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

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 $y + 11 > 11$ .

This leaves 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 $\boxed{\textbf{(B) }1}$ .

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2015 AMC 10A Problems/Problem 15

Problem

Solution

See Also