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 + \cdots + n?</math>
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<math>\text{(A) 8    } \text{(B) 12    } \text{(C) 16    } \text{(D) 17    } \text{(E) 21    }</math>
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== Solution ==
 
== Solution ==
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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!}
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{\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
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integer unless <math>n+1</math> is an odd prime. There are 8 odd primes less than or equal to 24, so there
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are <math>24 - 8 =
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\boxed{\text{(C)}16}</math> numbers less than or equal to 24 that satisfy the condition.
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==Video Solution==
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https://youtu.be/Ji5BR4SFkeE
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~savannahsolver
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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}}
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{{MAA Notice}}

Revision as of 23:17, 31 May 2021

Problem

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


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

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

Video Solution

https://youtu.be/Ji5BR4SFkeE

~savannahsolver

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

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