# Difference between revisions of "2015 AMC 10A Problems/Problem 15"

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<math>x</math> and <math>y</math> must be positive, so <math>x > 0</math> and <math>y > 0</math>, so <math>x - 10> -10</math> and <math>y + 11 > 11</math>. | <math>x</math> and <math>y</math> must be positive, so <math>x > 0</math> and <math>y > 0</math>, so <math>x - 10> -10</math> and <math>y + 11 > 11</math>. | ||

− | This leaves the factor pairs: <math>(-1, 110), (-2, 55), and (-5, 22)</math> | + | This leaves the factor pairs: <math>(-1, 110),</math> <math>(-2, 55),</math> and <math>(-5, 22).</math> |

But we can't stop here because <math>x</math> and <math>y</math> must be relatively prime. | But we can't stop here because <math>x</math> and <math>y</math> must be relatively prime. |

## Revision as of 18:23, 4 February 2015

## Problem

Consider the set of all fractions , where and are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by , the value of the fraction is increased by ?

## Solution

You can create the equation

Cross multiplying and combining like terms gives .

This can be factored into .

and must be positive, so and , so and .

This leaves the factor pairs: and

But we can't stop here because and must be relatively prime.

gives and . and are not relatively prime, so this doesn't work.

gives and . This doesn't work.

gives and . This does work.

We found one valid solution so the answer is .