8th Brazil 1986 Problem 2 Find the number of ways that a positive integer n can be represented as a sum of one or more consecutive positive integers.
The number of odd factors of 2n. Solution (m+1) + (m+2) + ... + (m+k) = k(2m+k+1)/2. So we require k(2m+k+1) = 2n. Note that k and 2m+k+1 have opposite parity and that k < 2m+k+1. Now suppose 2n = ab with a odd. Then b must be even. So a and b cannot be equal. Take k to be the smaller, then put m = |a-b-1|/2 and we have a solution. So the total number of solutions is just the number of odd factors of 2n. 8th Brazil 1986 © John Scholes jscholes@kalva.demon.co.uk 24 Oct 2003 Last corrected/updated 24 Oct 03