13th Chinese 1998 Problem B1 Find all positive integers n > 2 for which (1 + n + n(n-1)/2 + n(n-1)(n-2)/6 ) divides 2 2000 . Solution
: n = 3, 7 or 23. (1 + n + n(n-1)/2 + n(n-1)(n-2)/6 ) = n+1 + n(n-1)(3+n-2)/6 = (n+1)(n 2 -n+6)/6. If this divides 2 2000 , then (n+1)(n 2 -n+6) divides 3.2 2000 , so each of n+1 and n 2 -n+6 divide 3.2 2001 . There are two cases to consider. Case 1. n+1 = 3.2 k for some k > 0 (since n > 2). Then n 2 -n+6 = 2 h for some h. But n 2 -n+6 = 9·2 k - 9·2 k + 8. For k = 1, 2, 3 this gives 26, 116, 512. But 26 and 116 are not powers of 2. k = 3 works, giving n = 23 as a solution. If k > 3, then n 2 -n+6 = 8 mod 16, so it is not a power of 2. Thus n = 23 is the only solution with n+1 divisible by 3. Case 2. n+1 = 2 k for some k > 1. Then n 2 -n+6 = 2 2k -3.2 k + 8. For k = 2, 3, 4 this gives 12, 48, 216. If k > 4, then n 2 -n+6 = 8 mod 32, which is incompatible with being of the form 3·2 h . Thus the only solutions with n+1 a power of 2 are, n = 3, 7. 13th Chinese 1998 © John Scholes jscholes@kalva.demon.co.uk 19 June 2002