$30 \cdot 24$ Let $a_{n}=4^{2 n+1}+3^{n+2}(n=1,2,3, \cdots), p$ is the smallest prime that can divide infinitely many terms of $a_{1}, a_{2}$, $a_{3}, \cdots$, $q$ is the smallest prime that can divide every term of $a_{1}, a_{2}, a_{3}, \cdots$, then $p \cdot q$ is
(A) $5 \cdot 7$.
(B) $7 \cdot 13$.
(C) $13 \cdot 13$.
(D) None of the above answers.
(High School Mathematics Competition in Taiyuan City, Shanxi Province, China, 1993)
[Solution] Clearly, $p$ is not equal to 2 and 3. Let $n+2=4k (k \in \mathbb{N})$, then
\[
\begin{aligned}
a_{n} & =4 \cdot 16^{n}+3^{n+2} \equiv 4+3^{4k} \\
& \equiv 4+9^{2k} \equiv 4+(-1)^{2k} \equiv 0 \pmod{5}.
\end{aligned}
\]

Thus, $p=5$. This eliminates options $(B)$ and $(C)$.
Since $a_{1}=7 \cdot 13$, and
\[
a_{2}=4^{5}+3^{4}=4 \cdot 16^{2}+9^{2} \equiv 4 \cdot 2^{2}+2^{2} \equiv 20 \not\equiv 0 \pmod{7},
\]

it follows that $q \neq 7$, thus eliminating $(A)$.
Therefore, the correct choice is $(D)$.