Difference between revisions of "2024 AMC 10A Problems/Problem 5"
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| + | == 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) } 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> Since <math>23!</math> is a multiple of <math>2^3, 11,</math> and <math>23,</math> and <math>\gcd(2^3,11,23)=1,</math> we conclude that <math>23!</math> is a multiple of <math>2024.</math> Therefore, we have <math>n=\boxed{\textbf{(D) } 23}.</math> | ||
| + | |||
| + | ~MRENTHUSIASM | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC10 box|year=2024|ab=A|num-b=4|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Revision as of 15:42, 8 November 2024
Problem
What is the least value of 
such that 
is a multiple of 
?
Solution
Note that 
Since 
is a multiple of 
and 
and 
we conclude that is a multiple of 
Therefore, we have 
~MRENTHUSIASM
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
