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

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Cross multiplying and combining like terms gives <math>xy + 11x - 10y = 0</math>.
 
Cross multiplying and combining like terms gives <math>xy + 11x - 10y = 0</math>.
  
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This can be factored into <math>(x - 19)(y + 11) = -110</math>.
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This can be factored into <math>(x - 10)(y + 11) = -110</math>.
  
 
<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>.
 
<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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This leaves the factor pairs: <math>(-1, 110),</math> <math>(-2, 55),</math> and <math>(-5, 22).</math>
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Using the factors of 110, we can get the factor pairs: <math>(-1, 110),</math> <math>(-2, 55),</math> and <math>(-5, 22).</math>
  
 
But we can't stop here because <math>x</math> and <math>y</math> must be relatively prime.
 
But we can't stop here because <math>x</math> and <math>y</math> must be relatively prime.
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<math>(-5, 22)</math> gives <math>x = 5</math> and <math>y = 11</math>. This does work.
 
<math>(-5, 22)</math> gives <math>x = 5</math> and <math>y = 11</math>. This does work.
  
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We found one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math>
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We found one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math>.
  
 
==Solution 2==
 
==Solution 2==
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There is only one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math>
 
There is only one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math>
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==Solution 3==
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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 \(x = 5\).
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~POISONPOISSON
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==Video Solution==
 +
https://youtu.be/p7g0hTxE9I8
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 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 07:09, 18 December 2024

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

See Also

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

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