Belarus — Final Round

We have
$$\frac{p + q}{r} = p - q + r \Leftrightarrow q(r + 1) - p(r - 1) = r^2.$$
If $r > 2$, then $r$ is odd (since $r$ is prime). Thus the left-hand side of the latter equation is an even number while its right-hand side is an odd number, a contradiction. Therefore, $r = 2$. Then from the initial equality it follows that $p = 3q - 4$. So, if $q$ increases, then $p$ also increases. Thus, the product $pqr = (3q - 4) \cdot q \cdot 2$ has maximal value as $q$ has maximal value.

Since $p = 3q - 4$ and, by condition, $p + q < 111$, we have $3q - 4 + q = 4q - 4 < 111$, whence $q < 29$. Then $q$, as a prime, can admit only the following values $23$, $19$, $17$, $...$, $2$.

If $q = 23$, then $p = 3q - 4 = 3 \cdot 23 - 4 = 65$ is a composite number.
If $q = 19$, then $p = 3q - 4 = 3 \cdot 19 - 4 = 53$ is a prime number.

Thus, the largest possible value of the product $pqr = 53 \cdot 19 \cdot 2 = 2014$.