Difference between revisions of "2005 AMC 10B Problems/Problem 22"
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For how many positive integers <math>n</math> less than or equal to <math>24</math> is <math>n!</math> evenly divisible by <math>1 + 2 + \cdots + n?</math> | For how many positive integers <math>n</math> less than or equal to <math>24</math> is <math>n!</math> evenly divisible by <math>1 + 2 + \cdots + n?</math> | ||
| − | + | <math>\textbf{(A) } 8 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \ qquad \textbf{(E) } 21 </math> | |
| − | <math>\ | ||
== Solution == | == Solution == | ||
Revision as of 15:28, 16 December 2021
Contents
Problem
For how many positive integers 
less than or equal to 
is 
evenly divisible by 
Solution
Since 
, the condition is equivalent to having an integer value for 
. This reduces, when 
, to having an integer value for 
. This fraction is an
integer unless 
is an odd prime. There are 8 odd primes less than or equal to 24, so there
are 
numbers less than or equal to 24 that satisfy the condition.
Video Solution
~savannahsolver
See Also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
