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Counting Chicken Wings - posted January 21, 2002

At Annie's Home-Cooked Chicken Wings Restaurant, chicken wings are served by the bucket. The Biggest Bucket O' Wings is really big! Let's figure out how many wings are in it.

If they're removed two at a time, one wing will be left. If they're removed three at a time, two wings will remain. If they're removed four, five, or six at a time, then three, four, and five wings, respectively, will remain. If they're taken out seven at a time, no wings will be left over.

What's the smallest possible number of wings that could be in the bucket? How do you know?

Comments

Counting Chicken Wings was our first problem written with hints, ways to check your answer, examples of possible solutions, and a page with more ideas. We knew that this new format would take some getting used to! When we wrote back to you this time, we sometimes asked you to look at the hints that we had already written instead of giving you personal hints about how to improve your explanation. Thank you to those of you who really looked through everything and a special thank you to those students who gave us feedback by writing in the comment areas.

One thing agreed on by students solving the Counting Chicken Wings problem was that it certainly is a big bucket o' wings! There were a variety of ways that the facts given in the problem could be used. Many students started with the fact that the number of wings must be a multiple of 7 because when you took out the wings 7 at a time there would be nothing left in the bucket.

Sayuri Khandavilli of Woodrow Wilson Middle School and the team of Kayla G., Christi S, and Kanushri W. of Quest Academy started with that fact, wrote down a list of the multiples of 7, and then used each of the other facts to eliminate numbers from the list. Sathya Sankaran of Whitney Elementary School started by listing the multiples of 7 but then showed in a chart why each one didn't work until reaching 119. Isa Marin of National Cathedral School-Girls lets you hear her thinking as she works through the problem. She starts with a short list of multiples of 7 and tells you why she makes it longer as she tries to solve the problem. In comparison to these rather long explanations, there is the very efficient method written by Ayushman Ghosh of Bourne
Middle School. He thought of using the number facts that narrow down the possibilities the quickest.

Make sure to take a look at the solutions written by Andrei Lazanu of School No. 205, Natalie Krzyzanowski of Washington Junior High School, and Abhay Dang of Manav Sthali School. They are all examples of how using a little algebraic notation can also help to shorten the problem.

As we continue with problems this year, you'll notice many will be in this "new" format while a few will be in our previous format. Either way, we hope you enjoy the challenge.

-Suzanne

Highlighted solutions:

From:
Sayuri Khandavilli, age 12
School:
Woodrow Wilson Middle School, Edison, New Jersey

The smallest number of wings that could be in the bucket is 119.

The most important clue was that if 7 were taken at a time then none would be left. So, the number of
wings should be a multiple of 7.

I listed the multiples of 7 till 140

07 14 21 28 35 42 49 56 63 70 77 84 91 98 105 112 119 126 133 ................

From the problem:

When 2 wings are taken at a time 1 was left
(this gives the clue that the number must be odd)

So this eliminates all the even numbers

07 21 35 49 63 77

91 105 119 133 ................

When 3 wings are taken at a time 2 were left
(this gives the clue that it should not be the multiple of 3)

So this eliminates all the multiples of 3 :

07 35 49 77 91 119 133 ...

When 4 wings are taken at a time 3 were left :

So this eliminates the numbers which do not have a remainder of 3: 07 35 119

When 5 wings are taken at a time 4 were left :

So this give the clue that the number should not be the multiple of 5 and has a remainder of 4 .

So this eliminates the numbers which do not have a remainder of 4 and multiples of 5 : 119

When 6 wings are taken at a time 5 were left :

119 when divided by 6 leaves a remainder of 3

ANSWER : The smallest number of wings that could be in the bucket is 119.

From:
Sathya Sankaran, age 11
School:
Whitney Elementary School, Strongsville, Ohio

There are 119 chicken wings (at the least) in the biggest bucket of wings.

I started by making a list of multiples of seven, since it says if you take it out by sevens you get nothing left over. I checked each number to see if it was the number I was looking for. After a wrong number I will give the first number that is not compatible.

O = Right X = Wrong.

1. 7 X 3
2. 14 X 2
3. 21 X 3
4. 28 X 2
5. 35 X 4
6. 42 X 2
7. 49 X 3
8. 56 X 2
9. 63 X 3
10. 70 X 2
11. 77 X 4
12. 84 X 2
13. 91 X 3
14. 98 X 2
15. 105 X 5
16. 112 X 2
17. 119 O

So now I know that there are 119 chicken wings (at the least) in the biggest bucket of wings.

From:
Isa Marin, age 12

School:
National Cathedral School-Girls, Washington, DC

The smallest amount of chicken wings that could be in a bucket are 119.

Okay to solve this problem I am going to start with the multiples of 7:

7 14 21 28 35

One of the above must be the answer since if you remove wings 7 at a time you will end up with nothing. Now I ‘ll start from the beginning again. Since if I remove wings 2 at a time I will end up with one, The answer will be an odd number. This eliminates 14 and 28. So far these are our choices:
7 21 35

Next I know that the above must, if subtracted by 2, be a multiple of 3 because if I remove wings 3 at a time I will be left with 2.

This eliminates 7 and 21. So we have an answer! No, not yet at least.
First I must make sure. Oh ,and remember there are other multiples of 7 that are larger the 35, so if 35 does not work than we will check them.

The total amount of wings, if subtracted by 3 must be a multiple of 4. Lets check 35…… 35 – 3 = 32 and that is a multiple of 4! Hoorah!

Okay we are on a role! The answer, if subtracted by 4 must be a multiple of 5. Oh no! Oh no! 31 should be a multiple of 5!!!!!!!!

The noly thing left is to look beyond and start again:

42 49 56 63 70

To make this easier I will make a list of what rules the answer must abide by:

1. must be odd
2. must be a multiple of 3 if subtracted by 2
3. must be a multiple of 4 if subtracted by 3
4. must be a multiple of 5 if subtracted by 4
5. must be a multiple of 6 if subtracted by 5
6. must be a multiple of 7

42, 56, and 70 are eliminated by the first rule. And 63 drops out by the second. Again we have only one left let us not get out hopes up!

So, is 49 go by rule number 3 let’s check it out! 49 – 3 = 46
and 46 is not a multiple of 4. NOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!

So again we fail and the number go higher:

77 84 91 98 105

So let’s start with our rules again:

1) Must be odd
2) Must be a multiple of 3 if subtracted by 2
3) Must be a multiple of 4 if subtracted by 3
4) Must be a multiple of 5 if subtracted by 4
5) Must be a multiple of 6 if subtracted by 5
6) Must be a multiple of 7

So we can toss out 84 and 98 because of number 1 and 91 and 105 because of number 2. One is left again let us hope that 77 is the lucky number! But 77 fails at number 3!
How long this can take!!!

The next batch of numbers are

112 119 126 133 140

1) Must be odd
2) Must be a multiple of 3 if subtracted by 2
3) Must be a multiple of 4 if subtracted by 3
4) Must be a multiple of 5 if subtracted by 4
5) Must be a multiple of 6 if subtracted by 5
6) Must be a multiple of 7

112, 126, and 140 all end at number 1 and 133 stops at number 2.
119 is left. Please let this be the one. 119 passes number 3 and also number 4.
Come on we can do this!!!!!!!!! 119 – 5 = 114 and is a multiple of 6.
Hoooorah!!!!!!!!!! 119 is the answer!!!!!!!!!!!!!!!!!!!!!!!!! I know this because I have worked from smallest to largest meaning I will encounter the smallest possible number before the largest possible answer.
I hope you have been able to follow along!

From:
Andrei Lazanu, age 13
School:
School No. 205, Bucharest, Romania

The smallest possible number of wings is 119.

I noted with x the total number of wings. From the text of the
problem I wrote the following equations:

x = 2a + 1 (1)
x = 3b + 2 (2)
x = 4c + 3 (3)
x = 5d + 4 (4)
x = 6e + 5 (5)
x = 7f (6)

At equations (1), (2), (3), (4), (5), I added 1 obtaining:

x + 1 = 2(a + 1) = 3(b + 1) = 4(c + 1) = 5(d + 1) = 6(e + 1)

So, x + 1 must be a multiple of 2, of 3, of 4, of 5, of 6, in this
situation the smallest common multiple:

x + 1 = M2 = M3 = M4 = M5 = M6

so,

(x + 1) = M(3 * 4 * 5) = M60

so, x is a multiple of 60 - 1.

Now, I look for multiples of (multiples of 60) - 1. being also
multiples of 7.

Number | M7
-------+----
59 | No
119 | Yes

So, I found the smallest possible number of wings.

From:
Natalie Krzyzanowski, age 9
School:
Washington Junior High School, Naperville, Illinois

The least number of wings that can go in the bucket is 119.

Data: x - the number of wings in the bucket

Calculations: From the data in the question, I can make 6 equations:

1. 2a + 1 = x
2. 3b + 2 = x
3. 4c + 3 = x
4. 5d + 4 = x
5. 6e + 5 = x
6. 7f = x

From equation 1 and 6, I can derive that x is odd and a multiple of 7, so possible values of x are:

x = 7, 21, 35, 49, 63, 77, 91, 105, 119, 133, 147, ...

Next, I substitute the values for x, and if all equations will have integers as outcomes, that is the least possible value for x:

For x = 7, a = (7-1)/2 = 3
b = (7-2)/3 = 5/3 , which is not an integer

I do the same for the rest of the equations and find that the least integer possible to be x is 119. So 119 chicken wings are in the bucket.

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