(a,b,c) = (0,0,2), (1,2,3), (2,1,3) Solution It is easy to check that the solutions above are the only solutions for c ≤ 3. So suppose c ≥ 4. Then 2 c = 2 + 2 a + 2 b . Now rhs cannot be divisible by 4 unless a or b = 1 (or both a and b = 0, which is obviously impossible for c ≥ 4). So suppose a = 1. But then we need 2 b = 12 mod 16 which is impossible. So there are no other solutions. Thanks to Suat Namli 25th APMC 2002 © John Scholes jscholes@kalva.demon.co.uk 7 February 2004 Last corrected/updated 7 Feb 04
25th APMC 2002 Problem 1 Find all triples (a, b, c) of non-negative integers such that 2 a + 2 b + 1 is a multiple of 2 c - 1.