6th Iberoamerican 1991 Problem 4 Find a number N with five digits, all different and none zero, which equals the sum of all distinct three digit numbers whose digits are all different and are all digits of N. Solution
: 35964 There are 4.3 = 12 numbers with a given digit of n in the units place. Similarly, there are 12 with it in the tens place and 12 with it in the hundreds place. So the sum of the 3 digit numbers is 12.111 (a + b + c + d + e), where n = abcde. So 8668a = 332b + 1232c + 1322d + 1331e. We can easily see that a = 1 is too small and a = 4 is too big, so a = 2 or 3. Obviously e must be even. 0 is too small, so e = 2, 4, 6 or 8. Working mod 11, we see that 0 = 2b + 2d, so b + d = 11. Working mod 7, we see that 2a = 3b + 6d + e. Using the mod 11 result, b = 2, d = 9 or b = 3, d = 8 or b = 4, d = 7 or b = 5, d = 6 or b = 6, d = 5 or b = 7, d = 4 or b = 8, d = 3 or b = 9, d = 2. Putting each of these into the mod 7 result gives 2a - e = 4, 1, 5, 2, 6, 3, 0, 4 mod 7. So putting a = 2 and remembering that e must be 2, 4, 6, 8 and that all digits must be different gives a, b, d, e = 2,4, 7, 6 or 2, 7, 4, 8 or 2, 8, 3, 4 as the only possibilities. It is then straightforward but tiresome to check that none of these give a solution for c. Similarly putting a = 4, gives a, b, d, e = 3, 4, 7, 8 or 3, 5, 6, 4 as the only possibilities. Checking, we find the solution above and no others. 6th Ibero 1991 © John Scholes jscholes@kalva.demon.co.uk 11 July 2002