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 == | ||
Since <math>1 + 2 + \cdots + n = \frac{n(n+1)}{2}</math>, the condition is equivalent to having an integer value for <math>\frac{n!} | Since <math>1 + 2 + \cdots + n = \frac{n(n+1)}{2}</math>, the condition is equivalent to having an integer value for <math>\frac{n!} | ||
− | |||
{\frac{n(n+1)}{2}}</math>. This reduces, when <math>n\ge 1</math>, to having an integer value for <math>\frac{2(n-1)!}{n+1}</math>. This fraction is an | {\frac{n(n+1)}{2}}</math>. This reduces, when <math>n\ge 1</math>, to having an integer value for <math>\frac{2(n-1)!}{n+1}</math>. This fraction is an | ||
+ | integer unless <math>n+1</math> is an odd prime. There are <math>8</math> odd primes less than or equal to <math>24</math>, so there | ||
− | + | are <math>24 - 8 = | |
− | + | \boxed{\textbf{(C) }16}</math> numbers less than or equal to <math>24</math> that satisfy the condition. | |
− | \boxed{\ | ||
==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=XQQpjNuOL5E ~David | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2005|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:20, 17 September 2023
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 odd primes less than or equal to , so there
are numbers less than or equal to that satisfy the condition.
Video Solution
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=XQQpjNuOL5E ~David
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 |
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