Local Mathematical Competitions

All symbols in the sequel are denoting integer numbers. Let $n=2^{a}b$, $a\ge 0$, $b\ge 1$, $b$ odd. We want to have $n=(m+1)+(m+2)+\cdots +(m+k)$, with $m\ge 0$ and $k\ge 2$, hence $k(2m+k+1)=2^{a+1}b$.

If $b=1$, it follows $k=2^{\alpha }$, with $1\le \alpha \le a+1$, but then $2m+k+1>1$ is odd, so there are no solutions. Now, for $b>1$, we can exhibit the required writing.

If $b\ge 2^{a+1}+1$, then take $m=\frac{1}{2}(b-2^{a+1}-1)$ and $k=2^{a+1}$.

If $b\le 2^{a+1}-1$, then take $m=\frac{1}{2}(2^{a+1}-1-b)$ and $k=b$.