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

(→‎Solution 2 (very similar to Solution 1))
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~FRANKLIN2013
 
~FRANKLIN2013
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== Video Solution 1 by Power Solve ==
 
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https://youtu.be/j-37jvqzhrg?si=qwyiAvKLbySyDR7D&t=529
 
  
 
== Video Solution by Daily Dose of Math ==
 
== Video Solution by Daily Dose of Math ==
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~Thesmartgreekmathdude
 
~Thesmartgreekmathdude
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== Video Solution 1 by Power Solve ==
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https://youtu.be/j-37jvqzhrg?si=qwyiAvKLbySyDR7D&t=529
  
 
==See also==
 
==See also==

Revision as of 21:42, 8 November 2024

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 1

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 $n=\boxed{\textbf{(D) } 23}.$

~MRENTHUSIASM

Note: This is another example where knowing the prime factorization of the year is very useful

Solution 2 (very similar to Solution 1)

Again, note that the prime factorization of 2024 is $2^3\cdot11\cdot23$. We know that for a factorial to contain a prime number $n$, it must be at least $n!$. Therefore, we have $n=\boxed{\textbf{(D) } 23}.$

~FRANKLIN2013

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

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

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