$30 \cdot 41$ If $x$ is the cube of some positive integer, and $d$ represents the number of positive integer divisors of $x$. Then $d$ could be
(A) 200 .
(B) 201 .
(C) 202 .
(D) 203 .
(E) 204 .
(42nd American High School Mathematics Examination, 1991)
［Solution］Suppose $x$ is a perfect cube, then $x$ can be expressed as $x=p_{1}^{3 \alpha_{1}} p_{2}^{3 \alpha_{2}} \cdots p_{k}^{3 \alpha_{k}}$, where $p_{i}$ are prime numbers, and $\alpha_{i}$ are non-negative integers $(1 \leqslant i \leqslant k)$. Therefore, the number of its divisors is $\left(3 \alpha_{1}+1\right)\left(3 \alpha_{2}+1\right) \cdots\left(3 \alpha_{k}+1\right)$, which is an integer of the form $3 m+1$.
Obviously, by $2+0+2=4 \equiv 1(\bmod 3)$, we know $202 \equiv 1(\bmod 3)$. Therefore, the answer is $(C)$.