2024 AMC 10A Problem 5

Attempt Problem 5 of the 2024 AMC 10A 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 10A solutions, or check the answer key.

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

$11$

$21$

$22$

$23$

$253$

Answer: D

Concepts:factorial divisibility prime factorization

Difficulty rating: 1130

Solution:

Factor $2024 = 2^3 \cdot 11 \cdot 23.$ The prime $23$ is the bottleneck: for $23$ to divide $n!,$ we need $n \ge 23.$ At $n = 23,$ the product $23!$ already has $23,$ $11,$ and plenty of factors of $2,$ so $2024 \mid 23!.$ The least value is $23.$ Thus, D is the correct answer.

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