$30 \cdot 12$ three prime numbers $p, q, r$, satisfy $p+q=r$ and $1<p<q$, then $p$ equals
(A) 2.
(B) 3.
(C) 7.
(D) 13.
(E) 17.
(34th American High School Mathematics Examination, 1983)
[Solution] Let $p+q=r$, and $p, q, r$ are all prime numbers.
Except for 2, all other prime numbers are odd.
Clearly, $r \neq 2$, so $r$ is odd, which means $p, q$ must be one odd and one even.
Since $p<q$, then $p=2$.
Therefore, the answer is $(A)$.
Note: This proposition can be generalized and extended to:
Find the prime numbers $p, q, r$ that satisfy the equation $p^{q}+q^{p}=r$.
(53rd Moscow Mathematical Olympiad, 1990)
[Solution] If $p<q$, then there is only one solution: $p=2, q=3, r=17$.
Since $r$ is a prime number, one of $p, q$ must be even, so $p=2$. Furthermore, when $q$ is a prime number not less than 5, we have
\[
\begin{aligned}
2^{q}+q^{2} & =\left(2^{q}+1\right)+\left(q^{2}-1\right) \\
& =3\left(2^{q-1}-2^{q-2}+\cdots+1\right)+(q-1)(q+1),
\end{aligned}
\]

Since 3 divides $(q-1)(q+1)$, then 3 divides $\left(2^{q}+q^{2}\right)$, which contradicts the fact that $r$ is a prime number!