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

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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> | <math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 11 </math> | ||

− | ==Solution== | + | ==Solution 1== |

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

+ | |||

+ | == Solution 2 == | ||

+ | The sum of the first <math>n</math> positive integers is <math>\frac{n(n+1)}{2}</math>. If this is to divide <math>6n</math>, then there exists a positive integer <math>k</math> such that: | ||

+ | |||

+ | <math>k \cdot \frac{n(n+1)}{2} = 6n</math> | ||

+ | |||

+ | <math>k(n+1) = 12</math> | ||

+ | |||

+ | Therefore, <math>k</math> and <math>n+1</math> are divisors of <math>12</math>. There are <math>6</math> divisors of <math>12</math>, which are <math>1, 2, 3, 4, 6, 12</math>. The divisors which multiply to <math>12</math> can be assigned to <math>k</math> and <math>n+1</math> in either order. However, when <math>1</math> is assigned to <math>n+1</math>, then <math>n=0</math>, which is not possible, because <math>n</math> must be positive. Therefore, we have <math>6-1=5</math> values of <math>n</math> <math>\Rightarrow \boxed{B}</math> | ||

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

## Revision as of 03:15, 31 July 2008

## Contents

## Problem

For how many positive integers does evenly divide from ?

## Solution 1

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 positive 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

## Solution 2

The sum of the first positive integers is . If this is to divide , then there exists a positive integer such that:

Therefore, and are divisors of . There are divisors of , which are . The divisors which multiply to can be assigned to and in either order. However, when is assigned to , then , which is not possible, because must be positive. Therefore, we have values of