$133 a$ is a positive integer, $a<100$, and $a^{3}+23$ is divisible by 24. The number of such $a$ is ( )
A. 4
B. 5
C. 9
D. 10
133 B. From the given, we have $24 \mid\left(a^{3}-1\right)$. We have
$$
a^{3}-1=(a-1)\left(a^{2}+a+1\right)=(a-1)[a(a+1)+1] .
$$

Since $a(a+1)+1$ is odd, 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, it must be that $3 \mid(a-1)$. Thus, we have $24 \mid(a-1)$, i.e., $a=24 k+1 \quad(k \in \mathbf{Z})$.
From $24 k+1<100$, the possible values for $k$ are $0,1,2,3,4$, which means there are five such $a$ values: 1, 25, 49, 73, 97. Therefore, the answer to this problem is B.