2019 AMC 10B Problem 14

Because $19!$ is divisible by $1000$, its last three digits are zero, so $H=0$. Since $19!$ is divisible by $9$, its digit sum $33+T+M$ is divisible by $9$. Hence $T+M\equiv3\pmod9$, so $T+M$ is either $3$ or $12$. Divisibility by $11$ says the alternating digit sum $T-M-7$ is divisible by $11$, so $T-M\equiv7\pmod{11}$. Checking the digit possibilities from these two congruences gives $T=4$ and $M=8$. Therefore $T+M+H=4+8+0=12$. Thus, the answer is C .