, 2024

Solution:
By Vieta, the sum of the roots is $-10 \pmod{p}$. However, since the three roots are less than $p / 3$, it follows that the roots are $\left(p - a'\right) / 3$, $\left(p - b'\right) / 3$, and $\left(p - c'\right) / 3$, where there are finitely many choices $a' < b' < c'$. By pigeonhole, one choice, say $(u, v, w)$, must occur for infinitely many $p$. We then get that the roots of $P$ are $-u / 3$, $-v / 3$, and $-w / 3$. Moreover, we must have that $u, v, w$ are all $1 \pmod{3}$ or all $2 \pmod{3}$, and by Vieta, we have $u + v + w = 30$.

The polynomial is then uniquely determined by $u, v, w$. Thus, it suffices to count triples $u < v < w$ of positive integers such that $u, v, w$ are all $1 \pmod{3}$ or all $2 \pmod{3}$ and that $u + v + w = 30$. It's not very hard to list them all now.

When $u, v, w \equiv 1 \pmod{3}$, there are 7 triples: $(1, 4, 25)$, $(1, 7, 22)$, $(1, 10, 19)$, $(1, 13, 16)$, $(4, 7, 19)$, $(4, 10, 16)$, and $(7, 10, 13)$.

When $u, v, w \equiv 2 \pmod{3}$, there are 5 triples: $(2, 5, 23)$, $(2, 8, 20)$, $(2, 11, 17)$, $(5, 8, 17)$, and $(5, 11, 14)$.

Hence, the answer is $7 + 5 = 12$.