2024 AMC 10A Problems/Problem 5

Problem

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

$\textbf{(A) } 11\qquad\textbf{(B) } 21\qquad\textbf{(C) } 22\qquad\textbf{(D) } 23\qquad\textbf{(E) } 253$

Solution

Note that $2024=2^3\cdot11\cdot23.$ Since $23!$ is a multiple of $2^3, 11,$ and $23,$ and $\gcd(2^3,11,23)=1,$ we conclude that $23!$ is a multiple of $2024.$ Therefore, we have $n=\boxed{\textbf{(D) } 23}.$

~MRENTHUSIASM

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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2024 AMC 10A Problems/Problem 5

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