33rd Hellenic Mathematical Olympiad

We have: $$\left\{\begin{array}{l}pqr=n\\ (p+1)(q+1)r=n+138\end{array}\right\}\iff \left\{\begin{array}{l}pqr=n\\ pqr+(p+q)r+r=n+138\end{array}\right\}\iff \left\{\begin{array}{l}pqr=n\\ (p+q+1)r=138\end{array}\right\}.$$ From equation $$(p+q+1)r=138=2\cdot 3\cdot 23(1)$$ we will determine the possible values of $p$, $q$, $r$ and in the sequel from the equation $pqr=n$. We will find all the possible values of $n$. Since $p$, $q$, $r$ are primes, the possible values of $r$ are $2$ or $3$ or $23$, and therefore we have the cases: If $r=2$, then $p+q+1=69\iff p+q=68$, and hence, $(p,q$ primes$)$ we find the pairs $$(p,q)=(7,61),(p,q)=(61,7),(p,q)=(31,37),(p,q)=(37,31).$$ Hence for the product $pqr=n$ we obtain the values: $n=7\cdot 61\cdot 2=854$ or $n=31\cdot 37\cdot 2=2294$. If $r=3$, then $p+q+1=46\iff p+q=45$, and hence: $$(p,q)=(2,43),(p,q)=(43,2)\Rightarrow n=2\cdot 43\cdot 3=258.$$ * If $r=23$, then $$p+q+1=6\iff p+q=5\iff (p,q)=(2,3)\text{ or }(p,q)=(3,2),$$ and hence $n=2\cdot 3\cdot 23=138$. Therefore the possible values of $n$ are: 138, 258, 854 και 2294.