, 2022

Solution:

Since all of the coefficients are positive, any root $x$ must be negative. Moreover, by the rational root theorem, in order for $x$ to be an integer we must have either $x = -1$ or $x = -r$. So we must have either $p r^{2} - q r + r = 0 \Longleftrightarrow p r = q - 1$ or $p - q + r = 0$. Neither of these cases are possible if all three primes are odd, so we know that one of the primes is even, hence equal to $2$. After this we can do a casework check; the valid triples of $(p, q, r)$ are $(2,5,3), (2,7,5), (2,13,11), (2,19,17), (2,5,2), (2,7,3), (2,11,5), (2,23,11)$, allowing for $p$ and $r$ to be swapped. This leads to $15$ valid triples out of $1000$ (there are $10$ primes less than $30$).