Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AIME II Problems/Problem 5"
Line 1: Line 1:
 
Let <math>S</math> be the set of all positive rational numbers <math>r</math> such that when the two numbers <math>r</math> and <math>55r</math> are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of <math>S</math> can be expressed in the form <math>\frac{p}{q},</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q.</math>
 
Let <math>S</math> be the set of all positive rational numbers <math>r</math> such that when the two numbers <math>r</math> and <math>55r</math> are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of <math>S</math> can be expressed in the form <math>\frac{p}{q},</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q.</math>
 +
 +
==Solution==
 +
 +
Denote <math>r = \frac{a}{b}</math>, where <math>\left( a, b \right) = 1</math>.
 +
We have <math>55 r = \frac{55a}{b}</math>.
 +
Suppose <math>\left( 55, b \right) = 1</math>, then the sum of the numerator and the denominator of <math>55r</math> is <math>55a + b</math>.
 +
This cannot be equal to the sum of the numerator and the denominator of <math>r</math>, <math>a + b</math>.
 +
Therefore, <math>\left( 55, b \right) \neq 1</math>.
 +
 +
Case 1: <math>b</math> can be written as <math>5c</math> with <math>\left( c, 11 \right) = 1</math>.
 +
 +
Thus, <math>55r = \frac{11a}{c}</math>.
 +
 +
Because the sum of the numerator and the denominator of <math>r</math> and <math>55r</math> are the same,
 +
<cmath>
 +
\[
 +
a + 5c = 11a + c .
 +
\]
 +
</cmath>
 +
 +
Hence, <math>2c = 5 a</math>.
 +
 +
Because <math>\left( a, b \right) = 1</math>, <math>\left( a, c \right) = 1</math>.
 +
Thus, <math>a = 2</math> and <math>c = 5</math>.
 +
Therefore, <math>r = \frac{a}{5c} = \frac{2}{25}</math>.
 +
 +
Case 2: <math>b</math> can be written as <math>11d</math> with <math>\left( d, 5 \right) = 1</math>.
 +
 +
Thus, <math>55r = \frac{5a}{c}</math>.
 +
 +
Because the sum of the numerator and the denominator of <math>r</math> and <math>55r</math> are the same,
 +
<cmath>
 +
\[
 +
a + 11c = 5a + c .
 +
\]
 +
</cmath>
 +
 +
Hence, <math>2a = 5 c</math>.
 +
 +
Because <math>\left( a, b \right) = 1</math>, <math>\left( a, c \right) = 1</math>.
 +
Thus, <math>a = 5</math> and <math>c = 2</math>.
 +
Therefore, <math>r = \frac{a}{11c} = \frac{5}{22}</math>.
 +
 +
Case 3: <math>b</math> can be written as <math>55 e</math>.
 +
 +
Thus, <math>55r = \frac{a}{c}</math>.
 +
 +
Because the sum of the numerator and the denominator of <math>r</math> and <math>55r</math> are the same,
 +
<cmath>
 +
\[
 +
a + 55c = a + c .
 +
\]
 +
</cmath>
 +
 +
Hence, <math>c = 0</math>. This is infeasible.
 +
Thus, there is no solution in this case.
 +
 +
Putting all cases together, <math>S = \left\{ \frac{2}{25}, \frac{5}{22} \right\}</math>.
 +
Therefore, the sum of all numbers in <math>S</math> is
 +
<cmath>
 +
\[
 +
\frac{2}{25} + \frac{5}{121} = \frac{169}{550} .
 +
\]
 +
</cmath>
 +
 +
Therefore, the answer is <math>169 + 550 = \boxed{\textbf{(719) }}</math>.
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 16:46, 16 February 2023

Let $S$ be the set of all positive rational numbers $r$ such that when the two numbers $r$ and $55r$ are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of $S$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution

Denote $r = \frac{a}{b}$, where $\left( a, b \right) = 1$. We have $55 r = \frac{55a}{b}$. Suppose $\left( 55, b \right) = 1$, then the sum of the numerator and the denominator of $55r$ is $55a + b$. This cannot be equal to the sum of the numerator and the denominator of $r$, $a + b$. Therefore, $\left( 55, b \right) \neq 1$.

Case 1: $b$ can be written as $5c$ with $\left( c, 11 \right) = 1$.

Thus, $55r = \frac{11a}{c}$.

Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 5c = 11a + c . \]

Hence, $2c = 5 a$.

Because $\left( a, b \right) = 1$, $\left( a, c \right) = 1$. Thus, $a = 2$ and $c = 5$. Therefore, $r = \frac{a}{5c} = \frac{2}{25}$.

Case 2: $b$ can be written as $11d$ with $\left( d, 5 \right) = 1$.

Thus, $55r = \frac{5a}{c}$.

Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 11c = 5a + c . \]

Hence, $2a = 5 c$.

Because $\left( a, b \right) = 1$, $\left( a, c \right) = 1$. Thus, $a = 5$ and $c = 2$. Therefore, $r = \frac{a}{11c} = \frac{5}{22}$.

Case 3: $b$ can be written as $55 e$.

Thus, $55r = \frac{a}{c}$.

Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 55c = a + c . \]

Hence, $c = 0$. This is infeasible. Thus, there is no solution in this case.

Putting all cases together, $S = \left\{ \frac{2}{25}, \frac{5}{22} \right\}$. Therefore, the sum of all numbers in $S$ is \[ \frac{2}{25} + \frac{5}{121} = \frac{169}{550} . \]

Therefore, the answer is $169 + 550 = \boxed{\textbf{(719) }}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_II_Problems/Problem_5&oldid=189723"