N.B. In this exercise, and only this one, an answer without justification is requested.

Let $m>n>p$ be three (positive integer) prime numbers such that $m+n+p=74$ and $m-n-p=44$. Determine $m, n$, and $p$.

(A prime number is an integer strictly greater than one, and whose only divisors are one and itself.)
We find $m=59, n=13$ and $p=2$. Let's still provide a justification. We have $2 m=(m+n+p)+(m-n-p)=74+44=118$, so $m=59$. It follows that $n+p=74-59=15$. Since $n+p$ is odd, one of the two numbers is even. The only even prime number is 2, so $p=2$ and $n=13$.