$31 \cdot 45$ Given that $n$ is even, $m$ is odd, and the system of equations $\left\{\begin{array}{l}x-1988 y=n \\ 11 x+27 y=m\end{array}\right.$ has the solution $\left\{\begin{array}{l}x=p \\ y=q\end{array}\right.$ as integers, then
(A) $p, q$ are both even.
(B) $p, q$ are both odd.
(C) $p$ is even, $q$ is odd.
(D) $p$ is odd, $q$ is even.
("Zu Chongzhi Cup" Junior High School Mathematics Invitational Competition, 1988)
[Solution] Since $1988y$ is even, the first equation can be seen as: $x+$ even $=$ even, thus we know $p$ is even, so $11x$ is also even.

The second equation can be seen as: even $+27y=$ odd, thus we know $27y$ is odd, which means $q$ is odd.
Therefore, the answer is $(C)$.