2024 AMC 10A Problems/Problem 5: Difference between revisions
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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]]}} | |||
== Problem == | |||
What is the least value of <math>n</math> such that <math>n!</math> is a multiple of <math>2024</math>? | |||
(A) | <math>\textbf{(A) } 11\qquad\textbf{(B) } 21\qquad\textbf{(C) } 22\qquad\textbf{(D) } 23\qquad\textbf{(E) } 253</math> | ||
== Solution== | |||
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> | |||
<u><b>Remark</b></u> | |||
Memorizing the prime factorization of the current year is useful for the AMC 8/10/12 Exams. | |||
~MRENTHUSIASM | |||
<u><b>Remark</b></u> | |||
Note that <math>2024</math> is <math>2025 -1</math>, which is <math>45^2 -1 = (45+1)(45-1) = 46\cdot44</math>. | |||
~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 | |||
<i>~Education, the Study of Everything</i> | |||
== 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 | |||
==See Also== | |||
{{AMC10 box|year=2024|ab=A|before=[[2023 AMC 10B Problems]]|after=[[2024 AMC 10B Problems]]}} | |||
* [[AMC 10]] | |||
* [[AMC 10 Problems and Solutions]] | |||
* [[Mathematics competitions]] | |||
* [[Mathematics competition resources]] | |||
{{MAA Notice}} | |||
Latest revision as of 16:15, 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
such that
is a multiple of
?
Solution
Note that
in the prime factorization. Since
is a multiple of
and
we conclude that
is a multiple of
Therefore, we have
Remark
Memorizing the prime factorization of the current year is useful for the AMC 8/10/12 Exams.
~MRENTHUSIASM
Remark
Note that
is
, which is
.
~WISETIGERJ2
Video Solution by Central Valley Math Circle
~mr_mathman
Video Solution
Video Solution by Math from my desk
https://www.youtube.com/watch?v=fAitluI5SoY&t=3s
Video Solution (⚡️ 1 min solve ⚡️)
~Education, the Study of Everything
Video Solution by Pi Academy
https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW
Video Solution by Daily Dose of Math
~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
Video solution by TheNeuralMathAcademy
https://www.youtube.com/watch?v=4b_YLnyegtw&t=782s
See Also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by 2023 AMC 10B Problems |
Followed by 2024 AMC 10B Problems | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing