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2005 AMC 10A Problems/Problem 21: Difference between revisions

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==Solution==
==Solution==
If <math> 1+2+...+n </math> evenly divides <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]].  


Since <math> 1+2+...+n = \frac{n(n+1)}{2} </math> we may substitute the [[RHS]] in the above fraction.  
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 integers <math>n</math> is <math>\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}</math> an integer.  
So the problem asks us for how many [[postive 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 [[factor]] of <math>12</math>.  
<math>\frac{12}{n+1}</math> is an integer when <math>n+1</math> is a [[factor]] of <math>12</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>.
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>.


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>.  
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>.  


Therefore the number of possible values of <math>n</math> is <math>5\Rightarrow B</math>
Therefore the number of possible values of <math>n</math> is <math>5\Longrightarrow \mathrm{(B)}</math>


==See Also==
==See Also==

Revision as of 15:35, 2 August 2006

Problem

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

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

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 postive integers $n$ is $\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}$ an integer.

$\frac{12}{n+1}$ 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 $0$ isn't a positive integer, so only $1$, $2$, $3$, $5$, and $11$ are possible values of $n$.

Therefore the number of possible values of $n$ is $5\Longrightarrow \mathrm{(B)}$

See Also

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