5. If $1 \times 2 \times 3 \times \cdots \times 99 \times 100=12^{n} M$, where $M$ is a natural number, and $n$ is the largest natural number that makes the equation true, then $M$
(A) is divisible by 2, but not by 3;
(B) is divisible by 3, but not by 2;
(C) is divisible by 4, but not by 3;
(D) is not divisible by 3, nor by 2.
5. (A)

Solution In the prime factorization of $1 \times 2 \times 3 \times \cdots \times 100$, the number of factors of 2 is
$$
\begin{array}{l}
{\left[\frac{100}{2}\right]+\left[\frac{100}{2^{2}}\right]+\left[\frac{100}{2^{3}}\right]+\cdots+\left[\frac{100}{2^{6}}\right]} \\
=50+25+12+6+3+1=97 \text { (factors) ; }
\end{array}
$$
the number of factors of 3 is
$$
\begin{array}{l}
{\left[\frac{100}{3}\right]+\left[\frac{100}{3^{2}}\right]+\left[\frac{100}{3^{3}}\right]+\left[\frac{100}{3^{4}}\right]} \\
=33+11+3+1=48 \text { (factors). }
\end{array}
$$

Therefore, $1 \times 2 \times 3 \times \cdots \times 100=2^{97} \times 3^{48} \times P=12^{48} \times 2 P$, where 2 does not divide $P$, and 3 does not divide $P$, hence $M=2 P$.