In how many ways can 45 be written as a sum of two or more consecutive positive integers?

A run of k consecutive integers starting at a adds up to ka + k(k − 1)/2. Setting this equal to 45 and testing k = 2, 3, …: k = 2 gives 22 + 23, k = 3 gives 14 + 15 + 16, k = 5 gives 7 + 8 + 9 + 10 + 11, k = 6 gives 5 + 6 + … + 10 and k = 9 gives 1 + 2 + … + 9. Lengths 4, 7, 8 give no whole starting value, so there are 5 ways.