Slovenia — National Math Olympiad

The identities $qr = pqr - 15p - 7pq = p(qr - 15 - 7q)$ imply that $qr$ is divisible by $p$. Since $p$, $q$ and $r$ are prime numbers, there are only two possibilities: $p = q$ or $p = r$.

If $p = q$ we get $15 + 7q + r = qr$, which can be rewritten as
$$22 = qr - 7q - r + 7 = (q - 1)(r - 7).$$
Exactly one of the numbers $q-1$ and $r-7$ is odd, so one of the primes $q$ and $r$ is even. The only possible case is $q = 2$ and from this we get $r = 29$.

If $p = r$, then we can divide both sides by $r$ to get $15 + 8q = qr$ or $15 = q(r - 8)$. Now, $q$ is a prime and divides $15$. We conclude that $q$ is equal to either $3$ or $5$. If $q = 3$ then $r = 13$, if $q = 5$ then $r = 11$.

The solutions are $(p, q, r) = (2, 2, 29)$, $(13, 3, 13)$ and $(11, 5, 11)$.