Difference between revisions of "2024 AMC 10A Problems/Problem 5"
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Note: This is another example where knowing the prime factorization of the year is very useful | Note: This is another example where knowing the prime factorization of the year is very useful | ||
| − | == Solution 2 ( | + | == Solution 2 (the carbon copy of Solution 1)== |
Again, note that the prime factorization of 2024 is <math>2^3\cdot11\cdot23</math>. We know that for a factorial to contain a prime number <math>n</math>, it must be at least <math>n!</math>. Therefore, we have <math>n=\boxed{\textbf{(D) } 23}.</math> | Again, note that the prime factorization of 2024 is <math>2^3\cdot11\cdot23</math>. We know that for a factorial to contain a prime number <math>n</math>, it must be at least <math>n!</math>. Therefore, we have <math>n=\boxed{\textbf{(D) } 23}.</math> | ||
Revision as of 23:28, 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.
Contents
Problem
What is the least value of 
such that 
is a multiple of 
?
Solution 1
Note that 
in the prime factorization. Since 
is a multiple of 
and 
we conclude that is a multiple of 
Therefore, we have 
~MRENTHUSIASM
Note: This is another example where knowing the prime factorization of the year is very useful
Solution 2 (the carbon copy of Solution 1)
Again, note that the prime factorization of 2024 is 
. We know that for a factorial to contain a prime number , it must be at least . Therefore, we have
~FRANKLIN2013
Video Solution by Daily Dose of Math
~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. 
