AMC12 2020 A

First note that $n$ cannot have any prime factors greater than 7 because otherwise $\operatorname{lcm}(n, 5!)$ would contain such a factor but $5 \cdot \gcd(n, 10!)$ would not. Write $n = 2^a \cdot 3^b \cdot 5^c \cdot 7^d$. Observe that $5! = 2^3 \cdot 3 \cdot 5$ and $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$. In order that the power of 2 is the same on both sides of the given equation $\operatorname{lcm}(n, 5!) = 5 \cdot \gcd(n, 10!)$, it must be that $\max(a, 3) = \min(a, 8)$. This is true if and only if $3 \le a \le 8$. For powers of 3, it must be that $\max(b, 1) = \min(b, 4)$, which holds precisely when $1 \le b \le 4$. For powers of 5, it must be that $\max(c, 1) = 1 + \min(c, 2)$. This holds for $c = 0$, but $n$ is given to be a multiple of 5, so $c > 0$ and the only value that works is $c = 3$. For powers of 7, it must be that $\max(d, 0) = \min(d, 1)$, so $0 \le d \le 1$. There are 6 possible values for $a$, 4 for $b$, 1 for $c$, and 2 for $d$, so there are $6 \cdot 4 \cdot 1 \cdot 2 = 48$ such $n$.