$31 \cdot 5$ satisfies the equation $1983=1982 x-1981 y$ for a pair of natural numbers:
(A) $x=12785, y=12768$.
(B) $x=12784, y=12770$.
(C) $x=11888, y=11893$.
(D) $x=1947, y=1945$.
(China Fujian Junior High School Mathematics Competition, 1983)
［Solution］By the properties of the parity of numbers and since 1983 is odd, while 1982 is even, its product with any integer remains even. Therefore, $1981 y$ should be odd, meaning $y$ should be odd. Thus, options $(A),(B)$ can be ruled out.
Now, let's examine the last digits of $x, y$ in options $(C),(D)$:
$$
\begin{array}{l}
\langle 1982 \cdot 11888-1981 \cdot 11893\rangle=3, \\
\langle 1982 \cdot 1947-1981 \cdot 1945\rangle=9,
\end{array}
$$

Here $\langle a\rangle$ denotes the last digit of $a$.
Therefore, the correct choice is $(C)$.