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Difference between revisions of "2005 AMC 10A Problems/Problem 21"
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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>. Therefore the number of possible values of <math>n</math> is <math>5\Longrightarrow \boxed{\mathrm{(B)}}</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\Longrightarrow \boxed{\mathrm{(B)}}</math>.
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CHECK OUT Video Solution: https://youtu.be/WRv86DHa3zY
  
 
==See also==
 
==See also==

Revision as of 19:51, 30 October 2020

Problem

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

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

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 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 $0$ 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 $5\Longrightarrow \boxed{\mathrm{(B)}}$.

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

See also

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

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