20th VMO 1982 Problem B1 Find all positive integer solutions to 2 a + 2 b + 2 c = 2336.
{a,b,c} = {5,8,11} Solution Assume a ≥ b ≥ c (then we get other solutions by permuting a, b, c). 3·512 < 2336 < 4096, so a = 10 or 11. If a = 10, then since 512 + 512 < 2336 - 1024 = 1312, we must have b = 10. But that leaves 288, which is not a power of 2. Hence a = 11. So 2 b + 2 c = 288. But 128 + 128 < 288 < 512, so b = 8. Then c = 5. Thanks to Suat Namli 19th VMO 1981 © John Scholes jscholes@kalva.demon.co.uk 13 February 2004 Last corrected/updated 13 Feb 04