2022 AMC 10A Problem 7

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 $\boxed{\textbf{(B)}\ 6}$.

Both lcm and gcd act prime by prime: lcm takes the larger exponent, gcd the smaller. Each condition therefore becomes a small statement about a single exponent, and two conditions together pin every exponent down. Whenever a problem mixes gcd and lcm, factor and compare exponents rather than testing numbers.

Since $\gcd(n, 45) = 15$, $n$ is a multiple of $15$ but not of $9$; since $n \mid 180$, the candidates are $15, 30, 60$. Only $60$ is a multiple of $4$, which $\operatorname{lcm}(n, 18) = 180$ requires.

• Using only the lcm condition (n = 20, 60, or 180 all work) and guessing, instead of letting the gcd condition fix the power of 3.
• Taking $n = 180$ (digit sum $9$) or $n = 15$ (digit sum $6$, accidentally right) without checking both conditions.