Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "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? (A) 0% (B) 0% (C) 0% (D) 0% (E) 0%")
 
(→‎Video Solution by Central Valley Math Circle)
 
(37 intermediate revisions by 22 users not shown)
Line 1: Line 1:
−
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?
+
{{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) 0% (B) 0% (C) 0% (D) 0% (E) 0%
+
<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
 +
 
 +
==See also==
 +
{{AMC10 box|year=2024|ab=A|num-b=4|num-a=6}}
 +
{{AMC12 box|year=2024|ab=A|num-b=3|num-a=5}}
 +
{{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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_5&oldid=251259"