Exploring boxes of chicken nuggets

As a graduate student in Math Elementary Education, I often have to solve tasks that invite us to explore further a math concept. It is fun, and a good reminder of what students may experience, may feel in class. I usually solve these tasks through different ways, modeling, representation, and/or equations. This semester, I have to share some of the tasks on my blog. So here is the latest example of what I had to solve:

McDonald’s sells chicken nuggets in boxes of 6, 9, or 20. Obviously one could purchase exactly 15 nuggets by buying a box of 6 and a box of 9. Using only combinations of boxes of 6, 9, and/or 20 nuggets:

1. Could you purchase exactly 17 nuggets?
2. How would you purchase exactly 53 nuggets?
3. What is the largest number for which it is impossible to purchase exactly that number of nuggets?
4. Let’s say you could only buy the nuggets in boxes of 7, 11, and 17. What is the largest number for which it is impossible to purchase exactly that number of nuggets?

I started with a quick representation of the problem, and an equation. To buy nuggets, you can use any multiples of 6, 9 and 20, and any sums of these multiples.

Number of nuggets (N) = 6 x number of boxes of 6 (B6) + 9 x number of boxes of 9 (B9) + 20 x number of boxes of 9 nuggets (B20)

N = 6 x B6 + 9 x B9 + 20 x B20

1. Could you purchase exactly 17 nuggets?  I solve this task mentally.  Buying a box of 6 nuggets and a box of 9 would not provide enough nuggets (15 nuggets), and buying 2 boxes of 9 would provide too many (18 nuggets). Consequently, there is no combination possible to provide exactly 17 nuggets.

2. Could you purchase exactly 53 nuggets?
From the equation, I varied the number of boxes of 20, 9 and 6. I found 2 combinations to buy 53 nuggets

• 1 box of 20, 1 box of 9 and 4 boxes of 6
• 1 box of 20, 3 boxes of 9 and 1 box of 6

3. What is the largest number for which it is impossible to purchase exactly that number of nuggets ? I struggled with this question, mostly because I was not sure what to look for. I felt like a kid who started being anxious: “What in the world am I supposed to do ?”.

So, without being sure where I was going, I started by considering that any number of nuggets that is not

• a multiple of 6,
• a multiple of 9,
• a multiple of 20, or
• any sums of multiples of 6 and 9 and/or 20

would lead to an impossible exact purchase.

Based on that, I made a table, listing multiples of 6, 9 and 20, and sums of these multiples. I put the numbers on a number line, and noticed that the following numbers of nuggets could not be bought exactly: 0-5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 22, 23, 25, 28, 31, 34, 37, 43. Any number of nuggets higher than 43 appeared to be buyable exactly.

At this point, I realized that 43 was likely the answer expected i.e. the largest number for which it is impossible to purchase exactly that number of nuggets. I did another way of representation i.e. a graph representing each number of nuggets with, if possible, the boxes to buy that number exactly. And because my brain seems to work better in color, I colored the boxes of 6 nuggets are in pink, 9 nuggets in purple, and 20 nuggets in brown. After 43, any number of nuggets could be purchased exactly.

4. Let’s say you could only buy the nuggets in boxes of 7, 11, and 17. What is the largest number for which it is impossible to purchase exactly that number of nuggets?

I found the graphic representation I did with Question 3 quite useful (not to mention how impressed Rosie was with such large graph), so I solved Question 3 similarly to Question 2. I found out that the largest number of nuggets to be exactly purchased out of boxes of 7, 11 and 17 nuggets would be 37.

But then, I started wondering : why 43? Why 37?

Now, you may see the pattern right away, but it does not seem that obvious to me. From Question 2, I could see that once you reach 6 numbers of nuggets in a row that can be exactly purchased (e.g. 44 to 49), all the following numbers could be exactly purchase by just adding a box of 6 nuggets (44 +6 = 50, 45 + 6 = 51, etc). Same with Question 3. But there must be an equation that allows us to find out 43, or 37 without borrowing my kids’ Crayolas®.

My investigation is still ongoing 🙂