2024 AMC 12A Problem 4

Attempt Problem 4 of the 2024 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 2024 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

$11$

$21$

$22$

$23$

$253$

Answer: D

Concepts:prime factorization factorial

Difficulty rating: 1180

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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