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

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

@@ Line 1: / Line 1: @@
-You can create the equation  <math>\frac{x+1}{y+1}=(1.1)(\frac{x}{y})</math>
+==Problem==
+Consider the set of all fractions <math>\frac{x}{y}</math>, where <math>x</math> and <math>y</math> are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by <math>1</math>, the value of the fraction is increased by <math>10\%</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 1==
+You can create the equation <math>\frac{x+1}{y+1}=\frac{11x}{10y}</math>
+Cross multiplying and combining like terms gives <math>xy + 11x - 10y = 0</math>.
+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>.
+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.
+<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]]

2015 AMC 10A (Problems • Answer Key • Resources)
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