### Middle School PoW - Counting Chicken Wings

 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?

Hint 1:

A good strategy to use to start this problem would be to think about these questions:

1. Using the given information in the problem, is it possible to tell if the number of wings in the bucket is even or odd?
2. The number of wings in the bucket must be a multiple of what number?
3. If a number is divisible by 5, the units digit is either 0 or 5. Given this fact, when 4 remains after a number is divided by 5, what can we say about the units digit of that number?

Hint 2:
If you'd like some information about numbers, here are some sites you might want to visit.
Even and Odd Numbers - AAAMath.com
Odd and Even Numbers (On line Quiz) - Teachrkids.com
Divisibility Rules Tips - Math Forum
Divisibility Rules - Math Dept. Suffolk County Community College
Hint 3:
You can read questions other students have asked about numbers and responses from Dr. Math in the archives:
FAQ: Divisibility Rules
Integers, Rational & Irrational Numbers
Hint 4:

Try some of these simpler problems:

Simpler Problem 1: You have an amount of candy that you are dividing between kids. When you have 2 kids and divide the candy each kid gets an equal amount but there is 1 piece of candy left over. When you have 3 kids and divide the candy each kid gets an equal amount and there is no candy left over.

What do you know about the amount of candy that you were dividing?

Simpler Problem 2: You have a pizza that you are dividing so that each person will get an equal amount. You cut the pizza and give out the slices:

There would be 1 piece left over if it were for 2 people.
There would be 1 piece left over if it were for 3 people.
There would be 1 piece left over if it were for 4 people.
It would work perfectly for 5 people to each get the same amount.
How was the pizza cut? How many slices were there?

Ans 1:

Did you check the number that you got?

Is it divisible by 7?
When you divide it by 2, do you have a remainder of 1?
When you divide it by 3, do you have a remainder of 2?
When you divide it by 4, do you have a remainder of 3?
When you divide it by 5, do you have a remainder of 4?
When you divide it by 6, do you have a remainder of 5?
Ans 2:

When you divide a number by 7 and there is no remainder, the number is a multiple of 7. Is your answer one of these numbers?

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133
Ans 3:

If you know that there will be a remainder of 1 when you have divided your number by 2, that means that your original number is an odd number. So if we take the list of the multiples of 7 and only consider the odd multiples of 7, we'll have fewer numbers to check.

Is your number one of these?

7, 21, 35, 49, 63, 77, 91, 105, 119, 133
Solution:

The smallest possible number of wings that could have been in the bucket is 119.

Here are three different ways that were used to get that answer. Do any of them look like your solution? Did you learn anything from these solutions?

Solution 1:

I made a chart of numbers and as I checked each number to see if it worked I marked off the numbers that didn't work. I started with 1 but that was kind of silly because 1, 2, 3, 4, 5, and 6 can't be divided by 7 evenly. So, I quickly marked those off my chart.

7 works for some of the numbers:

7/2 = 3, remainder 1
but 7/3 = 2, remainder 1 and so that isn't right.

I kept going, checking each number until I got one part that didn't work. As I was working, I figured out that the number would have to be odd because none of the even numbers could be divided by 2 but still have a remainder of 1. Once I figured that out I crossed out all of the even numbers. Then I started thinking that if 7 went into the number with no remainder, the number would have to be a multiple of 7. I crossed out all of the numbers on my chart that weren't multiples of 7.

That made my work a little easier. When I finally got to 119, it worked for all of the numbers!

119/2 = 59, remainder 1
119/3 = 39, remainder 2
119/4 = 29, remainder 3
119/5 = 23, remainder 5
119/6 = 19, remainder 6
119/7 = 17, remainder 0

Solution 2:

Since the problem says that if you take the wings out seven at a time there won't be any wings left over, I know that the number of wings will be a multiple of 7. So, I started thinking of the multiples of 7:
7 works for 2 because 7/2 is 3 with a remainder of 1 but it doesn't work for 3 because 7/3 is 2 with a remainder of 1.
Next, I tried 14 but it didn't work for 2 because 14/2 is 7 with no remainder.
I tried 21 but it didn't work for 3 because 21/3 is 7 with no remainder.
I tried 28 but it didn't work for 2 because 28/2 is 14 with no remainder.
I tried 35 but it didn't work for 5 because 35/5 is 7 with no remainder.
I tried 42 but it didn't work for 2 because 42/2 is 21 with no remainder.
I tried 49 but it didn't work for 3 because 49/3 is 16 with 1 remainder.
I tried 56 but it didn't work for 2 because 56/2 is 18 with no remainder.
I tried 63 but it didn't work for 3 because 63/3 is 21 with no remainder.
I tried 70 but it didn't work for 2 because 70/2 is 35 with no remainder.
I tried 77 but it didn't work for 2 because 77/4 is 19 with 1 remainder.
I tried 84 but it didn't work for 2 because 84/2 is 42 with no remainder.
I tried 91 but it didn't work for 2 because 91/3 is 30 with 1 remainder.
I tried 98 but it didn't work for 2 because 98/2 is 49 with no remainder.
I tried 105 but it didn't work for 5 because 105/5 is 21 with no remainder.
I tried 112 but it didn't work for 2 because 112/2 is 56 with no remainder.
I tried 119 and it worked! Whew, I found the number. It's 119.

Solution 3:

Let N be the number of wings.
Since there was 1 left when dividing by 2, N must be one less than a multiple of 2.
Since there were 2 left when dividing by 3, N must be one less than a multiple of 3.
Since there were 3 left when dividing by 4, N must be one less than a multiple of 4.
Since there were 4 left when dividing by 5, N must be one less than a multiple of 5.
Since there were 5 left when dividing by 6, N must be one less than a multiple of 6.
Therefore we can say N is one less than a multiple of 60. And we know N is a multiple of 7.
60 - 1 = 59, not a multiple of 7.
120 - 1 = 119, Yes! A multiple of 7.

More:

If you worked the problem the first time using a method that took a long time, try using some of the ideas that you now know about numbers (the divisibility rules, for example) to work these similar problems.

Similar Problem 1:

I was counting the number of people waiting in line at the post office. If I counted them by threes, I got a remainder of 2. If I counted by fives I also got a remainder of 2. But if I counted by sevens I got a remainder of 5. How many people were waiting in line?

Similar Problem 2:

There is a number that produces remainder 4 when divided by 5, remainder 7 when divided by 9, and remainder 9 when divided by 11. Which number is it given that this number is less than 500?

Similar Problem 3:

The teacher has some apples to distribute to her students. If she gives all of the apples to the 13 students in her first class she will have 4 left. If she gives all of the apples to the 29 students in her second class she will have 9 left. If she gives all of the apples to the 37 students in her third class she will have 16 left. Whichever class she gives the apples to, she will give each student the same number of apples. What is the smallest number of apples the teacher can have?
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You also might find information about the Chinese Remainder Theorem interesting:

Cut-the-Knot: Chinese Remainder Theorem
MathWorld: Chinese Remainder Theorem
The Prime Glossary: Chinese Remainder Theorem
Dr. Math: Chinese Remainder Theorem and Modular Arithmetic
Dr. Math: Chinese Remainder Theorem