Jellybeans

This problem is a classic case of rethinking the problem to find the underlying mathematics.

From the list of heuristics, the most likely heuristics are #10, write an equation, or 3, restate the problem (in terms of perfect squares), and #2, guess and try.

No, I do not use real jellybeans. We read the problem carefully. If I need to help the students get started, I'll start by saying, "Okay, if there are 6 boys in the class, then how many jellybeans does each one get?" (answer: 6) "So how many jellybeans is that?" (36) "And if there are 7 boys, how many jellybeans total is that?" (49) "So what are we talking about here?" (squares) "Ah!"

"So the boys eat a total of some perfect square number of jellybeans, and the girls do too. And what do these two numbers add up to?" (400) "Huh-uh, you forgot something." (oh, 394). "So you need to find two perfect squares that add up to 394. Go for it."

The class will be guessing and trying for a while, but sooner or later they will come up with the answer.

As the students look for two perfect squares that total 394, they are actually finding b and g in the Diophantine¹ equation b² + g² = 394. They will soon find that the numbers 13 and 15 work, in which case there are 28 students in the class, 13 boys and 15 girls, or the other way around:

13² + 15² = 169 + 225 = 394

What if there were the same number of boys and girls in the class? How does this change things? Will Ms. Quigg necessarily have more jellybeans for herself?

¹A Diophantine equation is a polynomial equation with multiple integer solutions.