951) 6  &  15 (1952) 16.

15 (1952) 16  &  ??? (??NYS).  Three dimensional version by 'Nero'.

17 (1954) 13  &  ??? (??NYS).  Cross shaped, by 'Pythagoras'.

Philip E. Bath.  Fun with Figures.  Op. cit. in 5.C.  1959.  No. 100: A number crossword, pp. 38 & 61.  5 x 5  array with  17  clues.

P. E. Knight.  Oh, How I love thee, Dr. Pell.  In:  H. Phillips; Problems Omnibus II; Arco, London, 1962; pp. 163‑164 & 228‑229.  3 x 4  array, but complex, based on Pell's equation.

D. St. P. Barnard.  Fifty Observer Brain‑Twisters.  Op. cit. in 4.B.1.  1962.  Prob. 40: Crossnumber, pp. 46‑47, 65 & 98‑99.

D. St. P. Barnard.  Anatomy of the Crossword.  Bell, London, 1963.  On p. 30, he says  "The crossword ... has given rise recently to the Crossnumber Puzzle"  and he refers to his book above.

Birtwistle.  Math. Puzzles & Perplexities.  1971.

Cross-number puzzle 1, pp. 91 & 192.  8 x 8  grid,  32  clues.

Cross-number puzzle 2, pp. 92 & 192.  7 x 7  grid,  24  clues.

Cross-number puzzle 3, pp. 93 & 192-193.  9 x 9  grid,  36  clues.

Cross-number puzzle 4, pp. 94 & 193.  8 x 8  grid,  26  clues.

Birtwistle.  Calculator Puzzle Book.  1978. 

Prob. 15, pp. 14-15 & 75.  6 x 6  grid,  16  clues.

Prob. 33, pp. 24-25 & 82.  8 x 8  grid,  28  clues.

Prob. 63, pp. 46-47 & 103.  8 x 8  grid,  20  clues.

Prob. 90, pp. 63-64 & 123.  6 x 6  grid,  16  clues relating to a story.

K. Heinrich.  Zahlenkreuzrätsel.  In:  Johannes Lehmann; Kurzweil durch Mathe; Urania Verlag, Leipzig, 1980; pp. 38 & 138.  6 x 6  grid with  24  clues.

 

          7.AN. THREE ODDS MAKE AN EVEN, ETC.

 

Alcuin.  9C.  Prop. 43: Propositio de porcis.  Kill  300  (or  30)  pigs in three days, an odd number each day.  "[Haec ratio indissolubilis ad increpandum composita est.]  ...  Ecce fabula ....  Haec fabula est tantum ad pueros increpandos."  ([This unsolvable problem is set to cause confusion.]  This is a fable ....  This fable is posed to confuse children.)

William Leybourn.  Pleasure with Profit.  Richard Baldwin & John Dunton, 1694.  ??NYS _ described in Cunnington, op. cit. in 7.G.2, 1904, p. 151  and  in De Morgan, Rara, p. 633.  "How can you put five odd numbers to make twenty?"  "Write three nines upside down and two ones."  De Morgan says he does not recall ever seeing this problem, that Leybourn considers the answer a fallacy, but that he thinks "the question more than answered, viz. in very odd numbers."

Les Amusemens.  1749. 

P. xxv. 

          Quatre fois trois font quinze, il n'en faut rien rabattre;

          Neuf cing et un font douze, et rien de surplus;

          Deux sept et six font treize, et sur ce je conclus

          Par ce juste calcul que tout ne fait que quatre.

                    Solution is to take the number of letters in the words _ but  'et'  is printed like  '&'  which makes it a bit hard to recognize it as a two-letter word.

P. 52.  1^o:  Exprimer un nombre pair par  3  impairs.  General solution:   a a/a  =  a+1.  Examples:  7 7/7,  21 21/21.  Pp. 53-54 give other problems but they are not relevant, e.g.  33 3/3  _ see entry under 7.I.

Philip Breslaw (attrib.).  Breslaw's Last Legacy.  1784?  Op. cit. in 6.AF.  1795: 78-81.  'How to rub out Twenty Chalks at five Times rubbing out, every time and odd one.'  Set out marks numbered  1  to  20.  Then erase the last four, which are those starting at  17,  which is odd, etc. 

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Parlour magic, no. xx, p. 202 (1868: 203): How to rub out twenty chalks in five rubs, each time erasing an odd number.  "Begin at the bottom and rub out upwards, four at a time."  See Breslaw for clarification.

Magician's Own Book.  1857.  How to rub out twenty chalks at five times, rubbing out every time an odd one, p. 239.  = Boy's Own Conjuring Book, 1860, How to rub twenty chalks at five times rubbing out, every time an odd one, pp. 205‑206.  Cf Breslaw.

Magician's Own Book (UK version).  1871.  The Arabian trick, p. 313.  "To take up twenty cards, at five times, and each time an odd-numbered one", he lays out twenty cards, 1 ‑ 10, 1 - 10  (considered as  1 - 20)  and proceeds as in Breslaw.

Don Lemon.  Everybody's Pocket Cyclopedia.  Revised 8th ed., 1890.  Op. cit. in 5.A.  P. 136, no. 3.  "Place  15  sheep in  4  pens, so there will be the same number of sheep in each pen."  Though not of the same type as others in this section, and no solution is given, I think the solution is similar to other solutions here, namely to put the pens concentrically around the inner pen and put all the sheep inside the inner pen.

Pearson.  1907.  Part II.

No. 111, pp. 137 & 213.  Five odd figures to make  14.  Gives   1 + 1 + 1 + 11  =  14   and   1 + 1 + 1 + 1  =  4  with another  1  makes  14.

No. 115: What are the odds?, pp. 137 & 214.  Place  20  horses in three stalls with an odd number in each.  Puts  1, 3, 16  and says  16  is "an odd number to put into any stall".

Part III, no. 61: The shepherd's puzzle, p. 61.  Put  21  sheep in  4  pens, an odd number in each.  Uses concentric pens.

Loyd.  Cyclopedia.  1914.  The pig sty problem, pp. 37 & 343.  = MPSL2, prob. 7, pp. 6‑7 & 123.  = SLAHP: Pigs in pairs, pp. 51 & 104.  = Pearson's shepherd's puzzle.

Loyd.  Cyclopedia.  1914.  A tricky problem, p. 38.  = SLAHP: Torturing Dad, pp. 75 & 115.  Five odd figures to make  14.  Gives Pearson's first solution.

Smith.  Number Stories.  1919.  Pp. 126 & 146.  Put  10  pieces of sugar in three cups so each cup has an odd number.  Put  7  &  3  and put one cup inside another cup.

Blyth.  Match-Stick Magic.  1921.  The twenty game, p. 79.  As in Breslaw, etc., but more clearly expressed:  "The matchsticks have now to be removed in five lots, ....  Each time ... the last of the group must be an odd number."

Wood.  Oddities.  1927.  Prob. 50: The lumps of sugar, pp. 42-43.  Ten lumps of sugar into three cups so each cup contains an odd number of lumps.  Confusing solution, but lets one cup be inside another and lists all 15 possible solutions.

Evelyn August.  The Black-Out Book.  Op. cit. in 5.X.1.  1939.  Number, Please!, pp. 20 & 210.  Use the same odd figure five times to make  14.   1 + 1 + 1 + 11.

McKay.  Party Night.  1940.  Pill-taking extraordinary, p. 152.  "A man had a box holding  100  pills.  He took an odd number of pills on each of the seven days of the week, and at  the end of the week all the pills were gone.  How could he manage that?"  1  on each of the first  6  days and  94  on the last _ "you must admit that  94  is a very odd number of pills to take on any day."

Jerome S. Meyer.  Fun-to-do.  Op. cit. in 5.C. 1948.  Prob. 17: Pigs and pens, pp. 27 & 184.  Put nine pigs in four pens, an odd number in each pen.  Three pens of three with a big pen around them all.

The Little Puzzle Book.  Op. cit. in 5.D.5.  1955.  P. 19: Pigs in pens.  Same as Meyer.

John Paul Adams.  We Dare You to Solve This!.  Op cit. in 5.C.  1957?

Prob. 40: A pen pincher, pp. 24 & 40.  9  pigs in four pens, an odd number in each.  Three pens of three, contained in one large pen.

Prob. 169, pp. 60-61 & 120.  21  lambs in four pens, each with an odd number.  Same solution idea as in prob. 40.

Gibson.  Op. cit. in 4.A.1.a.  1963.

P. 68: Odd figuring.  Use seven odd figures to add up to  20.  Answer:  13 + 3 + 1 + 1 + 1 + 1.

P. 69 & 74: Cross-out groups.  Make a row of  20  labelled marks and cross out marks in consecutive groups so that the last mark is odd each time.  Easy _ just mark backward!

Joseph & Lenore Scott.  Master Mind Brain Teasers.  Op. cit. in 5.E.  1973.  Three short problems _ no. 1, pp. 41-42.  Have five odd numbers add to  14.   11 + 1 + 1 + 1  =  14.

 

          7.AO. DIVINATION OF A PERMUTATION

 

          There are simpler versions of this used to divine three numbers, e.g. to locate a ring on a person, finger and digit _ a common one uses:  x *2 +5 *5 +10 +y *10 +z  =  350 + 100x + 10y + z  _ cf Tagliente (1515).  The operations are performed from left to right, corresponding to the intructions given.  I am sure there are many earlier forms of this and I will replace Tagliente with an earlier form, but I will make no attempt to trace this very common but very dull type of problem.

          If the  a_i  are a permutation of  1, 2, ..., n,  the method of interest forms   P  =  a₁m₁ + a₂m₂ + ... a_nm_n   for multipliers  m_i.  By appropriate choice of the  m_i,  the value of  P  determines the permutation.  Generally  P  is subtracted from some convenient constant.  Sometimes the solution uses a division to yield  a₁  and  a₂.  Some formulae work even if  a_i  are not a permutation, but are digits or dice values.  If we have a permutation, one can ignore  a_n  since it is determined by the others, i.e. one can let  m_n = 0.

          NOTATION:   M  =  (m₁, m₂, ..., m_n)   denotes this process.

          A sequence of multipliers,  M,  is suitable if the products  P  are all distinct.  In the case that the  (a_i)  are a permutation, it is easy to see that the following processes preserve the suitability of  M.  (1) Permuting the  m_i.  (2)  Shifting all the  m_i  by a constant.  (3) Multiplying all the  m_i  by a non-zero constant.  Thus we can arrange the  m_i  in ascending order and make  m₁ = 0  and  m₂ = 1.  So for the case  n = 3,  any sequence can be brought to the form  (0, 1, m₃).  By subtracting from  m₃  and scaling, we see that this sequence is equivalent to  (0, 1, m₃/(m₃-1)).  A little work shows that either  m₃ ³ 2  or  m₃/(m₃-1) ³ 2,  so we can assume  m₃ ³ 2.   m₃ = 2  gives an unsuitable  M,  but  m₃ > 2  always gives a suitable  M.  So the simplest possible case is  (0, 1, 3),  which is equivalent to the most common case  (1, 2, 4)  and to  (3, 4, 6).  Cases equivalent to  (0, 1, 4),  (0, 1, 6)  and  (0, 1, 8)  also occur.

          For the case of permutations, subtracting  P  values from some constant  S  is equivalent to changing  m_i  to   S/Σa_i  -  m_i.

          As in the Josephus Problem, mnemonics were constructed.  For the case  n = 3,  objects were labelled by the vowels  a, e, i  and mnemonics were constructed consisting of words (or phrases) with these three vowels in all six permutations (or having just the first two vowels of each permutation).  See Bachet for an example.  Gardner also gives mnemonics using consonants.

          See Meyer for a slightly more complex multiplication process, which can be reduced to the usual form.

          See Tropfke 648‑651.

 

                    COMMON VOWEL MNEMONICS _ note that spelling and layout vary.

 

Angeli, Beati, Taliter, Messias, Israel, Pietas.  Baker;

Anger,  fear,  pain, may  be hid  with a  smile.  Magician's Own Book; Boy's Own Conjuring Book;

Aperi,  Premati,  Magister,  Nihil,  Femina,  Vispane,  Vispena.  Minguét.

Aperì  Prelati  Magister  Camille  Perina  Quid habes  Ribera.  Alberti 76-77;

Avec  éclat  L'Aï  brillant  devint  libre.  Labosne, under Bachet; Lucas;

Brave  dashing  sea, like a  giant  revives  itself.  Magician's Own Book; Boy's Own Conjuring Book;

Graceful  Emma,  charming she  reigns  in all  circles.  Magician's Own Book; Boy's Own Conjuring Book;

Il a  jadis  brillé  dans ce  petit  État.  Lucas;

James  Easy  admires now  reigning  with a  bride.  Magician's Own Book; Boy's Own Conjuring Book;

Par fer  César  jadis  devint  si grand  prince.  Bachet; Ozanam 1725, prob. 46; Alberti 76-77; Les Amusemens; Hooper; Magician's Own Book; Boy's Own Conjuring Book; Boy's Own Book; Magician's Own Book (UK version); Lucas;

Pare ella ai segni; Vita, Piè.  Alberti 76-77.

Salve  certa  anima  semita  vita  quies.  Van Etten; Ozanam 1725, prob. 46; Alberti 76-77; Manuel des Sorciers; Endless Amusement; Parlour Pastimes; Magician's Own Book; Boy's Own Conjuring Book; Magician's Own Book (UK version);

Take her  certain  anise  seedlings  Ida  quince.  Magician's Own Book (UK version);

 

Tabari.  Mift_h al-mu‘_mal_t.  c1075.  P. 109, part IV, no. 17.  ??NYS - described by Tropfke 648.  As in Fibonacci, below, with  a = 18  and  M = (2, 17, 18).

Fibonacci.  1202.  Pp. 307‑309.  If   a₁ + a₂ + a₃  =  a,   he uses  M = (2, a‑1, a)  and subtracts from  a²  to obtain   (a‑2) a₁ + a₂.   He also uses  M = (2, k, k+1)  and subtracts from a(k+1)  to get  (k‑1) a₁ + a₂.   These work whether the  a_i  are a permutation or not.  He uses  a = 6, k = 9,  hence  M = (2, 9, 10).  For permutations (?) of  2, 3, 4,  he uses  a = 9, k = 9.  No mnemonics.

Abbott Albert.  c1240.  Prob. 2, pp. 332‑333.  M = (2, 9, 10).  He then subtracts from  60  which produces  8a₁ + a₂.  This is identical to Fibonacci's first example, but Albert gives a complete table of all the partitions of  6  into  3  non‑negative summands and computes  2z + 9y + 10z  for each, showing that  P  determines the  a_i  if   a₁ + a₂ + a₃  =  6,   whether it is a permutation of 1, 2, 3 or not.  No mnemonics.  This can also be viewed as a form of 7.P.1.

BR.  c1305.

No. 38, pp. 56‑59.  Determine values of three dice from  a + b + c  and  2a + 8b + 9c,  so this is essentially Fibonacci's second case with  k = 8.

No. 39, pp. 58‑59.  Same using  a + b + c  and  3a + 9b + 10c.

No. 100, pp. 118‑119.  Determine values of two dice from  a + b  and  2a + 10b.

Munich 14684.  14C.  No. IX.  This is obscure, but is repeated more clearly as No. XIX.  M = (2, 9, 10).  IX is followed by a two line verse.  Curtze could make no sense of it, but I wonder if it might be a mnemonic??

AR.  c1450.  Prob. 269, p. 122, 180‑181, 227‑228.  M = (2, 9, 10).  Then   60 ‑ P  =  8a₁ + a₂.   Vogel cites several earlier appearances, but mentions no mnemonics.  ??NYS.

Chuquet.  1484.  Prob. 159.  M = (1, 2, 4).  Starts with  24,  but  6  are used to label the people, so this is really subtracting from  18.  Table, but no mnemonic.  FHM 232.

Tagliente.  Libro de Abaco.  (1515).  1541.  Prob. 142, f. 63v.  x*2+5*5+10+y*10+z  =  350 + 100x + 10y + z.

Cardan.  Practica Arithmetice.  1539.  Chap. 61, section 18, f. T.iv.v (p. 113).  Mentions divination of a permutation of three things by use of  18  counters, so this is probably  M = (1, 2, 4),  subtracting from  18,  as in Chuquet, Baker, etc.

Baker.  Well Spring of Sciences.  1562?  Prob. 4,  1580?: ff. 197r‑198r;  1646: pp. 310-312;  1670: pp. 354-355.  M = (1, 2, 4),  subtracting from  18.  Vowel mnemonic:  Angeli, Beati, Taliter, Messias, Israel, Pietas  with explanatory table.

John Wecker.  Op. cit. in 7.L.3.  (1582), 1660.  Book XVI _ Of the Secrets of Sciences: Chap. 20 _ Of Secrets in Arithmetick: To discover to one a thing that is hid, pp. 289‑290.  M = (1, 2, 4).  No mnemonics.  Cites Gemma Frisius, ??NYS.

Prévost.  Clever and Pleasant Inventions.  (1584), 1998. 

Pp. 177-180.  M = (1, 2, 4),  subtracting from 18.  No mnemonics. 

Pp. 180-182.  M = (1, 2, 4),  subtracting from 23.  No mnemonics. 

Bachet.  Problemes.  1612.  Prob. XXII: De trois choses et de trois personnes proposées deviner quelle chose aura été prise par chaque personne, 1612: 115-126.  Prob. XXV,  1624: 187-198;  1884: 127‑134.  M = (1, 2, 4).  Vowel mnemonic:  Par fer  César  jadis  devint  si grand  prince.  Gives a four person version,  S = 78,  M = (1, 4, 16, 0), referring to Forcadel as giving an erroneous method.  Labosne says Diego Palomino (1599) has studied the four person case.  [This must be Jacobo Palomino; Liber de mutatione aeris in quo assidua et mirabilis mutationis temporum historia cum suis caussis enarratur. _ Fragmentum quodam ex libro de inventionibus scientiarum; Madrid, 1597 or 1599, ??NYR.]  Labosne adds some explanation, another mnemonic:  Avec  éclat  L'Aï  brillant  devint  libre,  and expands on Bachet's work on the case of four objects, but eliminates reference to Forcadel. 

van Etten.  1624.  Prob. 8 (8), pp. 9‑11 (19‑22).  M = (1, 2, 4).  Vowel mnemonics:  Salve  certa  anima  semita  vita  quies;   Par fer  Cesar  Iadis  Devint  si grand  Prince.  Henrion's 1630 Notte, pp. 10‑11, says that Bachet has extended it to  4  objects.

Ozanam.  1694.  Prob. 28, 1696: 83;  1708: 74.  Prob. 32: 1725: 217-218.  Prob. 10, 1778: 154-155;  1803: 154-155;  1814: 136-137.  Prob. 9, 1840: 70.  M = (6, 4, 3).  No mnemonics.

Ozanam.  1725.  Prob. 46, 1725: 250-253.  Prob. 12, 1778: 158-161;  1803: 159-161;  1814: 140-142.  Prob. 11, 1840: 72.  M = (1, 2, 4).  Par fer  Cesar  jadis  devint  si grand  Prince   and   Salve certa anima semita vita quies.  1778 et seq. has  César  and  animæ.

Minguét.  Engaños.  1733.  Pp. 176-180 (1755: 127-129; not noticed in 1822).  M = (1, 2, 4)  as in Chuquet.  Vowel mnemonic:  Aperi,  Premati,  Magister,  Nihil,  Femina,  Vispane,  Vispena.

Alberti.  1747.  Part 2, p. ?? (69).  M = (6, 4, 3)  translated from Ozanam, 1725, prob. 32.

Alberti.  1747.  Part 2, pp. ?? (76-77).  M = (1, 2, 4)  as in Ozanam, 1725, prob. 46, but he first gives an Italian mnemonic:  Aperì  Prelati  Magister  Camille  Perina  Quid habes  Ribera.  He explains the usage using  4 = Camille  as example, but later notes that  4  never occurs!  Then gives  Salva certa anima semita vita quies;  Perfer Cesar Jadis devint sigrand Prince;  Pare ella ai segni; Vita, Piè.

Les Amusemens.  1749.  Prob. 13, pp. 134-135: Les trois Bijoux.  M = (1, 2, 4).  Par fer,  César  jadis  devint  si grand  Prince.

Hooper.  Rational Recreations.  Op. cit. in 4.A.1.  1774.  Recreation IX: The confederate counters, pp. 34-36.  M = (1, 2, 4).  Par fer  Cesar  jadis  devint  si grand  prince.

Badcock.  Philosophical Recreations, or, Winter Amusements.  [1820].  Pp. 85-86, no. 131: Three persons having each chosen privately one out of three things, to tell them which they have chosen.  Salve  certa  anima  semita  vita  quies.

Manuel des Sorciers.  1825.  ??NX 

Pp. 35-37, art. 12.  Salve  Certa  Anima  Semita  Vita  Quies.

Pp. 38-39, art. 13.  Extends to four objects.  Start with  88  counters and give  1, 2, 3, 4  to the people, leaving  78.  Then tell them to take  1, 4, 16, 0  times more, so this is the same as Bachet's four person version.

Endless Amusement II.  1826? 

P. 115:  "Three things being privately distributed to three Persons, ...."  c= Badcock.

P. 179:  "Three Cards being presented to Three Persons, ...."  M = (6, 4, 3)  using the inverse permutation.  No mnemonic.

Young Man's Book.  1839.  Pp. 198-199.  Identical to Endless Amusement II, p. 179.

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Parlour magic, no. xii, pp. 196-198 (1868: 209-210): How to discover the possessors of any articles taken from the table during your absence.  Salve  cesta  animæ  semita  vita  quies.

Magician's Own Book.  1857.

Prob. 36: The infallible prophet, pp. 23-24.  Mnemonics:  Salve  certa  animae  semita  vita  quies;   Par fer  Cesar  jadis  devint  si grand  prince.

The three graces, pp. 218-221.  Mnemonics:  James  Easy  admires now  reigning  with a  bride;   Anger,  fear,  pain, may  be hid  with a  smile;  Graceful  Emma,  charming she  reigns  in all  circles;   Brave  dashing  sea, like a  giant  revives  itself;   Salve ....

Boy's Own Conjuring Book.  1860.

Prob. 35: The infallible prophet, pp. 35-36.  Identical to Magician's Own Book.

The three graces, pp. 188-190.  Identical to Magician's Own Book.

Boy's Own Book.  Divination by cards.  1868: 637-638.  Basically  M = (1, 2, 4)  but set up as  (2, 3, 5).  Vowel mnemonic:  Par fer  César  jadis  devint  si grand  prince.

Magician's Own Book (UK version).  1871.  To tell which article each of three persons took, pp. 35-36.  Mnemonics:  Take her  certain  anise  seedlings  Ida  quince;  Salve  certa  animæe  servita  vita  quies;   Par-fer  Cæsar  jadis  devint  si-grand  prince.

Hanky Panky.  1872.  A new three-card trick, pp. 256-257.  M = (6, 4, 3),  which is equivalent to  (4, 2, 1).

Bachet-Labosne.  Problemes.  3rd ed., 1874.  Supp. prob. VII, 1884: 192-193.  M = (6, 4, 3),  which is the same as  (4, 2, 1).  No mnemonics.

Berkeley & Rowland.  Card Tricks and Puzzles.  1892.  Simple Tricks by Calculation No. I: To tell three persons which card each one has chosen, pp. 32-33.  M = (4, 2, 1).

É. Ducret.  Récréations Mathématiques.  Op. cit. in 4.A.1.  1892?  Pp. 107-?? (didn't xerox the following pages): Les trois Bijoux.  Seems to be usual form.

Lucas.  L'Arithmétique Amusante.  1895.  Prob. VIII-X, pp. 21-25.  M = (1, 2, 4)  with result subtracted from  18:   Par fer  César  jadis  devint  si grand  prince;   Avec  éclat  l'Aï  brillant  devint  libre.   M = (0, 1, 3)  with result subtracted from  12:   Il a  jadis  brillé  dans ce  petit  État. 

Meyer.  Big Fun Book.  1940.  The picture card trick, pp. 509-510.  Uses  K, Q, J.  Forms  P  =  (2a₁-1)m₁ + (2a₂-1)m₂ + (2a₃-1)m₃   for multipliers  (m_i) = (2, 3, 6).  Then examines the remainder of the deck, which has  49 - P = R  cards.  2 - [R/11]  is the person holding the  J.  A counting process, which is erroneously described, gives the person with the  Q  as congruent to  R + [R/11] (mod 3).  In fact  3 - R(mod 3)  is the position of the  K.

Gardner.  MM&M.  1956.  The purloined objects, pp. 57-59.  Gives several mnemonics using consonants.  E.g. when the objects are  Toothpick, Lipstick and Ring,  use:  tailor  altar  trail  alert  rattle  relate.  If the objects are denoted by  A, B, C,  use:  Abie's  bank  account  becomes  cash  club;  or if they are denoted  Small, Medium, Large,  use:  Sam  moves  slowly  (since)  mule  lost  limb.

 

          7.AP. KNOWING SUM VS KNOWING PRODUCT

 

          Two persons are told the sum and product of two integers.  They then have some conversation such as:  "I don't know what the numbers are."  "I knew that."  "I now know what they are."  "So do I."  What were the numbers?  This seems to be a recent problem and I have only a few references.  There are older versions, often called census-taker problems, where one knows the sum and product of three numbers (usually ages), but needs more information (often whether there is an oldest, which eliminates twins).

          Thanks to Leroy F. Meyers for many of these references.

 

W. T. Williams & G. H. Savage.  The Penguin Problems Book.  Penguin, 1940.  No. 96: The church afloat, pp. 53 & 135.  Three ages:  product = 840,  sum is twice the curate's age.  This is insufficient, but whether the eldest is older or younger than the vicar is sufficient to decide.

Lester R. Ford, proposer.  Problem E776.  AMM 54 (1947) 339 & 55:3 (Mar 1948) 159-160.  ??NYS/NX _ proposal is quoted and the location of the solution given in the Otto Dunkel Memorial Problem Book (= AMM 64:7, part II, (Aug-Sep 1957) 61 & 89) and the location is more specifically given by Meyers & See.  Four families of different sizes, not enough to form two baseball teams (i.e. the total is  < 18), and the product of the numbers is the host's house number.  The guest says he needs more information _ does the smallest family have just one child?  The host's answer allows him to determine the numbers _ what were they?

Meyers writes that he first heard of the census-taker problem in 1951.

AMM problem E1126, 1954?.  ??NYS.

W. A. Hockings, proposer; A. R. Hyde, solver.  Problem E1156 _ The dimensions of Jones's ranch.  AMM 62 (1955) 181 (??NX) & 63:1 (Jan 1956) 39-42.  This is a continuation of E1126 and involves four men and their ranches.

Hubert Phillips.  My Best Puzzles in Mathematics.  Dover, 1961.  Prob. 87: The professor's daughter, pp. 48 & 101.  Youngest daughter is at least three.  The product of their ages is  1200  and the sum is ten less than the wife's age.  Visitor computes and then makes two wrong guesses as to the age of the youngest daughter.  How old is the wife?  The fact that the visitor made two wrong guesses means there must be at least three sets of ages, all  ³ 3,  with product  1200  and the same sum, and indeed there is just one such situation, and this has three daughters.  Allowing younger children permits some more complicated possibilities since we have 

                              1 + 3 + 20 + 20  =  2 + 2 + 10 + 30  =  3 + 16 + 25  =  4 + 10 + 30 

          and               2 + 2 + 15 + 20  =  4 + 15 + 20  =  5 + 10 + 24  =  6 + 8 + 25. 

          (Phillips had most of these problems in his newspaper and magazine columns so it is likely that this will turn up in the period 1930-1950.)

M. H. Greenblatt.  Mathematical Entertainments, op. cit. in 6.U.2, 1968.  Chap. 1: "Census-taker" problems, pp. 1-7.  Says he believes these problems came from some wartime project at MIT.  Discusses three similar types.

                    1: The neighborhood census.  Product of three ages is  1296  and the sum is the house number.  Census taker asks if any of them are older than than the informant, who says 'no' and then the census taker knows the ages.

                    2: The priest and the banker.  Product of three ages of ladies with the banker is  2450  and the sum is the same as the priest's age.  The priest says this is insufficient and asks if any of the ladies is as old as the banker.  When he says 'no', the priest knows all the ages.  This even determines the banker's age!

                    3: The three Martians.  Product of three ages is  1252  and the sum is the age of the informant Martian's father.  The interrogator says this is insufficient and asks if any of the Martians is as old as the informant.  When he answers 'no', the interrogator knows the ages.  When you start on the problem, you find that  1252 = 2²·313  is not a suitable number.  However, the problem shows a drawing of the Martian and one see he only has three fingers on each hand!  Interpreting  1252  as a base  6  number gives the decimal  320  and the problem proceeds as before.

Gardner.  SA (Nov & Dec 1970)  c= Wheels, Chap. 3, prob. 10: The child with the wart.  Supplied by Mel Stover.  The product of the ages of three children is  36  and the sum is the questioner's house number.  When the questioner says the information is insufficient, the father says the oldest child has a wart, which is sufficient to determine the ages.  See Meyers & See for a generalization.

A. K. Austin.  A calculus for know/don't know problems.  MM 49:1 (Jan 1976) 12-14.  He develops a set-theoretic calculus for systematically solving problems involving spots on foreheads, etc. (see 9.D), including problems similar to the present section.  He gives the following problem considered by Conway and Patterson, but apparently unpublished.  Two persons each have a card on their back bearing a positive integer, visible to the other person, but not to the person.  They are told that the sum of the two integers is  6  or  7.  They are then asked, in turn, to state whether they know what their own integer is.  If they both have  3,  what is the sequence of responses?  Austin finds there are five 'no' answers, then the answers alternate  yes, no.

David J. Sprows, proposer;  Problem Solving Group, Bern, solver.  Prob. 977 _ Mr. P. and Ms. S.  MM 49:2 (Mar 1976) 96 & 50:5 (Nov 1977) 268.  P & S are given the product and sum of two integers greater than one, but the sum is  £ 100.  P says he doesn't know the numbers.  S says she knew that.  P replies that he now knows the values and S responds that so does she.  The unique answer is  4, 13.  Editor cites the above AMM problems, though they are not quite the same type.  See the discussion below.

Gardner.  SA 241:6 (Dec 1979) 20-24.  Problem 1: The impossible problem.  Gives a version of Sprows' problem and says the problem was sent by Mel Stover and had been circulating for a year or two.  This version assumes the numbers are greater than  1  but at most  20,  which gives the unique solution  4, 13.  However Gardner asserts that the solution remains the same if the bound is increased to  100 _ but I think there are further answers, e.g.  4, 61,  see the discussion below.  Stover says a computer program has checked and found no further solutions up to  2,000,000  and it may be that there is no further solution when the upper bound is removed _ this is a mistake of some sort, see the discussion below.  Further, Kiltinen & Young say they had a letter from Gardner conjecturing that there are infinitely many solutions.

John O. Kiltinen & Peter B. Young.  Goldbach, Lemoine, and a know/don't know problem.  MM 58:4 (Sep 1985) 195-203.  This discusses the Sprows problem, without the bound on  S  and various generalizations involving more stages of conversation.  This leads to use of Goldbach's Conjecture and a conjecture of Lemoine that every odd number  ³ 7  can be expressed as  2p + q,  where  p, q  are odd primes.

Friedrich Wille.  Humor in der Mathematik.  Vandenhoek & Ruprecht, Göttingen, (1984), 3rd ed., 1987.  Paul und Simon, pp. 62 & 121‑122.  This has solution  4, 13.  Note to the solution says there was much correspondence after the 1st ed., leading to a new solution and the computation of further examples with  Sum =  17,  65,  89,  127,  137,  163,  179,  185,  191,  233,  247,  269,  305.

 

                    I recently was sent a version of the Sprows problem, without a bound on  S,  by Adrian Seville and solved it before checking this section.  Dr. S (= Simon) and Dr. P (= Paul) are given the sum and product of two integers greater than  1.  Dr. S says he doesn't know the numbers, but he can tell that Dr. P cannot determine the numbers.  After a bit, Dr. P says he now knows the numbers.  After a bit more, Dr. S says he now knows the numbers also.  The version I received implies there is a unique solution, but there are more.  Wille's discussion notes that the possible values of the sum, after Dr. S's statement are the  {odd composites + 2},  which much simplifies the analysis.  I found some further values of  Sum:  343,  427,  457 and then extended considerably, finding:  547,  569,  583,  613,  637,  667,  673,  697,  733,  757,  779,  787,  817,  821,  853,  929,  967,  977,  989,  997,   giving  36  values less than  1000  and there are another  42  solutions between  1000  and  2000.  In looking at the earlier solutions, it appeared that one of the two numbers was always a power of 2, with the other number being odd, but for  S = 757,  P = 111756,  one has the numbers being  202  and  556.  Five more counterexamples to the initial appearance occur up to  2000.

                    While doing this investigation, I wondered what would happen if the conditon 'greater than  1'  was reduced to  'positive'.  After some calculation, it became apparent and is provable that all solutions have  P = S - 1  with the numbers being  1  and  S - 1.  Solutions occur for  S =  5,  9,  10,  16,  28,  33,  34,  36,  46,  50,  52,  66,  78,  82,  88,  92,  96,   giving  17  values less than  100,  and there are another  87  values between  100  and  1000,  making a total of  104  values less than  1000,  which is where I stopped.  If one assumes Goldbach's Conjecture, or part of it, one can show there are infinitely many solutions of the form  S = pq + 1,  where  p, q  are primes such that  p + q = r + 1  for a prime  r.  However, I have not been able to see if there are infinitely many solutions of the original form of the problem.

                    I later was referred to the MM problem 977 and was surprised to see that the solution is unique when  S £ 100  is imposed _ offhand, one might expect  S = 65 and 89  to also be solutions in this case, but the additional knowledge about  S  affects the intermediate stages of the deduction.  Rerunning an early version of my program with additional printout, I find that one needs  S < 107  to prevent  S = 65  being a solution.  I then found Gardner's 1979 version and examination of the same data indicates that  S = 65  is a solution when the numbers are bounded by  100.  All in all, I find the presence of a bound has subtle effects and I find it unsatisfying.

 

Tim Sole.  The Ticket to Heaven and Other Superior Puzzles.  Penguin, 1988.  Standard and Poor, pp. 116 & 126‑128.  Solution is  2, 15.

Leroy F. Meyers & Richard See.  The census-taker problem.  MM 63:2 (Apr 1990) 86-88.  The product of the three ages in the next house is  1296  and the sum is the house number.  The census-taker then asks if any of those people is older than the respondent.  When he says 'yes', the census-taker says he knows the ages.  This is a slight variant on Greenblatt's type 1.  Authors investigate what other values,  N,  can be used instead of  1296.  That is, we want  N = ABC = DEF  with  A + B + C = D + E + F  with just two such triples  A, B, C;  D, E, F.  They determine the simple forms of such  N  and list the  45  examples  £ 1296.  The first few are:  36,  40,  72,  96,  126,  176,  200,  234,  252,  280,  297,  320,  408,  520,  550,  576,  588,  600.  They give a number of references to similar questions.

 

          7.AQ. NUMBERS IN ALPHABETIC ORDER

 

          I recall the problem of figuring out the reason for the sequence:  8, 5, 4, 9, 1, 7, 6, 3, 2  from 1956 or 1957.

 

??  Alphabetic Number Tables, 0 ‑ 1000.  MIT, Cambridge, Massachusetts, 1972.  ??NYS.

"Raja" [= Richard & Josephine Andree].  Puzzle Potpourri #3.  Raja Books, Norman, Oklahoma, 1976.  No. 17: volumes  1, 2, ..., 12  shelved alphabetically.

Harvey & Robert Dubner.  A tabulation of the prime numbers in the range of one to one‑thousand, in English and in Roman numerals, in alphabetic order.  JRM 24 (1992) 89-93.

 

          7.AR. 1089

 

          New section.  Are there older examples?? 

          The basic process is to take a three digit number, find the difference between it and its reversal, then add that result to its reversal and one gets  1089.  The items by Meyer and Langford note that  9 * 1089 = 9801,  so this is connected to 7.AH.

 

??  Educational Times Reprints 53 (1890) 78.  ??NYS _ Ball, MRE, 1st ed., 1892, p. 9 cites this for the monetary version of the problem.

Ball.  MRE, 1st ed., 1892. 

P. 9.  Monetary version getting  £12 18s 11d.  "The rule can be generalized to cover any system of monetary units."  This version is dropped in the 12th ed., 1974.

P. 13.  Notes that a number minus its reversal is divisible by  9.

Berkeley & Rowland.  Card Tricks and Puzzles.  1892.  Card Puzzles No. X: A subtraction sum, p. 10.  A  3-digit number minus its reversal has the form  a9c,  where  a + c = 9.

Somerville Gibney.  An arithmetical flourish for drawing-room shows.  The Boy's Own Paper 15 (No. 734) (4 Feb 1893) 299.  Monetary version, getting  £12 18s 11d.  "An arithmetical curiosity which is quite new ...."

Somerville Gibney.  Re The arithmetical flourish for drawing-room shows.  The Boy's Own Paper 15 (No. 750) (27 May 1893) 556.  Explains why it works.  Correspondents from various places do it:  in francs and centimes, getting  99 99;  in thalers and grosschen  (30  grosschen = a thaler), getting  29 29;  and  in avoirdupois, getting  28cwt 2qr 27lb.

Ball‑FitzPatrick.  1st ed., 1898, pp. 14‑15.  Shows the property of  1089.  A remark of the translator shows the answer in base  b  is   (b‑1) * (b+1)².

L. Carroll.  The Lewis Carroll Picture Book, op. cit. in 5.B, 1899, p. 269 (Collins 194).  Monetary version, giving  £12 18s 11d.  Doesn't note problems when numbers of pounds and pence are equal.  Collingwood says he believes this was invented by Carroll.

Clark.  Mental Nuts.  1904: no. 81; 1916: no. 54.  Mental telegraphy.  As in Berkeley & Rowland.

Ball.  MRE, 4th ed., 1905, p. 9.  Adds a section on  1089  just before the monetary version.

Laisant.  Op. cit. in 6.P.1.  1906.  Chap. 18: Opérations curieuses: No. 1, p. 43.  1089.  Notes that the two end digits must be different.

E. N. Barisien.  ??  Suppl. al Periodico di Mat. 13 (1909) 20-21.  ??NYS _ cited by Dickson I 462, item 53, but the interesting material is cited in the following.

E. Nannei.  ??  Suppl. al Periodico di Mat. 14 (1910/11) 17-20.  ??NYS _ cited by Dickson I 462, item 55, as treating a problem of Barisen.  Takes a  6-digit number, subtracts its reversal and then adds the reversal of that to itself.  Result is one of  13  values:   0,  9900,  ...,  1099989.

Adams.  Indoor Games.  1912.  Magic money, 354-355.  Monetary version getting  £12 18s 11d.

Clark.  Mental Nuts.  1916: no. 53.  Foretelling the answer.  Gives the rule and an example, saying the answer will always be  1089,  though he doesn't make any restriction on the original number.

Dudeney.  AM.  1917.  Prob. 33: A puzzle in reversals, pp. 5 & 151.  Says most people know that the money process leads to  £12 18s 11d  _ provided you start with less than 12 in the  £  place and the number of pounds exceeds the number of pence.  If the number of pounds can be  12  or more, then what is the smallest sum for which the process fails and the largest sum for which it works?

T. O'Conor Sloane.  Rapid Arithmetic.  Quick and Special Methods in Arithmetical Calculation Together with a Collection of Puzzles and Curiosities of Numbers.  Van Nostrand, 1922.  [Combined into: T. O'Conor Sloane, J. E. Thompson & H. E. Licks; Speed and Fun With Figures; Van Nostrand, 1939.]  A mystery in money, pp. 178-180.  Uses US money to get  $10.89  and also gives the English result  £12 18s 11d.

Peano.  Giochi. 1924.

Prob. 48, p. 12.  1089.

Prob. 49, pp. 12-13.  As in Berkeley & Rowland.

A. B. Nordmann.  One Hundred More Parlour Tricks and Problems.  Wells, Gardner, Darton & Co., London, nd [1927 _ BMC].  No. 60: Mental arithmetic trick, pp. 51-52.  Monetary version.

Kraitchik.  Math. des Jeux.  Op. cit. in 4.A.2.  1930.  Chap. IV, no. 7, p. 55.  Notes that the first result is a multiple of  99,  but doesn't find  1089.  Not in his Math. Recreations.

Perelman.  1934??  Guessing a number without asking anything.  FFF (1934).  Not in 1957 ed.  1979: prob. 14, pp. 33-35.  = MCBF (1937), prob. 14, pp. 32-33.  Basic property of  1089.

Rohrbough.  Brain Resters and Testers.  c1935.  Brain Teaser, p. 15.  Assumes digits are in descending order.

Freda Holmdahl, ed.  The Rainy Day Book.  Nelson, Edinburgh, 1936.  Figure tricks, pp. 196‑197.  Monetary version.

Depew.  Cokesbury Game Book.  1939.  Descending numbers, p. 205.  1089.

"Willane".  Willane's Wizardry.  Academy of Recorded Crafts, Arts and Sciences, Croydon, 1947.  Curious figures, p. 45.  Version with English money, with  £  less than  12.

P. M. Seeviour & M. Keates.  Note 3077:  Fibonacci again.  MG 48 (No. 363) (Feb 1964) 78‑79.  Cites Kraitchik, Math. des Jeux.  Determines all the possible results of the process for  n‑digit numbers in base  b,  as being  (b² ‑ 1)  times certain  n-digit numbers formed of  0s  and  1s,  corresponding to the borrows in the subtraction.  They find the number  t(n)  of such values is given by  t(2n)  =  t(2n+1)  =  F_2n+1  where  F_m  are the Fibonacci numbers.  They do this via the case when the first and last digits are distinct for which it is stated that a certain recurrence holds, but this is not immediate _ I spent about half a day before I could see it and it would still require some work to make a careful proof.  For convenience, I list the first solutions for the case of different end digits.

                    n = 2.  99.

                    n = 3.  1089.

                    n = 4.  10989,  9999,  10890.

                    n = 5.  109989,  99099,  109890.

                    n = 6.  1098900,  1099989,  1090089,  1099890,  1089990,  999999,  990099,  991089.

                    n = 7.  10998900,  10999989,  10891089,  10999890,  10890990,  10008999,  9900099,  9901089.

C. Dudley Langford.  Note 3102:  A party puzzle extended (Cf Note 3077).  MG 48 (No. 366) (Dec 1964) 432.  If  a + b = 10,  then  1089*a  and  1089*b  are reversals.  1089  can be replaced by  99 * 11...1  and he asserts that these, up to trailing  0s,  are the numbers obtained in Note 3077.  However, his numbers are:  99,  1089,  10989,  109989,  1099989, ..., 109...989, ...,  and these do not immediately yield the above solutions.

Jerome S. Meyer.  Arithmetricks.  Scholastic Book Services, NY, 1965.  The  2178  trick, pp. 3-4.  Gets  1089,  but he doubles the result at the end to produce  2178.

Michael Holt.  What is the New Maths?  Anthony Blond, London, 1967.  Pp. 88-89.  Gives a four digit version, saying to choose the least significant two digits as  < 5  and the most significant as  ³ 5,  then the result is always  10890.

Walter Gibson.  Big Book of Magic for All Ages.  Kaye & Ward, Kingswood, Surrey, 1982.  Magic Math  &  Super Math, pp. 108-109.  After doing  1089,  he considers doing the same process with five digit numbers.  For convenience, he assumes the digits are all distinct.  Then only two possible answers can arise:  99099  or  109890.  No proof given.

 

          7.AS.  CIGARETTE BUTTS

 

          New section.  There must be older examples.  The usual form uses  b  butts to make a cigarette and the person has found  B = b²  butts which give him  b+1  smokes.  The Scotts are the only ones to realise that  B  can more generally be such that  b-1  divides  B-1,  giving  (B‑1)/(b-1)  smokes.  I have just found the general result that the number of smokes is  ë(B‑1)/(b-1)û.  Taking  b = 2  gives the result that a knockout tennis tournament with  B  players has  B-1  matches.

 

Adams.  Puzzles That Everyone Can Do.  1931.  Prob. 192, pp. 74 & 158: Scrooge the miser.  Starts with  125  cigarettes, but saves all his butts, using  5  to make a cigarette, so he gets  31  further cigarettes.

Rudin.  1936.  No. 49, pp. 16 & 85.  36  butts, using  6  to make a cigarette.

J. R. Evans.  The Junior Week‑End Book.  Op. cit. in 6.AF.  1939.  Prob. 47, pp. 266 & 272.  Tramp has  49  butts and uses seven to make a cigarette.  How many cigarettes can he make?

Harold Hart.  The World's Best Puzzles.  Home Recreation Library, NY, 1943.  Smoke gets in your eyes, pp. 22 & 54.  125  butts and five make a cigarette.

Jules Leopold.  At Ease!  Op. cit. in 4.A.2.  1943.  Iffs and Butts, pp. 10-11 & 196.  36  butts, six butts to a cigarette.

Sullivan.  Unusual.  1943.  Prob. 9: A bum cigarette.  As in Rudin.

Jonathan Always.  Puzzles to Puzzle You.  Op. cit. in 5.K.2.  1965.  No. 131: Cigarettes this way, pp. 41 & 89.  As in Evans.

Joseph & Lenore Scott.  Master Mind Brain Teasers.  Op. cit. in 5.E.  1973.  Hard times made easier, pp. 35-36.  41  butts, six butts to a cigarette.

David Singmaster.  Cigarette butts.  MS 31:2 (1998/9) 40.  The number of smokes is  ë(B‑1)/(b-1)û.  Taking  b = 2  gives the result that a knockout tennis tournament with  B  players has  B-1  matches.

 

          7.AT. BOOKWORM'S DISTANCE

 

          New section.  This must appear in the 19C??

 

Loyd.  Cyclopedia.  1914.  Pp. 327 & 383.  = SLAHP, Timing a bookworm, pp. 69 & 112.  2  volumes.

Dudeney.  AM.  1917.  Prob. 420: The industrious bookworm, pp. 143-144 & 248-249.  3  volumes.

Peano.  Giochi.  1924.  Prob. 10, p. 4.  3  volumes.

King.  Best 100.  1927.  No. 3, pp. 8 & 39.  = Foulsham's, no. 1, pp. 5 & 10.  3  volumes.

William P. Keasby.  The Big Trick and Puzzle Book.  Whitman Publishing, Racine, Wisconsin, 1929.  Pp. 37 & 63.

Streeter & Hoehn.  Op. cit. in 7.AE.  Vol. 1, 1932, p. 300, no. 10:  "Brain twister".

Evelyn August.  The Black-Out Book.  Op. cit. in 5.X.1.  1939.  Bookworm's pilgrimage, pp. 163 & 215.  4  volumes.

McKay.  At Home Tonight.  1940.  Prob. 18: The bookworm, pp. 66 & 79-80.  2  volumes.

The Home Book of Quizzes, Games and Jokes.  Op. cit. in 4.B.1, 1941.  P. 148, prob. 4.  2  volumes.

Jules Leopold.  At Ease!  Op. cit. in 4.A.2.  1943.  Way of a worm, pp. 9 & 195.

Owen Grant.  Popular Party Games.  Universal, London, nd [1940s?].  Prob. 10, pp. 37 & 50.  3  volumes.

Birtwistle.  Math. Puzzles & Perplexities.  1971.  Second, pp. 145 & 196.  4  volumes.

 

          7.AU. NUMBER OF CUTS TO MAKE  N  PIECES

 

          New section.  A typical problem is that it takes twice as long to make three pieces as it does to make two pieces.  This will include other situations where one has to change from  n  to  n-1.

          If one can overlay the material, then one can reduce the number of cuts.  If one can fold the material, then one can reduce to one cut as in 6.BG.

          I've just added the problem of cutting a cube into 27 pieces here _ this must be much older??

          See also 7.AV.

 

Alcuin.  9C.  Prob. 15: De homine.  "How many furrows has a man made ..., when he has made three turning at each end of the field?"  The Alcuin text gives six, but Bede, Folkerts and Folkerts & Gericke give seven.

Child.  Girl's Own Book.  Arithmetical puzzles.  1832: No. 10, pp. 171 & 179;  1833: No. 10, pp. 185 & 193 (answer numbered 8);  1839: No. 10, pp. 156 & 173;  1842: No. 10, pp. 284 & 291;  1876: No. 8, pp. 232 & 244.  "If you cut thirty yards of cloth into one-yard pieces, and cut one yard every day, how long will it take you?"  The 1839, 1842 & 1876 texts omit the hyphen in one-yard.  I didn't copy the exact text from the 1832.

The Sociable.  1858.  Prob. 46: A dozen quibbles: part 9, pp. 300 & 318.  "If you cut thirty yards of cloth into one yard pieces, and cut one yard every day, how long will it take?"  = Book of 500 Puzzles, 1859, prob. 46: part 9, pp. 18 & 36.  c= Depew; Cokesbury Game Book; 1939; Cutting up, p. 216.  c= Magician's Own Book (UK version), 1871, Paradoxes [no. 5], p. 38.  = Wehman, New Book of 200 Puzzles, 1908, p. 50.

Hoffmann.  1893.  Chap. IX, no. 24: The draper's puzzle, pp. 318 & 326.

Lucas.  L'Arithmétique Amusante.  1895.  Prob. XII: La coupe du tailleur, p. 26.  How long to cut a piece of length  16  into lengths of  2?  [This makes the problem more deceptive.]

Benson.  1904.  The tailor's puzzle, p. 227.

King.  Best 100.  1927.  No. 24, pp. 15 & 44.  Cut  90  yards of cloth into  90  lengths.

Meyer.  Big Fun Book.  1940. 

Sissy's scissors, pp. 120 & 739.  Three sheets of paper,  20 x 40,  to be cut into  2 x 20,  without folding.  He does it in  19  cuts by stacking the three sheets and cutting the side of  40  into  20  pieces.  One can easily reduce this to  10  by first cutting the  20 x 40  into two  20 x 20,  stacking them and then cutting the stack into  10  parts.  One can reduce further by cutting the  10  edge in the middle, getting  5, 5.  Stack these and cut the  5  into  1, 2, 2  with two cuts.  Then stack the  2s  and cut once more.  This takes a total of  5  cuts, which seems minimal to me.  There is a problem in that the last  2 x 10  pack is  24 sheets thick! 

No. 9, pp. 162 & 752.  Draw four equidistant horizontal lines and then four equidistant verticals.  How many squares are formed?  This gives a  3 x 3  array of squares, but he counts all sizes of squares, as in 5.X.2, getting  9 + 4 + 1 = 14.

Birtwistle.  Math. Puzzles & Perplexities.  1971.  Fifth, pp. 58 & 190.  How many cuts to make  27  cubes from a cube?

Putnam.  Puzzle Fun.  1978.  Nos. 93-96: Cube and  27  cubes, pp. 13 & 37.  How many cuts to make  27  cubes from a cube?  Other questions deal with colourings.

Barbara Lee.  Six short problems.  M500 162 (?? 1998)  ??NYS _ cited by solver.  A, J. Welton.  Solutions to 'Six short problems'.  M500 163 (Aug 1998) 12-13.  No. 3: "A clock strikes six in 5 seconds.  How long does it take to strike twelve?"  "Six hours and five seconds."

 

          7.AV. HOW LONG TO STRIKE TWELVE?

 

          New section.  I have now started adding rate problems of this type.

          See also 7.AU.

 

King.  Best 100.  1927.  No. 36, pp. 19 & 47.  If a clock takes  8  sec to strike  8,  how long does it take to strike  12?

Adams.  Puzzles That Everyone Can Do.  1931.  Prob. 186, pp. 72 & 157: Big Ben.  "If it takes Big Ben ten seconds to strike five o'clock, how long will it take to strike twelve o'clock?"

Evelyn August.  The Black-Out Book.  Op. cit. in 5.X.1.  1939.  Is your brain working? _ no. 2, pp. 148 & 215.  If Big Ben takes  4  seconds to strike  4,  how long does it take to strike midnight?

Depew.  Cokesbury Game Book.  1939.  The clock, p. 217.  If it takes  7  seconds to strike  7,  how long does it take to strike  11.

Sullivan.  Unusual.  1943.  Prob. 25: Caution.  If it takes  8  minutes to pass  8  stop lights, how long does it take to pass  12?

W. A. Bagley.  Puzzle Pie.  Op. cit. in 5.D.5.  1944.  No. 25: The patriot's paradox, pp. 30-31.  60  shots in an hour is not the same as one shot per minute, since the latter requires  61  shots to be fired in a  60  minute interval from the first shot.

 

          7.AW.          28/7  =  13

 

          There are three (or more) ways of demonstrating this result _ e.g.  7  doesn't go into  2,  but it goes into  8  once, leaving us with  21  and  7  goes into  21  3  times.  The other ways are based on multiplying  13  by  7  and on  adding seven  13s.  This is the basis of a vaudeville routine which was used in the Abbott and Costello movie 'In the Navy', 1949.  I can only recall seeing it in the following.

 

Rohrbough.  Brain Resters and Testers.  c1935.  Tricks for the Toastmaster (no. 2), pp. 12-13. 

W. A. Bagley.  Paradox Pie.  Op. cit. in 6.BN.  1944.  No. 2: Financial skullduggery, pp. 7‑8.

William R. Ransom.  Op. cit. in 6.M.  1955.  Fake arithmetic, pp. 9‑11.  Shows there are  22  examples of this phoney arithmetic.

Robert Harbin.  Party Lines.  Op. cit. in 5.B.1.  1963.  The Christmas club, p. 62.

 

          7.AX. SUM  =  PRODUCT,  ETC.

 

          The archetypal example is to determine  A, B  such that  A + B  =  A * B,  usually with some additional condition.  I have not noticed early examples of such problems until recently reading Riese's Die Coss and later Muscarello, but I recall other later examples such as Briggs & Bryan.

 

Muscarello.  1478.  Ff. 69r-69v, p. 180.  A + B  =  A * B.  He gives examples:  6 & 6/5;  7 & 7/6;  8 & 8/7;  and says it continues.

Calandri.  Arimethrica.  1491.  F. 97v.  A + B  =  A * B.  He finds  5/2, 5/3.

Riese.  Die Coss.  1524.

No. 38, p. 45.  A + B  =  A/B,  with  A/B  =  3/2.

No. 41, p. 46.  A ‑ B  =  A/B,  with  A/B  =  4/3.

Ozanam.  1694.  Prob. 5, question 14, 1696: 13;  12.  Prob. 8, quest. 14, 1725: 27-28.  A + B  =  A * B.  Gives  (a+b)/a  and  (a+b)/b,  e.g.  5/2  &  5/3.

Eadon.  Repository.  1794. 

P. 299, no. 8.  A + B  =  A * B.  Gives  B  =  A/(A-1).

P. 299, no. 9.  A - B  =  A² - B².  Gives  B = 1-A.

P. 299, no. 10.  A + B  =  A² - B².  Gives  B = A-1.

Manuel des Sorciers.  1825.  ??NX  P. 85.  A + B  =  A * B.

Walter Taylor.  The Indian Juvenile Arithmetic ....  Op. cit. in 5.B.  1849.  Pp. 193-194.  Solves  A + B  =  A*B  as  A = (a+b)/a,  B = (a+b)/b  for any  a, b.

Briggs & Bryan.  The Tutorial Arithmetic, Part II.  Op. cit. in 7.H.  1898.  Exercises X, prob. 16, pp. 124 & 579.  A + B  =  A * B  =  A² ‑ B².

Nathan Altshiller Court.  Mathematics in Fun and in Earnest.  Op. cit. in 5.B.  1961.  Prob. e, pp. 189-190.  A - B  =  A/B.

Jerome S. Meyer.  Arithmetricks.  Scholastic Book Services, NY, 1965.  Pp. 35-36.  Considers all four cases.

          A * B  =  A + B.    (n+1)/n  and  n+1.

          A * B  =  A - B.    n  and  n/(n+1).

          A / B  =  A + B.    n + 1/(n+2)  and  (n+1)/(n+2).

          A / B  =  A - B.    n + 1/(n-2)  and  n-1.

Bronnie Cunningham.  Funny Business.  An Amazing Collection of Odd and Curious Facts with Some Jokes and Puzzles Too.  Puffin, 1978.  Pp. 37 & 142.  A - B  =  A/B  = 4.

 

          7.AY. SUM OF POWERS OF DIGITS

 

          New section.  A  PDI  (Perfect Digital Invariant) is an  n  digit number which is equal to the sum of the  k-th powers of its digits.  If  k = n,  it is called a  PPDI  (PluPerfect Digital Invariant).  More generally, one can consider the function  f_k(N) = sum of the  k-th powers of the digits of  N  (in base  10  usually)  and study its iterates.  It is not hard to see that  f_k(N) < N  from some point on, so the iterates must lead to cycles.  See 7.BB for other iterated functions of integers.

 

F. Hoppenot.  Courrier du "Sphinx".  Sphinx 7:4 (Apr 1937) 72.  Gives  153, 371, 407, 8208, 9474  as PPDIs.

M. Fistié.  Courrier du "Sphinx".  Sphinx 7:5 (May 1937) 87.  Adds  370  to Hoppenot's list and says these are all the PPDIs for  k = 3.

Ball.  MRE, 11th ed., 1939, p. 13.  Gives the four PPDI's for  k = 3:  153, 370, 371, 407,  citing Sphinx.

G. H. Hardy.  A Mathematician's Apology.  CUP, 1940.  "There are just four numbers (after  1)  which are the sums of the cubes of their digits ....  These are odd facts, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in them which appeals to a mathematician."  He cites Ball for them.

H. Steinhaus.  One Hundred Problems in Elementary Mathematics.  (As:  Sto Zada_, PWN _ Polish Scientific Publishers, Warsaw, 1958.)  Pergamon Press, 1963.  With a Foreword by M. Gardner, Basic Books, NY, 1964.  Prob. 2: An interesting property of numbers, pp. 11-12 & 55-58.  Considers iterates of  f₂(n).  He shows this always leads to a cycle and find there are two cycles,  1  of length  1  starting from  1  and  1  of length  8  starting from  4.

Kiyoshi Iseki and coworkers extended Steinhaus's idea to  k = 3, 4, 5  in 5 papers during 1960‑1963.  See Leveques's Reviews in Number Theory, A64-12 _ A64-16 for details.  For  k = 3,  there are  9  cycles  (5  of which have period  1:  1  and the four PPDI's for  k = 3).  For  k = 4,  there are  6  cycles  (4  of which have length  1:  1,  1634,  8208,  9474).  For  k = 5,  there are  15  cycles, the longest having length  28.  There are  5  cycles of length  1:  1,  54748,  93084,  92727,  194979.

Max Rumney.  Digital invariants.  RMM 12 (Dec 1962) 6-8.  Quotes Hardy.  General introduction to the ideas.  Interested in repeating the process of summing the  n-th powers of an  n-digit number, so his PDI is what is now called a PPDI.  He also considers cases where the process cycles through a few values, e.g. for the sum of the cubes of the digits, we get the cycle:  133,  055,  250,  133.  There are such cycles starting with  133,  136,  217,  919.

Gardner.  SA (Jan 1963) c= Magic Numbers, chap. 3.  Surveys what is known with numerous references up through 1973.  In particular, D. St. P. Barnard showed that the number of PPDIs is finite by noting that  61 * 9⁶¹ < 10⁶⁰,  so that a PPDI must have  < 61  digits. 

Harry L. Nelson.  More on PDI's.  Publication UCRL-7614, Univ. of California.  1 Dec 1963.  ??NYS _ cited by Schwarz and others below, who say it proves and improves Barnard's result and that it makes the distinction between PDI and PPDI as now generally used.

Azriel Rosenfeld, proposer;  Nathan J. Fine, solver.  Problem E1651 _ Sums of squares of digits.  AMM 71 (1964) 90  &  1042-1043.  "Prove that no multidigit integer is equal to the sum of the squares of its digits."

Joseph S. Madachy.  Op. cit. in 6.D.  1966.  Narcissistic numbers: Digital invariants, pp. 163‑165.  Based on Rumney and Nelson.  Carries on to consider other types of 'narcissistic numbers' which are numbers which are some function of their digits _ various forms are discussed on pp. 165-177.

Benjamin L. Schwarz.  Finiteness of a set of self-generating integers.  JRM 2:2 (Apr 1969) 79‑83.  Proves Barnard's result and improves it to  < 60  digits.

Benjamin L. Schwarz.  Finite bounds on digital invariants _ some conjectures.  JRM 3:2 (Apr 1970) 88-92.  General discussion of whether the number of PDIs is infinite.  He gives a heuristic argument that there are infinitely many.

George D. Poole.  Integers and the sum of the factorials of their digits.  MM 44:5 (Nov 1971) 278-279.  The only integers equal to the sum of the factorials of their digits are:  1,  2,  145,  40585.  He has to check up to  2,000,000  before a general method can be used.

Joseph S. Madachy.  Some new narcissistic numbers.  Fibonacci Quarterly 10:3 (Apr 1972) 295-298.  "[R]eports on various narcissistic numbers other than digital invariants."  Gives various forms and some references.

Benjamin L. Schwarz.  Self-generating integers.  MM 46:3 (May 1973) 158‑160.  Gives a simple condition for functions of the digits so that the number of PPDIs (= SGIs in his notation) must be finite.  This applies to the sum of the factorials of the digits.

Lionel E. Deimel Jr. & Michael T. Jones.  Finding pluperfect digital invariants:  techniques, results and observations.  JRM 14:2 (1981‑82) 87-108.  Cites work from 1962.  All PPDIs for bases  2  through  10  were found, using months of Data General MV/8000 time, checking primality for bases  < 10.  There are  89  PPDIs in base  10  (but this includes  0  as an example for exponent  1,  though it is not clearly a  1-digit number, so perhaps  88  is a better count). 

Mary T. Whalen & Gordon L. Miller.  Prime PPDI's.  JRM 25:2 (1993) 118‑123.  Says PPDIs are also known as Armstrong numbers.  The present authors have checked the above results and looked for examples of primes in base  10.  The base  10  results took a DEC 6310 just over  14  days and three primes (beyond the exponent  1  cases) were found.  Results were checked with Mathematica on a NeXT.  They give  88  PPDIs for base  10.

Tony Forbes.  Recurring digital invariants.  M500 165 (Dec 1998) 4-7.  After some comments and results on PDIs in base  10  and loops in the iteration process, he decides to examine base  3  and finds there are many examples and gives some very large examples.

 

          7.AZ. DIVINATION OF A PAIR OF CARDS FROM ITS ROWS

 

          A spectator mentally chooses a pair from ten pairs of cards.  You then lay them out in a  5  by  4  array and ask him to tell you in which rows his pair appears, whereupon you point out his pair.  This is done by laying out the cards in a particular pattern traditionally given by the Latin words:  MUTUS  DEDIT  NOMEN  COCIS.  In fact, such patterns are easily found for  r(r+1)/2  pairs laid out in  r  rows _ but finding a memorable pattern takes some doing.  Since the person's response can be either one row or two rows, the number of responses is  r + BC(r, 2)  =  r(r+1)/2  =  BC(r+1, 2).

          Unger,  Ball,  Ahrens  and  Indoor Tricks and Games  deal with triples.  With  r  rows, there are  r + BC(r, 2) + BC(r, 3)  =  r (r² + 5)/6  possible responses.  However, this does not always give a possible situation.  E.g., for  r = 2,  there are  3  possible responses, but one cannot layout  3  triples in two rows, and the same problem occurs when  r  is even.  For  r = 3,  there is an easy pattern:  AAADDEG  BBBDFFG  CCCEEFG  and I imagine it can be easily done whenever  r  is odd.  However, we are getting to more responses than we can handle and it might be easier to use fewer of them.  See Unger for an approach.

          This is actually a combinatorial problem and I will probably shift this section into Chapter 5.

 

Bachet.  Problemes.  1612.  Prob. XVI: De plusieurs cartes disposées en divers rangs deviner laquelle on aura pensée: 1612: 87-92.  Prob. 18, 1624: 143‑151;  1884: 72-83.  This problem covers several card divinations.  The present topic has been added to "cette seconde [impression de ce livre]".  He discusses the problem for  r  rows and shows the simplest pattern, but gives no mnemonics.  For  10  pairs, the simplest pattern has rows:  AABCD,  BEEFG,  CFHHI,  DGIJJ.

Wingate/Kersey.  1678?.  Prob. 5, pp. 539-542.  Does  10  and  15  pairs as in Bachet.

Hooper.  Rational Recreations.  Op. cit. in 4.A.1.  1774.  Recreation XXVII: The ten duplicates, pp. 70-71.  MUTUS  DEDIT  NOMEN  COCIS.

Endless Amusement; A Collection of Nearly 400 Entertaining Experiments ....  Thomas Boys, London, 1818?;  2nd ed., 1819?;  3rd ed., 1825.  P. 106: The ten duplicates.  As in Hooper.

Badcock.  Philosophical Recreations, or, Winter Amusements.  [1820].  Pp. 122-123, no. 184: The ten duplicates.  c= Endless Amusement.

Manuel des Sorciers.  1825.  Pp. 48-51, art. 21.  ??NX  MUTUS  NOMEN  DEDIT  COCIS.  Extends to 15 pairs, in a different way than Bachet, by bordering the usual array of letters with a left hand column  5, 4, 3, 2, 1  and adding a bottom row which continues from the  1  just added with  1, 2, 3, 4, 5.

Unger.  Arithmetische Unterhaltungen.  1838.  Pp. 233-235, nos. 897-901.  He first gives the 10 pair case, using  ABCBD  EFAGE  HGHIC  KFKID,  which is isomorphic to  MUTUS  NOMEN  DEDIT  COCIS,  but he never gives any mnemonics.  He then asks which other arrays can be done and sees that  r  rows can be used with  r(r+1)/2  pairs.  He gives a scheme for  r = 3:  ABAC  BDDE  EFCF,  and a scheme for  r = 5:  ABCDEA  BFGHFI  KGLLCM  ONNDHK  EIOPMP.  He then considers triples and gives a scheme for 8 triples:  AAA123  BBB124  CCC143  DDD423.  This has the feature that the responses are either one row or three rows, giving   r + BC(r, 3) = r (r² ‑ 3r + 8)/6  responses.  He shows an easy method for extending a scheme to the next number of rows getting  AAA123567  BBB124589  CCC143860  DDD423907  EEE567890  with 15 triples.  He notes that one may want to rearrange the columns to make it less obvious when the cards are laid out.  He really should have started with  AAA1  BBB1  CCC1  and then constructed  AAA123  BBB124  CCC134  DDD234  in order to make the construction clearer.

Young Man's Book.  1839.  Pp. 205-206.  The Ten Duplicates.  Identical to Endless Amusement.

Parlour Pastime, 1857.  = Indoor & Outdoor, c1859, Part 1.  = Parlour Pastimes, 1868.  Parlour magic, no. xxii, pp. 202-203 (1868: 213-214): The ten duplicates.  MUTUS  DEDIT  NOMEN  COCIS.

Magician's Own Book.  1857.  No. 21: The pairs re-paired, pp. 65-66.  MUTUS  DEDIT  NOMEN  COCIS  ["Mutus gave a name to the Coci," a people who have yet to be discovered.]  = Boy's Own Conjuring Book, 1860, p. 73.

The Secret Out.  Op. cit. in 4.A.1.  1871?  To tell cards thought of, p. 4.  MUTUS  DEDIT  NOMEN  COCIS.

Elliott.  Within‑Doors.  Op. cit. in 6.V.  1872.  Chap. 3, no. 9: The pairs repaired, pp. 83-84.  MUTUS  DEDIT  NOMEN  COCIS.

Boy's Own Book.  The ten pairs of cards distinguished.  1868: 638.  NOMEN  MUTUS  DEDIT  RORIS.

Hanky Panky.  1872.  To tell two cards out of twenty, p. 254.  MISAI  TATLO  NEMON  VESUL.

Berkeley & Rowland.  Card Tricks and Puzzles.  1892.  Simple Tricks by Calculation No. XII: To name several different pairs of cards chosen by one or more persons, pp. 44-46.  SHEER  CHAFF  USUAL  COLOR;   MUTUS  DEDIT  NOMEN  COCIS;   MISAI  TATLO  NEMON  VESUL (where  V = U).

Ball.  MRE.  1st ed., 1892.  Determination of a selected pair of cards out of  ½n(n+1)  given pairs, pp. 98-101.  Gives general rule for pairs.  MATAS  DEDIT  NOMEN  COCIS.  Mentions  8  triples:  LANATA  LEVETE  LIVINI  NOVOTO.

Ahrens.  MUS I.  1910.  Pp. 148-152.  MUTUS  DEDIT  NOMEN  COCUS.  Discusses generalizations, including examples with triples, etc.  Cites Unger, 1838, ??NYR, for the first idea of finding pairs and quotes Ball's mnemonic.

Indoor Tricks and Games.  Success Publishing, London, nd [1930s??]. 

The wandering pairs, pp. 38-40.  MUTUS  DEDIT  NOMEN  COCIS.

The re-united triplets, pp. 40-41.  LIVINI  LANATA  LEVETE  NOVOTO.

Phyllis Fraser & Edith Young.  Puzzles Quizzes and Games.  Bantam Books, NY, 1947.  No. 2: Card trick, pp. 97-99.  THIGH  ATLAS  GOOSE  BIBLE.

Stewart Judah.  The divertisements of Stewart Judah.  The New Phoenix, No. 319 (Nov 1954) 83.  UNDUE  GOANO  TETRA  RIGID.

Will Dexter.  The Illustrated Book of Magic Tricks.  Abbey Library, London, nd, but Introduction dated 1957.  The Romans had a word for it!, pp. 73-74.  DAVID  LOVEL  IN YON  ABBEY;   MUTUS  NOMEN  COCIS  DEDIT ("Mutus gave a name to Cocis").

Martin Gardner.  Mutus Nomen.  Pallbearers Review 5:7 (May 1970) 338.  Reprinted in his:  Martin Gardner Presents; Richard Kaufman & Alan Greenberg, 1993, pp. 237-238.  For 30 cards:  LIVELY  RHYTHM  MUFFIN  SUPPER  SAVANT.  For 20 cards:  BIBLE  ATLAS  GOOSE  THIGH.

Walter Gibson.  Big Book of Magic for All Ages.  Kaye & Ward, Kingswood, Surrey, 1982.  Twenty Cards  &  New Deal Twenty Card Trick, pp. 162-165.  Mentions  MUTUS ...  and says there are many English possibilities and gives  RUFUS  STEEL  TIARA  FOLIO.  The New Deal version uses a more complicated laying-out process starting with ten piles of two using  MAGIC  STORE,  then laying out according to  AGORA  METER  SCOTS  GIMIC.

Karl Fulvesl.  Mutus Nomen.  The author, 1998.  A 90pp booklet on variations of this classic trick.  ??NYR.

 

          7.BA. CYCLE OF NUMBERS WITH EACH CLOSER TO TEN THAN THE

                              PREVIOUS

 

          New section, due to discovering Lewis Carroll's version.  I have no idea if this is older than Carroll.

 

Lewis Carroll.  A Tangled Tale.  (1885) = Dover, 1965.  The pigs.  Knot VIII: De omnibus rebus, pp. 52-57 & 132‑134.  (In the answers, this part of the Knot is denoted §1. The pigs.)  "... place twenty‑four pigs in those four sties, so that, as she goes round the court, she may always find the number in each sty nearer to ten than the number in the last."  Answer:  8, 10, 0, 6  _ the key is that "nothing is nearer ten than  10".

 

          7.BB. ITERATED FUNCTIONS OF INTEGERS

 

          See 7.T and 7.AY for some special forms.

          Let  f(n)  be a function of an integer  n.  The problems in this section concern the behaviour of the sequence of iterations of  f:  n, f(n), f²(n) = f(f(n)), f³(n), ....  There are a great number of behaviors that can occur.

          1.       The sequence can diverge to infinity.

          2.       The sequence can converge to a fixed point.

          3.       The sequence can become cyclic.

          4.       The sequence can behave chaotically.

Further the behaviour may be different for different starting points. 

          If  f(n) > n  in all cases, then the iteration will diverge to infinity and is not very interesting.  Hence we are normally only interested in functions which have  f(n) < n  for some non-zero proportion of the integers.

          If  f(n) £ n  in all cases, then the sequence must converge to a fixed point, though different starting points may lead to different fixed points.  Then the problem is to identify the fixed points and the sets they attract.

          So the most interesting cases have  f(n)  sometimes greater and sometimes smaller than  n.  Two claasic examples are the following.

          The proper sum of divisors,  Σ(n),  is  σ(n) - n,  where  σ(n)  is the classic number theoretic function of the sum of all divisors.  The iterates of this have been studied since the 19C as the fixed points are the perfect numbers and 2-cycles are amicable numbers _ see 7.AB.  Some larger cycles have been found _ I recall a 5 and a 17 cycle.  No provably infinite sequence has been found.

          The Syracuse function is the following.  If  n  is even, halve it;  if  n  is odd,  form  3n + 1.  Some of the sequences have been computed for hundreds of thousand of terms without termination.

          Problems of this sort are genuine mathematical recreations but are a bit too elaborate for me to include here.  What I will include are functions whose sequences are always finite.  These generally are of two types.  Functions such that  f(n) < n  for all  n  above some limit _ see, e.g., 7.AY, and functions such that  f(n)  has the same number of digits as  n.

 

Roger Cook.  Reversing digits.  MS 31:2 (1998/9) 35-37.  The initial problem was to take a four digit number, arrange the digits in ascending order and in descending order and subtract the first from the second.  In all cases except  aaaa,  the iterations converge to  6174  (7641 - 1467 = 6174).  He discusses other number of digits and other bases.

 

          7.BC. UNUSUAL DIFFICULTY IN GIVING CHANGE

 

          The typical problem here is that a customer offers a note to pay a bill and the shopkeeper says he can't give change, but when the customer offers a larger note, then the shopkeeper can give change.

          New section.  I have seen a 1935 version somewhere.

 

Clark.  Mental Nuts.  1916: no. 4.  The conductor's money.  Passenger tenders a $1 note to pay 5¢ (= $.05) fare, but conductor cannot give change, but he can give change for a $5 note.  The answer is that he can give a $2.50 gold piece, a $2 note and 45¢ in coins.

David Singmaster.  No change!  Written up as a problem for my Telegraph column in summer 1999.  Shopkeeper cannot give change for a £5 note, but can for two £5 notes.  How can this happen?  How about in the US?