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 $\boxed{\textbf{(B) }1}$

Solution 3

So from this question, we can get $\frac{x+1}{y+1} = \frac{11x}{10y}$. We can transform this equation into $x + 11 \cdot \left( \frac{x}{y} \right) = 10$. Two numbers are added to get 10, and one of them, $x$, is a positive and prime integer. So the other number also has to be a positive integer. Therefore, $11 \cdot \left( \frac{x}{y} \right)$ is a positive integer. The only possibility of this being true is when $y$ and 11 cancel out, leaving a singular $x$. So $y = 11$ and $x + x = 10$. Therefore, $y = 11$ and $x = 5$.

~POISONPOISSON

Video Solution

https://youtu.be/p7g0hTxE9I8

~savannahsolver

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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