Difference between revisions of "2024 AMC 10A Problems/Problem 5"
MRENTHUSIASM (talk | contribs) |
MRENTHUSIASM (talk | contribs) (→Solution) |
||
| Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
| − | Note that <math>2024=2^3\cdot11\cdot23 | + | 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> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 15:43, 8 November 2024
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 
~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. 
