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

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<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many}</math>
 
<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many}</math>
  
==Solution==
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==Solution 1==
  
 
You can create the equation <math>\frac{x+1}{y+1}=\frac{11x}{10y}</math>
 
You can create the equation <math>\frac{x+1}{y+1}=\frac{11x}{10y}</math>
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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>.
 
<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>.
  
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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We found one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math>.
 
We found one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math>.
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==Solution 2==
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The condition required is <math>\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}</math>.
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Observe that <math>x+1 > \frac{11}{10}\cdot x</math> so <math>x</math> is at most <math>9.</math>
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By multiplying by <math>\frac{y+1}{x}</math> and simplifying we can rewrite the condition as <math>y=\frac{11x}{10-x}</math>.  Since <math>x</math> and <math>y</math> are integer, this only has solutions for <math>x\in\{5,8,9\}</math>.  However, only the first yields a <math>y</math> that is relative prime to <math>x</math>.
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There is only one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math>
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==Video Solution==
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https://youtu.be/p7g0hTxE9I8
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 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=14|num-a=16}}
 
{{AMC10 box|year=2015|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category: Introductory Algebra Problems]]

Latest revision as of 14:45, 27 September 2020

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

Video Solution

https://youtu.be/p7g0hTxE9I8

~savannahsolver

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 10 Problems and Solutions

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