7th Mexico 1993 Problem A2 Find all numbers between 100 and 999 which equal the sum of the cubes of their digits.
153, 407, Solution This seems to require one to look at a large number of cases. There are two basic approaches: to deal with each value of a in turn (for example a = 9, so 171 < b3+c3 < 271, so one of b,c is 5 or 6. If one is 5 etc); or to deal with each value of c in turn. The latter requires more cases but less thought: The idea is that if we fix c and one other digit, then the last digit is fixed by the requirement that a 3 +b 3 +c 3 = c mod 10. Thus, if c = 0 or 1, then the other two digits must be 1,9 giving 73x, or 2,8 giving 52x, or 3,7 giving 40x, or 4,6 giving 28x, or 5,5 giving 25x, none of which work. If c = 2, then the other two digits must be 0,4 giving 72, or 1,7 giving 353, or 2,6 giving 232, or 3,3 giving 62, or 5,9 giving 862, or 8,8 giving 1032, none of which work. If c = 3, then the other two digits must be 0,6 giving 243, or 1,5 giving 153, or 2,2 giving 43, or 3,9 giving 783, or 4,8 giving 603, or 7,7 giving 713. Only 153 works. If c = 4, then 1,9 →794, 2,8 →584, 3,7 →464, 4,6 →344, 5,5 →314, none of which work. If c = 5, then 1,9 →855, 2,8 → 645, 3,7 → 525, 4,6 → 405, 5,5 →375, none of which work. If c = 6, then 1,9 → 946, 2,8 → 736, 3,7 → 616, 4,6 →496, 5,5 → 466, none of which work. If c = 7, then 1,7 → 687, 2,6 → 567, 3,3 → 397, 4,0 → 407, 5,9 → 1197, 8,8 → 1367, and only 407 works. If c = 8, then 0,6 → 728, 1,5 → 638, 2,2 → 528, 3,9 → 1268, 4,8 → 1088, 7,7 → 1198, none of which work. If c = 9, then 1,9 and 2,8, and 3,7, and 4,6 give results which are too big and 5,5 → 979, which does not work. 7th Mexican 1993 © John Scholes jscholes@kalva.demon.co.uk 29 February 2004 Last corrected/updated 29 Feb 04