Difference between revisions of "2024 AMC 10A Problems/Problem 5"
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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 == | == Problem == | ||
What is the least value of <math>n</math> such that <math>n!</math> is a multiple of <math>2024</math>? | What is the least value of <math>n</math> such that <math>n!</math> is a multiple of <math>2024</math>? | ||
| Line 4: | Line 5: | ||
<math>\textbf{(A) } 11\qquad\textbf{(B) } 21\qquad\textbf{(C) } 22\qquad\textbf{(D) } 23\qquad\textbf{(E) } 253</math> | <math>\textbf{(A) } 11\qquad\textbf{(B) } 21\qquad\textbf{(C) } 22\qquad\textbf{(D) } 23\qquad\textbf{(E) } 253</math> | ||
| − | == Solution == | + | == 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> | 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 | ~MRENTHUSIASM | ||
| + | |||
| + | == 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}} | ||
Latest revision as of 08:55, 15 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.
Contents
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
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
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. 
