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Difference between revisions of "2024 AMC 10A Problems/Problem 5"
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A user attempted to look at 2024 AMC 10A Problem 5. What is the probability that they can see the question before the actual test date?
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{{duplicate|[[2024 AMC 10A Problems/Problem 5|2024 AMC 10A #5]] and [[2024 AMC 12A Problems/Problem 4|2024 AMC 12A #4]]}}
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== Problem ==
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What is the least value of <math>n</math> such that <math>n!</math> is a multiple of <math>2024</math>?
  
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(A) 0% (B) 0% (C) 0% (D) 0% (E) 0% (F) skibidi
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<math>\textbf{(A) } 11\qquad\textbf{(B) } 21\qquad\textbf{(C) } 22\qquad\textbf{(D) } 23\qquad\textbf{(E) } 253</math>
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== Solution==
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Note that <math>2024=2^3\cdot11\cdot23</math> in the prime factorization. Since <math>23!</math> is a multiple of <math>2^3, 11,</math> and <math>23,</math> we conclude that <math>23!</math> is a multiple of <math>2024.</math> Therefore, we have <math>n=\boxed{\textbf{(D) } 23}.</math>
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<u><b>Remark</b></u>
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Memorizing the prime factorization of the current year is useful for the AMC 8/10/12 Exams.
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~MRENTHUSIASM
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<u><b>Remark</b></u>
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Note that <math>2024</math> is <math>2025 -1</math>, which is <math>45^2 -1 = (45+1)(45-1) = 46\cdot44</math>.
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~WISETIGERJ2
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==Video Solution by Central Valley Math Circle==
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https://youtu.be/Hc5DxRT-DOU
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~mr_mathman
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==Video Solution==
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https://youtu.be/l3VrUsZkv8I
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== Video Solution by Math from my desk ==
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https://www.youtube.com/watch?v=fAitluI5SoY&t=3s
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== Video Solution (⚡️ 1 min solve ⚡️) ==
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 +
https://youtu.be/FD6rV3wGQ74
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<i>~Education, the Study of Everything</i>
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== Video Solution by Pi Academy ==
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https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW
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== Video Solution by Daily Dose of Math ==
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https://youtu.be/DXDJUCVX3yU
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~Thesmartgreekmathdude
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== Video Solution 1 by Power Solve ==
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https://youtu.be/j-37jvqzhrg?si=qwyiAvKLbySyDR7D&t=529
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=6SQ74nt3ynw
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==Video Solution by TheBeautyofMath==
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For AMC 10: https://youtu.be/uKXSZyrIOeU?t=1168
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For AMC 12: https://youtu.be/zaswZfIEibA?t=984
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~IceMatrix
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==Video Solution by Dr. David==
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https://youtu.be/uhpWASWW2ns
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==See also==
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{{AMC10 box|year=2024|ab=A|num-b=4|num-a=6}}
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{{AMC12 box|year=2024|ab=A|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 17:16, 16 June 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

See also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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