$29-35$ Let $a$ be a positive integer, $a<100$, and $a^{3}+23$ is divisible by 24. Then, the number of such $a$ is
(A) 4.
(B) 5.
(C) 9.
(D) 10.
(China High School Mathematics League, 1991)
[Solution] We only need to consider $24 \mid \left(a^{3}-1\right)$, we have
$$
a^{3}-1=(a-1)\left(a^{2}+a+1\right)=(a-1)[a(a+1)+1] \text {. }
$$

Since $a(a+1)+1$ is odd, for $24 \mid \left(a^{3}-1\right)$, it must be that $2^{3} \mid (a-1)$.
If $a-1$ is not divisible by 3, then $3 \mid a(a+1)$, which means $a(a+1)+1$ cannot be divisible by 3.
Therefore, for $24 \mid \left(a^{3}-1\right)$, it must be that $3 \mid (a-1)$. This leads to
$24 \mid (a-1)$, i.e., $a=24k+1 \quad(k \in \mathbb{Z})$.
Given $24k+1<100$, $k$ can take all possible values of $0,1,2,3,4$, which means such $a$ values are $1,25,49,73,97$—five in total.
Thus, the answer is $(B)$.