I've been looking at the five triangles site recently: fivetriangles.blogspot.com and I like alot of the problems there. I'm now considering whether the most recent one will work for the club.
Given a pair of positive integers a, b, the operation a∗b is defined as 23 × a + 31 × b.
For example, 20∗15 = 23 × 20 + 31 × 15 = 925.
Determine both pairs of positive integers x, y such that x∗y = 2015.
For me this problem is just a regular Diophantine equation. So I find the standard form (via Euclid's method or just checking some of the simple combinations)
3 * 31 - 4 * 23 = 1
From there you multiple everything by 2015:
6045 * 31 - 8060* 23 = 2015
which gets you a solution but not with 2 positive integers.
You can then adjust both terms by adding and subtracting 31 * 23 or 713 to find other solutions
So add and subtract 261 * 23 * 31 == 8091 * 23 == 6003 * 31
and you get
42 * 31 + 31 * 23 = 2015 and you're found the first pair.
Scale down once again to
19 * 31 + 62 * 23 and you have the next pair
As you can see the next scaling will cause the first term to go negative so there really are only 2 pairs of positive tuples.
However none of the kids know number theory so the question is whether this is really solvable for them?
One thought is that you could notice that 2015 is divisible by 31
31 * 65 = 2015 and then perhaps stumble onto scaling to go down to the 2 solutions. From there if we explored scaling a bit with some easier equations first this might be more obvious. Alternatively, I could just ask them to find the standard form via trial and error and go from there.
I'm, still brainstorming if I can think a sequence of problems to bridge this gap.