67th Romanian Mathematical Olympiad

The number $m+n$ is prime and, obviously, $m+n\ge 5$. If $m+n=5$, the largest value for $N=2^m{3}^n(m+n)$ is $2^1\cdot 3^4\cdot 5=810$, which has only three digits. Suppose now that $m+n\ge 11$. Then $N\ge 2^{10}\cdot 3\cdot 11>10000$, hence $N$ cannot have four digits. So, $m+n=7$. In this case, the four-digit numbers are: $2^6\cdot 3^1\cdot 7=1344$, $2^5\cdot 3^2\cdot 7=2016$, $2^4\cdot 3^3\cdot 7=3024$, $2^3\cdot 3^4\cdot 7=4536$ and $2^2\cdot 3^5\cdot 7=6804$. The equality $N=\frac{p(p+1)}{2}$ can be fulfilled only for $N=2016$, when $p=63$.