2024 AMC 10A Problems/Problem 5

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

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

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 3	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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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