6. Determine all integers $n \geq 3$ such that

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

Remark: The expression on the right-hand side denotes the product over all sums of two distinct primes less than or equal to $n$. For $n=6$, this is equal to $(2+3)(2+5)(3+5)$.
Solution:For a fixed $n \geq 3$, let us denote the product on the right by $P(n)$ and let $r$ be the largest prime less than or equal to $n$. We now observe that

$$
r|n!| P(n)=P(r)
$$

and therefore there are primes $p<q \leq r$ with $r \mid p+q$. But $0<p+q<2 r$ implies $p+q=r$. Since $r \geq 3$ is an odd prime, we must have $p=2$ and $q=r-2$ is an odd prime. But now we also have

$$
r-2|n!| P(n)=P(r)=P(r-2) \prod_{\substack{s \leq r-2 \\ s \text { prime }}}(r+s)
$$

Hence either $r-2 \mid P(r-2)$ and similar to the above case we find that $r-4$ is prime, or there is a prime $s \leq r-2$ with $r-2 \mid r+s$. Since $r-2<r+s<3(r-2)$, we must have $r+s=2(r-2)$ and again $s=r-4$ is prime. We conclude that all of $r, r-2, r-4$ are prime and they have distinct residues modulo 3 , it follows that $r-4=3$ and thus $r=7$. It remains to check the values $n=7,8,9$ and 10 .

Observing that

$$
7!=2^{4} \cdot 3^{2} \cdot 5 \cdot 7 \mid 2^{6} \cdot 3^{3} \cdot 5^{2} \cdot 7=P(7)
$$

we find that $n=7$ is a solution. However, since $P(7)=P(8)=P(9)=P(10)$ and since all of $8!, 9$ !, 10 ! are divisible by $2^{7}$, we conclude that $n=7$ is the only solution.