## Problem 2.

Find the primes $p, q, r$, given that one of the numbers $p q r$ and $p+q+r$ is 101 times the other.


Solution. We may assume $r=\max \{p, q, r\}$. Then $p+q+r \leq 3 r$ and $p q r \geq 4 r$. So the sum of the three primes is always less than their product. The only relevant requirement thus is $p q r=101(p+q+r)$. We observe that 101 is a prime. So one of $p, q, r$ must be 101. Assume $r=101$. Then $p q=p+q+101$. This can be written as $(p-1)(q-1)=102$. Since $102=1 \cdot 102=2 \cdot 51=3 \cdot 34=6 \cdot 17$, the possibilities for $\{p, q\}$ are $\{2,103\},\{3,52\},\{4,35\},\{7,18\}$ The only case, where both the numbers are primes, is $\{2,103\}$. So the only solution to the problem is $\{p, q, r\}=\{2,101,103\}$.

