2024 AMC 12A Problem 4

Below is the professionally curated solution for Problem 4 of the 2024 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12A solutions, or check the answer key.

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Concepts:prime factorization factorial

Difficulty rating: 1180

What is the least value of $n$ such that $n!$ is a multiple of $2024?$

$11$

$21$

$22$

$23$

$253$

Solution:

Factoring, $2024=2^3\cdot11\cdot23.$ The factorial $n!$ contains the prime $23$ only when $n\ge23.$ At $n=23,$ the product $23!$ already includes $23,\ 11,$ and plenty of factors of $2,$ so $23!$ is a multiple of $2024.$ Thus, the correct answer is D.

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Problem 4 in Other Years

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