[Number Theory]
Problem: Find all positive integers $n>2$ such that

$$
n ! \mid \prod_{\substack{p<q \leqslant n \\ p, q \text { primes }}}(p+q)
$$

Solution: Assume that $n$ satisfies $n ! \mid \prod_{p<q \leqslant n}(p+q)$ and let $2=p_{1}<p_{2}<\cdots<p_{m} \leqslant n$ be the primes in $\{1,2, \ldots, n\}$. Each such prime divides $n$ !. In particular, $p_{m} \mid p_{i}+p_{j}$ for some $p_{i}<p_{j} \leqslant n$. But

$$
0<\frac{p_{i}+p_{j}}{p_{m}}<\frac{p_{m}+p_{m}}{p_{m}}=2
$$

so $p_{m}=p_{i}+p_{j}$ which implies $m \geqslant 3, p_{i}=2$ and $p_{m}=2+p_{j}=2+p_{m-1}$.

Similarly, $p_{m-1} \mid p_{k}+p_{l}$ for some $p_{k}<p_{l} \leqslant n$. But

$$
0<\frac{p_{l}+p_{k}}{p_{m-1}} \leqslant \frac{p_{m}+p_{m-1}}{p_{m-1}}=\frac{2 p_{m-1}+2}{p_{m-1}}<3
$$

so either $p_{m-1}=p_{l}+p_{k}$ or $2 p_{m-1}=p_{l}+p_{k}$. As above, the former case gives $p_{m-1}=2+p_{m-2}$. If $2 p_{m-1}=p_{l}+p_{k}$, then $p_{m-1}<p_{k}$, so $k=m$ and

$$
2 p_{m-1}=p_{l}+p_{m-1}+2 \Rightarrow p_{m-1}=p_{l}+2=p_{m-2}+2
$$

Either way, $p_{m-2}>2$ and 3 divides one of $p_{m-2}, p_{m-1}=p_{m-2}+2$ and $p_{m}=p_{m-2}+4$. This implies $p_{m-2}=3$ and thus $p_{m}=7$, giving $7 \leqslant n<11$.

Finally, a quick computation shows that $7 ! \mid \prod_{p<q \leqslant 7}(p+q)$ but 8 ! $\nmid \prod_{p<q \leqslant 7}(p+q)$, so neither does 9 ! and 10 !.

Final answer: 7