2022 AMC 12A Problem 4

Since $180=2^2\cdot3^2\cdot5$ and $18=2\cdot3^2,$ the condition $\operatorname{lcm}(n,18)=180$ forces $n$ to contribute $2^2$ and $5,$ with its power of $3$ at most $2.$ From $\gcd(n,45)=\gcd(n,3^2\cdot5)$ $=15=3\cdot5,$ the power of $3$ in $n$ is exactly $1$ and the power of $5$ is at least $1.$ Therefore $n=2^2\cdot3\cdot5=60,$ whose digits sum to $6.$ Thus, the correct answer is B.