On the last day of school, Ms. Quigg gave her class jellybeans. She gave each boy as many jellybeans as there were boys in the class. She gave each girl as many jellybeans as there were girls in the class. She had bought 400 jellybeans and when she finished giving them out she had six left; these she ate herself. How many students were in her class?
Solution
Discussion: This problem is a classic case of re-thinking the problem to find the underlying math.
Solving the Problem: It's a matter of finding and in the Diophantine equation . A bit of guess-and-try, and it turns out that the numbers are and . There are boys and girls, or the other way around.
Heuristics: #10, write an equation, or simply #3, re-state the problem (in terms of perfect squares); then #2, guess and try.
Using the Problem with Students: No, I don't use real jellybeans. We read the problem carefully. If I need to help them get going on solving it, I'll start by saying, "OK, if there are six 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 seven 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 those two numbers add up to? (400). Uh-uh, you forgot something." (oh--394). "So you need to find two perfect squares that add up to 394. Go for it."
This'll be guess-and-try for awhile, and sooner or later the class will come up with 13 and 15. .
Now, which is the boys and which is the girls? It doesn't matter--it could be either way.
You could mention the equation at some point: . You could even start with it, though once it's on the board there's no way to solve it except by guess-and-try.
Modifying or Extending the Problem: What if there are the same number of boys and girls in the class? How does this change things? Will Ms. Quigg necessarily have more jellybeans for herself?