2022 AMC 12A Problem 4

Below is the professionally curated solution for Problem 4 of the 2022 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12A solutions, or check the answer key.

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Concepts:least common multiple greatest common divisor prime factorization

Difficulty rating: 1200

The least common multiple of a positive integer $n$ and $18$ is $180,$ and the greatest common divisor of $n$ and $45$ is $15.$ What is the sum of the digits of $n?$

$3$

$6$

$8$

$9$

$12$

Solution:

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.

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