Difference between revisions of "2005 AMC 10B Problems/Problem 22"
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== 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!}{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 8 odd primes less than or equal to 25, so there are <math>24 - 8 = \boxed{16}</math> numbers less than or equal to 24 that satisfy the condition. | ||
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== See Also == | == See Also == | ||
*[[2005 AMC 10B Problems]] | *[[2005 AMC 10B Problems]] | ||
Revision as of 19:27, 14 August 2010
Problem
For how many positive integers n less than or equal to 24 is n! evenly divisible by 1 + 2 + ... + n?
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 25, so there are 
numbers less than or equal to 24 that satisfy the condition.
