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Difference between revisions of "2024 AMC 10A Problems/Problem 5"

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Revision as of 13:06, 18 August 2025

The following problem is from both the 2024 AMC 10A #5 and 2024 AMC 12A #4, so both problems redirect to this page.

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

Remark

Memorizing the prime factorization of the current year is useful for the AMC 8/10/12 Exams.

~MRENTHUSIASM

Remark

Note that $2024$ is $2025 -1$, which is $45^2 -1 = (45+1)(45-1) = 46\cdot44$.

~WISETIGERJ2

Video Solution by Central Valley Math Circle

https://youtu.be/Hc5DxRT-DOU

~mr_mathman

Video Solution

https://youtu.be/l3VrUsZkv8I

Video Solution by Math from my desk

https://www.youtube.com/watch?v=fAitluI5SoY&t=3s

Video Solution (⚡️ 1 min solve ⚡️)

https://youtu.be/FD6rV3wGQ74

~Education, the Study of Everything

Video Solution by Pi Academy

https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW

Video Solution by Daily Dose of Math

https://youtu.be/DXDJUCVX3yU

~Thesmartgreekmathdude

Video Solution 1 by Power Solve

https://youtu.be/j-37jvqzhrg?si=qwyiAvKLbySyDR7D&t=529

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

Video Solution by TheBeautyofMath

For AMC 10: https://youtu.be/uKXSZyrIOeU?t=1168

For AMC 12: https://youtu.be/zaswZfIEibA?t=984

~IceMatrix

Video Solution by Dr. David

https://youtu.be/uhpWASWW2ns

Video solution by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=782s

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