4th Junior Balkan 2001 Problem 1 Find all positive integers a, b, c such that a 3 + b 3 + c 3 = 2001.
10,10,1 Solution Recollect that the cubes are 1 3 = 1, 2 3 = 8, 3 3 = 27, 4 3 = 64, 5 3 = 125, 6 3 = 216, 7 3 = 343, 8 3 = 512, 9 3 = 729, 10 3 = 1000, 11 3 = 1331, 12 3 = 1728, 13 3 = 2197. 8 3 +8 3 +8 3 = 1536  2001, so the largest of a, b, c must be 9, 10, 11 or 12. Assume a ≥ b ≥ c. If a = 9, then b 3 + c 3 = 1272, but 8 3 + 8 3 = 1024, so b = 9, leaving c 3 = 543, which fails. If a = 10, then b 3 + c 3 = 1001, so b = 10, giving the solution 10,10,1 or b = 9, giving c 3 = 272, which fails or b = 8 giving c 3 = 489 which fails. If a = 11, then b 3 + c 3 = 670, so b = 8 giving c 3 = 158, which fails, or b = 7, giving c 3 = 327, which fails. If a = 12, then b 3 + c 3 = 273, so b must be 6, giving c 3 = 57, which fails. Thanks to Suat Namli 5th Junior Balkan 2001 © John Scholes jscholes@kalva.demon.co.uk 7 February 2004 Last corrected/updated 7 Feb 04