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

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==Solution== | ==Solution== | ||

− | If <math> 1+2+...+n </math> evenly | + | 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 | + | 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 | + | 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\ | + | Therefore the number of possible values of <math>n</math> is <math>5\Longrightarrow \mathrm{(B)}</math> |

==See Also== | ==See Also== |

## Revision as of 16:35, 2 August 2006

## Problem

For how many positive integers does evenly divide ?

## Solution

If evenly divides , then is an integer.

Since we may substitute the RHS in the above fraction.

So the problem asks us for how many postive integers is an integer.

is an integer when is a factor of .

The factors of are , , , , , and .

So the possible values of are , , , , , and .

But isn't a positive integer, so only , , , , and are possible values of .

Therefore the number of possible values of is