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Difference between revisions of "2015 AMC 10A Problems/Problem 15"

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This can be factored into <math>(x - 10)(y + 11) = -110</math>.
 
This can be factored into <math>(x - 10)(y + 11) = -110</math>.
  
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Either <math>x - 10</math> has to be positive and <math>y + 11</math> has to be negative or vice versa, but if <math>y + 11</math> as negative, then <math>y</math> would be negative, so this halves our search space.
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<math>x</math> and <math>y</math> must be positive, so <math>x > 0</math> and <math>y > 0</math>, so <math>x - 10> -10</math> and <math>y + 11 > 11</math>.
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Similarly, <math>x > 0</math>, so <math>x - 10> -10</math> and <math>y > 0</math>, so <math>y + 11 > 11</math>.
 
  
 
This leaves the factor pairs: <math>(-1, 110), (-2, 55), and (-5, 22)</math>.
 
This leaves the factor pairs: <math>(-1, 110), (-2, 55), and (-5, 22)</math>.

Revision as of 18:09, 4 February 2015

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

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