AMC10 2016 10B, Problem 18

In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?

Solution: Factor $345=3\cdot 5\cdot 23$. Suppose we take an odd number $k$ of consecutive integers, with the median as $m$. Then $mk=345$ with $\tfrac12k<m$. Looking at the factors of $345$, the possible values of $k$ are $3,5,15,23$ with medians as $115,69,23,15$ respectively. Suppose instead we take an even number $2k$ of consecutive integers, with median being the average of $m$ and $m+1$. Then $k(2m+1)=345$ with $k\le m$. Looking again at the factors of $345$, the possible values of $k$ are $1,3,5$ with medians $(172,173),(57,58),(34,35)$ respectively. Thus the answer is $\fbox{7}$.

Answer: 7