20th VMO 1982

Find all positive integer solutions to 2^a + 2^b + 2^c = 2336.

Answer

{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