2022 AMC 12A Problem 4

Attempt Problem 4 of the 2022 AMC 12A below, then check your answer against the professionally curated solution 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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4.

The least common multiple of a positive integer nn and 1818 is 180,180, and the greatest common divisor of nn and 4545 is 15.15. What is the sum of the digits of n?n?

33

66

88

99

1212

Answer: B
Concepts:least common multiplegreatest common divisorprime factorization
Difficulty rating: 1200
Solution:

Since 180=22325180=2^2\cdot3^2\cdot5 and 18=232,18=2\cdot3^2, the condition lcm(n,18)=180\operatorname{lcm}(n,18)=180 forces nn to contribute 222^2 and 5,5, with its power of 33 at most 2.2.

From gcd(n,45)=gcd(n,325)\gcd(n,45)=\gcd(n,3^2\cdot5) =15=35,=15=3\cdot5, the power of 33 in nn is exactly 11 and the power of 55 is at least 1.1.

Therefore n=2235=60,n=2^2\cdot3\cdot5=60, whose digits sum to 6.6.

Thus, the correct answer is B.

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