# 2015 AMC 10A Problems/Problem 15

## Problem

Consider the set of all fractions $\frac{x}{y}$, where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1$, the value of the fraction is increased by $10\%$? $\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many}$

## Solution 1

You can create the equation $\frac{x+1}{y+1}=\frac{11x}{10y}$

Cross multiplying and combining like terms gives $xy + 11x - 10y = 0$.

This can be factored into $(x - 10)(y + 11) = -110$. $x$ and $y$ must be positive, so $x > 0$ and $y > 0$, so $x - 10> -10$ and $y + 11 > 11$.

Using the factors of 110, we can get the factor pairs: $(-1, 110),$ $(-2, 55),$ and $(-5, 22).$

But we can't stop here because $x$ and $y$ must be relatively prime. $(-1, 110)$ gives $x = 9$ and $y = 99$. $9$ and $99$ are not relatively prime, so this doesn't work. $(-2, 55)$ gives $x = 8$ and $y = 44$. This doesn't work. $(-5, 22)$ gives $x = 5$ and $y = 11$. This does work.

We found one valid solution so the answer is $\boxed{\textbf{(B) }1}$.

## Solution 2

The condition required is $\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}$.

Observe that $x+1 > \frac{11}{10}\cdot x$ so $x$ is at most $9.$

By multiplying by $\frac{y+1}{x}$ and simplifying we can rewrite the condition as $y=\frac{11x}{10-x}$. Since $x$ and $y$ are integer, this only has solutions for $x\in\{5,8,9\}$. However, only the first yields a $y$ that is relative prime to $x$.

There is only one valid solution so the answer is $\boxed{\textbf{(B) }1}$

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