$32 \cdot 51$ For the equation $6 x^{2}=(2 m-1) x+m+1$ with respect to $x$, there is a root $\alpha$ that satisfies the inequality: $-1988 \leqslant \alpha \leqslant 1988$, and makes $\frac{3}{5} \alpha$ an integer, then the number of values $m$ can take is
(A) 6627 .
(B) 3977 .
(C) 2385 .
(D) 1989 .
(China Guangzhou, Wuhan, Fuzhou, and other five cities junior high school mathematics league, 1988)
[Solution] The original equation can be transformed into $(3 x-m-1)(2 x+1)=0$.
Thus, $\alpha=\frac{1}{3}(m+1)$.
Substituting $-1988 \leqslant \alpha \leqslant 1988$, and noting that $\frac{3}{5} \alpha$ is an integer, we have
$$
-1192 \frac{4}{5} \leqslant \frac{3}{5} \alpha=\frac{1}{5}(m+1) \leqslant 1192 \frac{4}{5},
$$

Then $\frac{3}{5} \alpha$ can take 2385 values from $-1192$ to $1192$, thus $m$ can also take 2385 values. Therefore, the answer is $(C)$.