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Difference between revisions of "2005 AMC 10A Problems/Problem 21"

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For how many positive integers <math>n</math> does <math> 1+2+...+n </math> evenly divide from <math>6n</math>?  
 
For how many positive integers <math>n</math> does <math> 1+2+...+n </math> evenly divide from <math>6n</math>?  
  
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<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 11 </math>
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<math> \textbf{(A) } 3\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 9\qquad \textbf{(E) } 11 </math>
  
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==Solution==
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== Solution ==
 
If <math> 1+2+...+n </math> evenly [[divide]]s <math>6n</math>, then <math>\frac{6n}{1+2+...+n}</math> is an [[integer]].  
 
If <math> 1+2+...+n </math> evenly [[divide]]s <math>6n</math>, then <math>\frac{6n}{1+2+...+n}</math> is an [[integer]].  
  
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Since <math> 1+2+...+n = \frac{n(n+1)}{2} </math> we may substitute the [[RHS]] in the above [[fraction]].  
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Since <math> 1+2+...+n = \frac{n(n+1)}{2} </math> we may substitute the [[RHS]] in the above [[fraction]]. So the problem asks us for how many [[positive integer]]s <math>n</math> is <math>\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}</math> an integer, or equivalently when <math>k(n+1) = 12</math> for a positive integer <math>k</math>.  
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So the problem asks us for how many [[positive integer]]s <math>n</math> is <math>\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}</math> an integer.  
 
  
 
<math>\frac{12}{n+1}</math> is an integer when <math>n+1</math> is a [[divisor |factor]] of <math>12</math>.  
 
<math>\frac{12}{n+1}</math> is an integer when <math>n+1</math> is a [[divisor |factor]] of <math>12</math>.  
  
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The factors of <math>12</math> are <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>6</math>, and <math>12</math>.
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The factors of <math>12</math> are <math>1, 2, 3, 4, 6,</math> and <math>12</math>, so the possible values of <math>n</math> are <math>0, 1, 2, 3, 5,</math> and <math>11</math>.
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So the possible values of <math>n</math> are <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, and <math>11</math>.
 
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But <math>0</math> isn't a positive integer, so only <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, and <math>11</math> are possible values of <math>n</math>.  
 
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Therefore the number of possible values of <math>n</math> is <math>5\Longrightarrow \mathrm{(B)}</math>
 
  
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==See Also==
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But since <math>0</math> isn't a positive integer, only <math>1, 2, 3, 5,</math> and <math>11</math> are the possible values of <math>n</math>. Therefore the number of possible values of <math>n</math> is <math>\boxed{\textbf{(B) }5}</math>.
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*[[2005 AMC 10A Problems]]
 
  
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*[[2005 AMC 10A Problems/Problem 20|Previous Problem]]
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==Video Solution==
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CHECK OUT Video Solution: https://youtu.be/WRv86DHa3zY
  
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*[[2005 AMC 10A Problems/Problem 22|Next Problem]]
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==See also==
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{{AMC10 box|year=2005|ab=A|num-b=20|num-a=22}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 11:54, 14 December 2021

Problem

For how many positive integers $n$ does $1+2+...+n$ evenly divide from $6n$?

$\textbf{(A) } 3\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 9\qquad \textbf{(E) } 11$

Solution

If $1+2+...+n$ evenly divides $6n$, then $\frac{6n}{1+2+...+n}$ is an integer.

Since $1+2+...+n = \frac{n(n+1)}{2}$ we may substitute the RHS in the above fraction. So the problem asks us for how many positive integers $n$ is $\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}$ an integer, or equivalently when $k(n+1) = 12$ for a positive integer $k$.

$\frac{12}{n+1}$ is an integer when $n+1$ is a factor of $12$.

The factors of $12$ are $1, 2, 3, 4, 6,$ and $12$, so the possible values of $n$ are $0, 1, 2, 3, 5,$ and $11$.

But since $0$ isn't a positive integer, only $1, 2, 3, 5,$ and $11$ are the possible values of $n$. Therefore the number of possible values of $n$ is $\boxed{\textbf{(B) }5}$.

Video Solution

CHECK OUT Video Solution: https://youtu.be/WRv86DHa3zY

See also

2005 AMC 10A (Problems • Answer Key • Resources)
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Problem 20 		Followed by
Problem 22
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All AMC 10 Problems and Solutions

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