For those of you who read last night's post and are curious about what I consider an interesting math problem, here's one in particular that I liked. (I'm recording it from memory, so it may not be exactly the same as the one in Saturday's relay.)
Find n, the number of right triangles of integer side lengths (i.e., Pythagorean triples) that have a leg of length 45 units.
It got me to thinking (while I was driving, no less!) about the formulas for deriving Pythagorean triples... One of the kids at the contest said he thought it could be solved by setting 45^2 = c^2-b^2, factoring the right hand side of this, and finding the number of factors of 45^2 (that's easy, remember how?) to find the number of triangles. Maybe he's onto something, but I have another (though related) way that seems simpler to me. (Or maybe I'm just quite rusty on contest problems and there is an obvious answer...)