$32 \cdot 49$ Let the roots of the quadratic equation $x^{2}+b x+c=0$ be 98 and 99. In the quadratic function $y=x^{2}+b x+c$, if $x$ takes the values $0,1,2,3 \cdots, 100$, then the number of times $y$ is divisible by 6 is 
(A) 33.
(B) 34.
(C) 65.
(D) 67.
(China Sichuan Junior High School Mathematics Competition, 1991)
[Solution] Given $y=(x-98)(x-99)$, to make $6 \mid y$, there are only the following two cases:
(1) $x-98=3 k$, and $\quad x-99=3 k-1 \quad(k \in Z)$.
By

we have $0 \leqslant x \leqslant 100$,

i.e., $\square$ $-98 \leqslant 3 k \leqslant 2$,
$-32 \leqslant k \leqslant \frac{2}{3}$,

thus $k$ can take 33 values.
$$
\text { (2) } x-99=3 k \text {, and } \quad x-98=3 k+1 \quad(k \in Z) \text {. }
$$

By
we have
$$
\begin{array}{c}
0 \leqslant x \leqslant 100, \\
-99 \leqslant 3 k \leqslant 1,
\end{array}
$$

i.e., $\square$
$$
-33 \leqslant k \leqslant \frac{1}{3},
$$

thus $k$ can take 34 values.
In summary, $k$ can take a total of $\quad 33+34=67$ values. Therefore, the answer is $(D)$.