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

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== Problem ==
 
== Problem ==
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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 + \ldots + 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>\text{(A) 162    } \text{(B) 180    } \text{(C) 324    } \text{(D) 360    } \text{(E) 720    }</math>
+
<math>\textbf{(A) } 8 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 21 </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!}{\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 8 odd primes less than or equal to 25, so there are <math>24 - 8 = \boxed{\text{(C)}16}</math> numbers less than or equal to 24 that satisfy the condition.
+
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  
 +
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.
 +
 
 +
==Video Solution==
 +
https://youtu.be/Ji5BR4SFkeE
 +
 
 +
~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 $n$ 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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