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

(Solution)
(Solution)
 
Line 8: Line 8:
 
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!}
 
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  
 
{\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 24, so there are <math>24 - 8 =  
+
integer unless <math>n+1</math> is an odd prime. There are 8 odd primes less than or equal to 24, so there  
 +
 
 +
are <math>24 - 8 =  
 
\boxed{\text{(C)}16}</math> numbers less than or equal to 24 that satisfy the condition.
 
\boxed{\text{(C)}16}</math> numbers less than or equal to 24 that satisfy the condition.
  

Latest 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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