3. 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
(B)
3. 【Analysis and Solution】Consider the $a$ that makes $24 \mid a^{3}-1$.
$$
a^{3}-1=(a-1)\left(a^{2}+a+1\right)=(a-1)[a(a+1)+1] \text {. }
$$

Since $a(a+1)+1$ is odd,
if $24 \mid\left(a^{3}-1\right)$, it must be that $2^{3} \mid(a-1)$.
If $a-1$ is not divisible by 3, then $3 \mid a(a+1)$,
thus $a(a+1)+1$ cannot be divisible by 3.
Therefore, if $24 \mid\left(a^{3}-1\right)$, it must be that $3 \mid(a-1)$, which means
$24 \mid(a-1)$, i.e., $a=24 k+1$,
by $24 k+1<100, k$ can take all possible values of $0,1,2,3,4$, which means such $a$ are $1,25,49,73,97$ five.