2022 AMC 10A Problem 7

The least common multiple and greatest common divisor both work one prime at a time, so break everything into primes first.

1. Prime factors. $18 = 2 \cdot 3^2$, $\ 180 = 2^2 \cdot 3^2 \cdot 5$, $\ 45 = 3^2 \cdot 5$, $\ 15 = 3 \cdot 5$.
2. What n can contain. $180$ is a multiple of $n$, so $n$ is built only from $2$s, $3$s and $5$s, and has no more of each than $180$ has.
3. The lcm condition. For each prime, the lcm uses the larger count from $n$ and $18$.
- Twos: $18$ has one $2$, but $180$ has two. The extra $2$ must come from $n$, so $n$ has exactly two $2$s.
- Fives: $18$ has no $5$, but $180$ has one, so $n$ has exactly one $5$.
- Threes: $18$ already has two $3$s, so this condition tells us nothing about the $3$s in $n$.
4. The gcd condition. For each prime, the gcd uses the smaller count. $45$ has two $3$s, but the gcd $15$ has only one. So $n$ has exactly one $3$. (With no $3$s the gcd would have none; with two $3$s it would have two.)
5. Put it together. $n = 2^2 \cdot 3 \cdot 5 = 60$. Check: $\operatorname{lcm}(60, 18) = 180$ and $\gcd(60, 45) = 15$.
6. The digit sum of $60$ is $6 + 0 = 6$.

The answer is $\boxed{\textbf{(B)}\ 6}$.

The lcm takes the larger power of each prime and the gcd takes the smaller one. So when a problem mixes the two, write every number in primes and settle one prime at a time. Here each condition fixed a different prime, and together they pinned $n$ down.

List the candidates. Since $\gcd(n, 45) = 15$, $n$ is a multiple of $15$ but not a multiple of $9$. The multiples of $15$ that divide $180$ are $15, 30, 45, 60, 90, 180$; removing the multiples of $9$ leaves $15, 30, 60$. For the lcm with $18$ to be $180$, $n$ must be a multiple of $4$, and only $60$ is.

Factor everything: $18 = 2 \cdot 3^2$, $180 = 2^2 \cdot 3^2 \cdot 5$, $45 = 3^2 \cdot 5$, $15 = 3 \cdot 5$. Write $n = 2^a 3^b 5^c$ (no other primes can appear, since $n$ divides $180$).

From $\operatorname{lcm}(n, 18) = 180$: the exponent of $2$ in the lcm is $\max(a, 1) = 2$, so $a = 2$; the exponent of $5$ is $\max(c, 0) = 1$, so $c = 1$; and $\max(b, 2) = 2$, so $b \le 2$.

From $\gcd(n, 45) = 15$: the exponent of $3$ is $\min(b, 2) = 1$, so $b = 1$ (and $\min(c, 1) = 1$ confirms $c = 1$).

Therefore $n = 2^2 \cdot 3 \cdot 5 = 60$, whose digit sum is $6 + 0 = 6$.

The answer is $\boxed{\textbf{(B)}\ 6}$.