Mualpha7 wrote: Consider all sums that add up to $2015$. In each sum, the addends are consecutive positive integers, and all sums have less than $10$ addends. How many such sums are there? So $\sum_{k=1}^n(a+k)=2015$ with $a\ge 0$ and $n\in[1,9]$ So $n(2a+n+1)=4030$ and so $n$ is a divisor of $4030$ in $[1,9]$ : and so $n\in\{1,2,5\}$ $n=1$ gives $a=2014$ and the sequebce $\boxed{\text{S1 : }\{2015\}}$ $n=2$ gives $a=1006$ and the solution $\boxed{\text{S2 : }\{1007,1008\}}$ $n=5$ gives $a=400$ and the solution $\boxed{\text{S3 : }\{401,402,403,404,405\}}$