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

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== Video Solution ==
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https://www.youtube.com/watch?v=XQQpjNuOL5E  ~David
  
 
== See Also ==
 
== See Also ==
 
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{{AMC10 box|year=2005|ab=B|num-b=21|num-a=23}}
 
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Latest revision as of 18:20, 17 September 2023

Problem

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

$\textbf{(A) } 8 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 21$

Solution

Since $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$, the condition is equivalent to having an integer value for $\frac{n!} {\frac{n(n+1)}{2}}$. 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 $24$, so there

are $24 - 8 = \boxed{\textbf{(C) }16}$ numbers less than or equal to $24$ that satisfy the condition.

Video Solution

https://youtu.be/Ji5BR4SFkeE

~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
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All AMC 10 Problems and Solutions

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