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"
m (Reverted edits by Maa is stupid (talk) to last revision by MRENTHUSIASM)
(Tag: Rollback)
Line 1: Line 1:
−
==Problem==
+
{{duplicate|[[2024 AMC 10A Problems/Problem 5|2024 AMC 10A #5]] and [[2024 AMC 12A Problems/Problem 4|2024 AMC 12A #4]]}}
−
Andrea is taking a series of several exams. If Andrea earns <math>61</math> points on her next exam, her average score will decrease by <math>3</math> points. If she instead earns <math>93</math> points on her next exam, her average score will increase by <math>1</math> point. How many points should Andrea earn on her next exam to keep her average score constant?
+
== Problem ==
 +
What is the least value of <math>n</math> such that <math>n!</math> is a multiple of <math>2024</math>?
  
−
<math>\textbf{(A)}~80 \qquad\textbf{(B)}~82 \qquad\textbf{(C)}~83 \qquad\textbf{(D)}~85 \qquad\textbf{(E)}~86</math>
+
<math>\textbf{(A) } 11\qquad\textbf{(B) } 21\qquad\textbf{(C) } 22\qquad\textbf{(D) } 23\qquad\textbf{(E) } 253</math>
  
−
==Solution 1==
+
== Solution==
−
Let Andrea's current average be <math>a</math>, and assume there have been <math>n</math> tests already. The <math>(n + 1)</math>th test must have a score of <math>a + k(n + 1)</math> in order to increase her average by <math>k</math> and <math>a - k(n + 1)</math> in order to decrease her average by <math>k</math>. Hence, we see that the average must be <math>\tfrac{3}{4}</math> of the way from <math>61</math> to <math>93</math>, which is <math>\boxed{\textbf{(D)}~85}</math>.
+
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>
  
−
~joshualiu315
+
<u><b>Remark</b></u>
  
−
==Solution 2==
+
Memorizing the prime factorization of the current year is useful for the AMC 8/10/12 Exams.
−
Let <math>s</math> be the sum of Andrea's scores currently and let <math>n</math> be the number of tests. We have s+61n+1=sn−3s+93n+1=sn+1 Equate <math>s</math> to get a quadratic in <math>n</math> to find <math>n = 7</math>, <math>s = 595</math> so <math>\frac{595}{7} = \boxed{\textbf{(D)}~85}</math>.
 
  
−
~eg4334
+
~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
 +
 
 +
<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
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=4|num-a=6}}
 
{{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}}
 
{{MAA Notice}}

Revision as of 20:08, 21 March 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

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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_5&oldid=245763"