Difference between revisions of "2005 AMC 10B Problems/Problem 22"

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== Problem ==
 
== Problem ==
For how many positive integers n less than or equal to 24 is n! evenly divisible by 1 + 2 + ... + n?
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For how many positive integers n less than or equal to <math>24</math> is <math>n!</math> evenly divisible by <math>1 + 2 + \ldots + n</math>?
  
 
== Solution ==
 
== Solution ==
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== See Also ==
 
== See Also ==
*[[2005 AMC 10B Problems]]
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{{AMC10 box|year=2005|ab=B|num-b=21|num-a=23}}

Revision as of 16:46, 18 July 2011

Problem

For how many positive integers n less than or equal to $24$ is $n!$ evenly divisible by $1 + 2 + \ldots + n$?

Solution

Since $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$, the condition is equivalent to having an integer value for $\frac{n!}{n(n+1)/2}$. This reduces, when $n\ge 1$, to having an integer value for $\frac{2(n-1)!}{n+1}$. This fraction is an integer unless $n+1$ is an odd prime. There are 8 odd primes less than or equal to 25, so there are $24 - 8 = \boxed{16}$ numbers less than or equal to 24 that satisfy the condition.

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
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