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 .
3. (B)

Consider $a$ such that $24 \mid a^{3}-1$.
$$
a^{3}-1=(a-1)\left(a^{2}+a+1\right)=(a-1)[a(a+1)+1] \text {. }
$$

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 $3 \mid a(a+1)$,
which means $a(a+1)+1$ is not divisible by 3.
Therefore, for $24 \mid\left(a^{3}-1\right)$, it must be that $3 \mid(a-1)$. This implies
$24 \mid(a-1)$, i.e., $a=24 k+1, \quad(k \in Z)$.
Given $24 k+1<100$, $k$ can take the values $0,1,2,3,4$, which means there are five such $a$: $1,25,49,73,97$.