21. D.

From the problem, we know that $[n, 5!]=5(n, 10!)$, which means $\left[5 \times 2^{3} \times 3, n\right]=5\left(2^{8} \times 3^{4} \times 5^{2} \times 7, n\right)$.
Therefore, the prime factors of $n$ are $2, 3, 5, 7$.
Let $n=2^{\alpha} 3^{\beta} 5^{\gamma} 7^{\tau}$, where $\alpha \in\{3,4, \cdots, 8\}$, $\beta \in\{1,2,3,4\}$, $\gamma \in\{3\}$, $\tau \in\{0,1\}$.
Thus, the number of possible $n$ is $6 \times 4 \times 1 \times 2=48$.
21. Given a positive integer $n$ that is a multiple of 5, and the least common multiple of $n$ and 5! is 5 times the greatest common divisor of $n$ and 10!. Then the number of $n$ that satisfies the condition is ( ).
(A) 12
(B) 24
(C) 36
(D) 48
(E) 72