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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>
  
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==Solution==
+
==Solution 1==
  
−
You can create the equation
+
You can create the equation <math>\frac{x+1}{y+1}=\frac{11x}{10y}</math>
−
<math>\frac{x+1}{y+1}=(1.1)(\frac{x}{y})</math>
 
  
−
<math>\frac{x+1}{y+1}=\frac{1.1x}{y}</math>
+
Cross multiplying and combining like terms gives <math>xy + 11x - 10y = 0</math>.
  
−
<math>(x+1)(y)=(1.1x)(y+1)</math>
+
This can be factored into <math>(x - 10)(y + 11) = -110</math>.
  
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<math>xy+y=1.1xy+1.1x</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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<math>y=.1xy+1.1x</math>
+
Using the factors of 110, we can get the factor pairs: <math>(-1, 110),</math> <math>(-2, 55),</math> and <math>(-5, 22).</math>
  
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<math>10y=xy+11x</math>
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But we can't stop here because <math>x</math> and <math>y</math> must be relatively prime.
 +
 
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<math>(-1, 110)</math> gives <math>x = 9</math> and <math>y = 99</math>. <math>9</math> and <math>99</math> are not relatively prime, so this doesn't work.
 +
 
 +
<math>(-2, 55)</math> gives <math>x = 8</math> and <math>y = 44</math>. This doesn't work.
 +
 
 +
<math>(-5, 22)</math> gives <math>x = 5</math> and <math>y = 11</math>. This does work.
 +
 
 +
We found one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math>.
 +
 
 +
==Solution 2==
 +
 
 +
The condition required is <math>\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}</math>.
 +
 
 +
Observe that <math>x+1 > \frac{11}{10}\cdot x</math> so <math>x</math> is at most <math>9.</math>
 +
 
 +
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>.
 +
 
 +
There is only one valid solution so the answer is <math>\boxed{\textbf{(B) }1}</math>
 +
 
 +
==Video Solution==
 +
https://youtu.be/p7g0hTxE9I8
 +
 
 +
~savannahsolver
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2015|ab=A|num-b=14|num-a=16}}
 +
{{MAA Notice}}
 +
 
 +
[[Category: Introductory Algebra Problems]]

Latest revision as of 13:41, 14 September 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 $\boxed{\textbf{(B) }1}$

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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