$30 \cdot 20$ If $p$ is a prime number, and the equation $x^{2}+p x-444 p=0$ has two integer roots, then
(A) $1<p \leqslant 11$.
(B) $11<p \leqslant 21$.
(C) $21<p \leqslant 31$.
(D) $31<p \leqslant 41$.
(E) $41<p \leqslant 51$.
(38th American High School Mathematics Examination, 1987)
［Solution］From the given conditions and Vieta's formulas, we have (assuming the two roots of the equation are $x_{1}, x_{2}$):
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-p \\
x_{1} x_{2}=-444 p=-4 \cdot 3 \cdot 37 p .
\end{array}\right.
$$

Since $p>1$,
it is impossible for $x_{1}=37 p, x_{2}=-12$, as we only need to note that $37 p-12 \neq-$ $p$.

Also, when $x_{1}= \pm 12 p, x_{2}=\mp 37$, by $12 p-37 \neq-p, 37-12 p$ $\neq-p$, it is also impossible.
When $x_{1}= \pm 3 p, \quad x_{2}=\mp 4 \cdot 37$,

from $4 \cdot 37-3 p=-p$, i.e., $4 \cdot 37=2 p$, we know $p$ is a composite number, which is not appropriate; while $3 p-4 \cdot 37=-p$, i.e., $4 p=4 \cdot 37$, we know $p=37$.
When $x_{1}= \pm 4 p, x_{2}=\mp 3 \cdot 37$,

from $4 p-3 \cdot 37=-p$, i.e., $5 p=3 \cdot 37$ is impossible;
while $3 \cdot 37-4 p=-p$, i.e., $3 p=3 \cdot 37$, we get $p=37$.
When $x_{1}= \pm p, x_{2}=\mp 444$,

from $p-444=-p$ we know $p=222$ is a composite number, which is not appropriate,
while $444-p=-p$, it is a contradiction!
In summary, $p=37$.
Therefore, the answer is $(D)$.