$30 \cdot 19$ If $p, q$ are natural numbers, and the equation
$$
p x^{2}-q x+1985=0
$$

has two prime roots, then the value of $12 p^{2}+q$ is
(A) 404 .
(B) 1998 .
(C) 414 .
(D) 1996 .
(China Beijing Junior High School Mathematics Competition, 1985)
[Solution] Let the two roots of $p x^{2}-q x+1985=0$ be $x_{1}, x_{2}$, then
$$
x_{1} x_{2}=\frac{1985}{p}=\frac{5 \cdot 397}{p} \text {. }
$$

Since $x_{1}, x_{2}$ are both prime numbers, we have $p=1$, and the two roots $x_{1}, x_{2}$ are 5 and 397 respectively.
By $x_{1}+x_{2}=\frac{q}{p}, q=x_{1}+x_{2}=5+397=402$,
thus
$$
12 p^{2}+q=12+402=414 \text {. }
$$

Therefore, the answer is $(C)$.