2015 AMC 10A Problems/Problem 15

Revision as of 18:06, 4 February 2015 by Warrenwangtennis (talk | contribs) (Solution)

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

Either $x - 10$ has to be positive and $y + 11$ has to be negative or vice versa, but if $y + 11$ as negative, then $y$ would be negative, so that halves our search space.

Similarly, $x > 0$, so $x - 10> -10$ and $y > 0$, so $y + 11 > 11$.

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

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