2015 AMC 10A Problem 15

Question

Consider the set of all fractions

\dfrac{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\%?

0

1

2

3

\text{infinitely many}

Turn the fraction condition into a factorization

Only negative factor pairs with

y+11>11

can work

The condition is

\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}.

Cross-multiplying gives

10y(x+1)=11x(y+1)

, or

xy+11x-10y=0

.

Factoring by grouping after subtracting

110

gives

(x-10)(y+11)=-110.

Since

x,y

are positive, the useful negative factor pairs are

(-1,110)

,

(-2,55)

, and

(-5,22)

, producing

(x,y)=(9,99),(8,44),(5,11)

.

Only

\frac{5}{11}

has relatively prime numerator and denominator, so exactly one fraction works.

Thus, B is the correct answer.

Problem 15 in other years