Problem 3. Determine the natural numbers $p$ with the property that the sum of the first $p$ non-zero natural numbers is a four-digit natural number with prime factorization $2^{m} \cdot 3^{n} \cdot (m+n)$, where $m, n \in \mathbb{N}^{*}$.
Solution. Clearly, $m+n \geq 5$. If $m+n=5$, the largest value that the number $N=2^{m} \cdot 3^{n} \cdot (m+n)$ can take is $2^{1} \cdot 3^{4} \cdot 5=810$, which does not have four digits, contradiction $\ldots \mathbf{2 p}$

Assuming that $m+n \geq 11$, then $N \geq 2^{10} \cdot 3 \cdot 11>10000$, so $N$ cannot have four digits, contradiction $2 p$

Therefore, $m+n=7$. 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 $2^{2} \cdot 3^{5} \cdot 7=6804$.

Considering that $N=\frac{p(p+1)}{2}$, analyzing the cases, only $N=2016$ fits, for which we get $p=63$

$3 \mathrm{p}$