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HOW TO SOLVE WORD DIVISION CODES!
NOTE:
-(1) == Optional carry to be subtracted
+(10) == Optional carry to be added as a ten
Problem is the image on the right of each coded example!
CODED WORD #1
Because B=0: a factor, then sUb = 5 or ruN = 5
And since allR <> 0, 5 thus ruN <> 5 and thus U=5.
Since sUb*RUN=CONO and 'run' is 3 digits
and 'cono' is 4 digits, then C is less than R
similarily A is less than R but R = A + 1 or A.R in column 1.
And thus A > C or C...A.R and thus S > U as factors of
products 'cono' and 'allr'.
Also O is greater than N and O = N + 1 or N.O in column 3
From relationships, on the right,
one can easily set up a table of values, below.
L
N
= 2 * L
O
= N + 1
R
= 10 - N
A
= R - 1
S
= R / N
1 2 3 8 7 4, 9
2 4 5 6 5
not 5=N -
3 6 7 4 3
not 3=L -
4 8 9 2 1 1
not 1=A
5 10
not 0=B - - - -
6 12 3 8 7 4, 9
7 14 5 6 5
not 5=O -
8 16 7 4 3 2
9 18 9
not 9=L - - -
Since B +(10) -(1) - L = O then L + O = 10 -(1) = 9, 0 in column 3.
Only row 6 satisfies this equation.
Thus: L=6 ; N=2 ; O=3 ; R=8 ; A=7 ; S=4, 9 ; N.O ; A.R !
Since a carry is required for 'R-A=0' in column 1,
I is less than L & C in column 2.
And since I-(1)+(10)-l=C than C=I+3,4 in column 2
Sofar one has:
S=4,9 > U
I < C < A = 7
C=I+3,4
B . N O . . L A R S
0 1 2 3 4 5 6 7 8 9
Left overs are i ; c ; u with I = 1 on inspection!
The word BINOCULARS can easily be deduced from this,
without doing extra work.
CODED WORD #2
For M*T=T:
Since 6*2=12 ; 6*4=24 ; 6*8=48
thus M = 6, 1 or T=5 but M <> 1
And since tihaD <> 5,0 & tewiW <> 5,0 - columns 6 & 7
Thus T <> 5 so M=6
Since T - T = H then H = 0 in column 5
Since T is greater than I or T = I + 1 or I.T in column 1
And since a carry is required for 'IT' in column 1,
O is less than M & T in column 2.
D is greater than I & S in column 3.
Since S is greater than T or S = T + 1 or I.T.S in column 3
From relationships, on the right,
one can easily set up a table of values, below.
M
T
S = T + 1
I = T - 1
W = WT = M * ST
6 2 3 1 9
6 4 5 3 2
6 8 9 7 8 ; not 8=T
Since S +(10) - W = I in column 4:
Only the middle row of the table satisfies this equation
thus M=6 ; T=4 ; S=5 ; I=3 ; W=2 ; I.T.S ;
Since O=I+D=3+8=11 then O=1 in column 6.
Since D=W+M=6+2=8 then D=8 in column 7.
Sofar one has:
H O W I T S M . D .
0 1 2 3 4 5 6 7 8 9
Left overs are a ; e the word HOWITSMADE can easily
be deduced from this, without doing extra work.
CODED WORD #3
Can = 1 x IDEA = IDEA.
S is greater than I & E in column 1.
A is greater than O & O = 0 in column 2.
and since no carry was needed D is greater than N
and D=2*N+(1) column 3
A is greater than E and A = E + 1 or E.A - column 2.
Since a carry is required for 'A-E=0' in column 2,
N is less than R & I in column 3.
From relationships, on the right,
one can easily set up a table of values, below.
A
S = A * A
E = A - 1
I = S -(1) - E
D = E + I -(10)
1 1 : not 1=C - - -
2 4 1 : not 1=C - -
3 9 2 7,6 9,8 : not 9=S
4 16 3 3,2 : not 3=E 5
5 25 : not 5=A - - -
6 36 : not 6=A - - -
7 49 6 3,2 9,5
8 64 7 : not S>E - -
9 81 : not 1=C - - -
Since 3*I + ? = E*10 + 0 = E*10 = 20 - row 5, columns 1,2
Only 3*6+2=20 in row 3 satisfies this equation.
Thus 3 * 6823 = 20469
And A=3 ; S=9 ; E=2 ; I=6 ; D=8 ; N=4
Since B = N + (10) - S then B=14-9=5 in column 5.
Sofar one has:
O C E A N B I . . S
0 1 2 3 4 5 6 7 8 9
Left overs are r ; d the word OCEANBIRDS can easily be deduced from this,
without doing extra work.
CODED WORD #4
Factor N=0 because there is no division in column 6.
Since R is greater than F and R=F+1 or F.R in column 2.
Since T-T=0 requires no carry then A is greater than T & R
in column 2.
Since X -(1) +(10) -X=N then N=0,9 in column 3.
Since 'R-F=0' requires a carry then N < Y or N=0 column 1
and N -(1) +10 -Y=0 then Y=10 -(1) = 9 in column 3.
As before atE=6,1 and Fin & fIn =2,4,8, rows 3 & 4 - products => F & I.
If E=6 than:
E=6
I=2,4,8
R=4,2,8 :R=E-I: not 8=I
F=3,1,- :R=F-1: this refutes F=2,4,8, thus E<>6
And thus E=1
Since N=0 there is no carry 'X-X=0' in column 3,
I is greater than F & X in column 4.
And also from their respective products
F is greater than T since 'ate' is 3 digits and 'ttxf' is 4 digits .
ie.:
N.E---T---F.R---I----Y : X < I & T,R < A
So far one has:
N E . . . . . . . Y
0 1 2 3 4 5 6 7 8 9
Left overs are x ; d ; a with X = 2 giving NEXT on inspection!
The word NEXTFRIDAY can easily be deduced from this,
without doing extra work.
CODED WORD #5
Factor O=1 because EYES is repeated in row 2.
Since there is no carry in K-K=0 in column 1
then K=2*O+(1)=2*1+(1)=2,3 in column 2.
Since E+(10)-(1)-E=H then H=0,9 in column 5
Since A*S=K & K*S=A then S=9 in rows 4 & 6
and thus leaving H=0.
NOTE: example 7*9=63 &3*9=27
Since C+(10)-S=K in column 4
then C=S+K-(10)=9+2,3-(10)=2,3-1=1,2
Since O=1 then C=2 and leaves K=3.
Since K-(1)+(10)-E=S in column 5
then E=K+(1)+(10)-S=3+(1)+(10)-9=4+(1)=4,5
If E.K or E..K is true then S=0 , in column 4,
but since S=9 then K.E or K..E is true.
F is greater than K in column 3.
Sofar one has:
H . C.K.E . . . . S
0 1 2 3 4 5 6 7 8 9
Since K is greater than O in column 2
then O=1 by inspection.
Left overs are y ; f ; a ; n with O = 1 giving HOCKEY on inspection!
The word HOCKEYFANS can easily be deduced from this,
without doing extra work.
NOTE:
1. It is easy to make a list and eliminate the value of zero,
from each unknown, because of the greater than properties
in the division!
2. Knowing which is 0 or 9 or 5 or 6 or 1 goes a long way
to solving the problem sometimes!
3. Verification can be made by substituting real values
for letters in each division code problem!
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CODED WORD #1
Because B=0: a factor, then sUb = 5 or ruN = 5
And since allR <> 0, 5 thus ruN <> 5 and thus U=5.
Since sUb*RUN=CONO and 'run' is 3 digits
and 'cono' is 4 digits, then C is less than R
similarily A is less than R but R = A + 1 or A.R in column 1.
And thus A > C or C...A.R and thus S > U as factors of
products 'cono' and 'allr'.
Also O is greater than N and O = N + 1 or N.O in column 3
From relationships, on the right,
one can easily set up a table of values, below.
| L | N = 2 * L | O = N + 1 | R = 10 - N | A = R - 1 | S = R / N |
|---|---|---|---|---|---|
| 1 | 2 | 3 | 8 | 7 | 4, 9 |
| 2 | 4 | 5 | 6 | 5 not 5=N | - |
| 3 | 6 | 7 | 4 | 3 not 3=L | - |
| 4 | 8 | 9 | 2 | 1 | 1 not 1=A |
| 5 | 10 not 0=B | - | - | - | - |
| 6 | 12 | 3 | 8 | 7 | 4, 9 |
| 7 | 14 | 5 | 6 | 5 not 5=O | - |
| 8 | 16 | 7 | 4 | 3 | 2 |
| 9 | 18 | 9 not 9=L | - | - | - |
Only the middle row of the table satisfies this equation thus M=6 ; T=4 ; S=5 ; I=3 ; W=2 ; I.T.S ; Since O=I+D=3+8=11 then O=1 in column 6. Since D=W+M=6+2=8 then D=8 in column 7. Sofar one has: H O W I T S M . D . 0 1 2 3 4 5 6 7 8 9 Left overs are a ; e the word HOWITSMADE can easily be deduced from this, without doing extra work.
Only 3*6+2=20 in row 3 satisfies this equation. Thus 3 * 6823 = 20469 And A=3 ; S=9 ; E=2 ; I=6 ; D=8 ; N=4 Since B = N + (10) - S then B=14-9=5 in column 5. Sofar one has: O C E A N B I . . S 0 1 2 3 4 5 6 7 8 9 Left overs are r ; d the word OCEANBIRDS can easily be deduced from this, without doing extra work.
CODED WORD #4
Factor N=0 because there is no division in column 6.
Since R is greater than F and R=F+1 or F.R in column 2.
Since T-T=0 requires no carry then A is greater than T & R
in column 2.
Since X -(1) +(10) -X=N then N=0,9 in column 3.
Since 'R-F=0' requires a carry then N < Y or N=0 column 1
and N -(1) +10 -Y=0 then Y=10 -(1) = 9 in column 3.
As before atE=6,1 and Fin & fIn =2,4,8, rows 3 & 4 - products => F & I.
If E=6 than:
E=6
I=2,4,8
R=4,2,8 :R=E-I: not 8=I
F=3,1,- :R=F-1: this refutes F=2,4,8, thus E<>6
And thus E=1
Since N=0 there is no carry 'X-X=0' in column 3,
I is greater than F & X in column 4.
And also from their respective products
F is greater than T since 'ate' is 3 digits and 'ttxf' is 4 digits .
ie.:
N.E---T---F.R---I----Y : X < I & T,R < A
So far one has:
N E . . . . . . . Y
0 1 2 3 4 5 6 7 8 9
Left overs are x ; d ; a with X = 2 giving NEXT on inspection!
The word NEXTFRIDAY can easily be deduced from this,
without doing extra work.
CODED WORD #5
Factor O=1 because EYES is repeated in row 2.
Since there is no carry in K-K=0 in column 1
then K=2*O+(1)=2*1+(1)=2,3 in column 2.
Since E+(10)-(1)-E=H then H=0,9 in column 5
Since A*S=K & K*S=A then S=9 in rows 4 & 6
and thus leaving H=0.
NOTE: example 7*9=63 &3*9=27
Since C+(10)-S=K in column 4
then C=S+K-(10)=9+2,3-(10)=2,3-1=1,2
Since O=1 then C=2 and leaves K=3.
Since K-(1)+(10)-E=S in column 5
then E=K+(1)+(10)-S=3+(1)+(10)-9=4+(1)=4,5
If E.K or E..K is true then S=0 , in column 4,
but since S=9 then K.E or K..E is true.
F is greater than K in column 3.
Sofar one has:
H . C.K.E . . . . S
0 1 2 3 4 5 6 7 8 9
Since K is greater than O in column 2
then O=1 by inspection.
Left overs are y ; f ; a ; n with O = 1 giving HOCKEY on inspection!
The word HOCKEYFANS can easily be deduced from this,
without doing extra work.
NOTE:
1. It is easy to make a list and eliminate the value of zero,
from each unknown, because of the greater than properties
in the division!
2. Knowing which is 0 or 9 or 5 or 6 or 1 goes a long way
to solving the problem sometimes!
3. Verification can be made by substituting real values
for letters in each division code problem!