$33 \cdot 43$ If $1 \cdot 2 \cdot 3 \cdot 4 \cdot \cdots \cdot 100=12^{n} \cdot M$, where $M$ is a natural number, and $n$ is the largest natural number that makes the equation true. Then $M$
(A) can be divided by 2, but not by 3.
(B) can be divided by 3, but not by 2.
(C) can be divided by 4, but not by 3.
(D) cannot be divided by 3, nor by 2.
(China Junior High School Mathematics League, 1991)
[Solution] In the prime factorization of $1 \cdot 2 \cdot 3 \cdot 4 \cdots \cdot 99 \cdot 100$, the number of factors of 2 is
$$
\begin{aligned}
& {\left[\frac{100}{2}\right]+\left[\frac{100}{2^{2}}\right]+\left[\frac{100}{2^{3}}\right]+\left[\frac{100}{2^{4}}\right]+\left[\frac{100}{2^{5}}\right]+\left[\frac{100}{2^{6}}\right]+\left[\frac{100}{2^{7}}\right]+\cdots } \\
= & 50+25+12+6+3+1=97
\end{aligned}
$$
The number of factors of 3 is
$$
\begin{array}{l}
\quad\left[\frac{100}{3}\right]+\left[\frac{100}{3^{2}}\right]+\left[\frac{100}{3^{3}}\right]+\left[\frac{100}{3^{4}}\right]+\left[\frac{100}{3^{5}}\right]+\left[\frac{100}{3^{6}}\right]+\cdots \\
=33+11+3+1=48 .
\end{array}
$$

Therefore, $1 \cdot 2 \cdot 3 \cdot 4 \cdot \cdots \cdot 99 \cdot 100=2^{97} \cdot 3^{48} \cdot P=12^{48}(2 P)$, where $P$ cannot be divided by 2 and 3.
Thus,
$$
M=2 P .
$$

Hence, the answer is $(A)$.