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"
(→‎Solution)
 
(7 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 +
==Problem==
 +
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==
  
−
--Solution 1--
+
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>.
  
−
Since 55r is greater than r, we know 55r can be simplified. Call r = <math> \frac{a}{b}</math>. This means 55r is <math>\frac{55a}{b}</math>. Since 55r can clearly be simplified, let's look at the possibilities, where b is divisible by factors of 55.
+
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>.
  
−
Case 1: B is divisible by 55.
+
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>.
  
−
If this is the case, the sum of the numerator and denominator of 55r is a + <math>\frac{b}{55}</math>. This clearly cannot equal a+b, so this case is invalid.
+
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>.
  
−
Case 2: B is divisible by 11.
+
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>
  
−
If this is the case, the sum of the numerator and denominator of 55r is 5a + <math>\frac{b}{11}</math>. This is equal to a+b. Solving this equation gives <math>\frac{a}{b}</math> = r = <math>\frac{5}{22}</math>.
+
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: B is divisible by 5.
+
Case 3: <math>b</math> can be written as <math>55 e</math>.
  
−
In this case, the sum of the numerator and denominator of 55r is 11a + <math>\frac{b}{5}</math>. This is equal to a+b, and solving that equation gives <math>\frac{a}{b}</math> = <math>\frac{2}{25}</math> = r.
+
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>
  
−
Clearly, B is divisible by 1 doesn't make sense, so these are the only possibilities. Summing these fractions give <math>\frac{169}{550}</math>, so the answer is 169 + 550 = 719.
+
Hence, <math>c = 0</math>. This is infeasible.
 +
Thus, there is no solution in this case.
  
−
~Anonymus
+
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>
 +
\[
 +
\frac{2}{25} + \frac{5}{22} = \frac{169}{550} .
 +
\]
 +
</cmath>
 +
 
 +
Therefore, the answer is <math>169 + 550 = \boxed{\textbf{(719) }}</math>.
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==Video Solution by The Power of Logic==
 +
https://youtu.be/qUJtReB_9sU
 +
 
 +
== See also ==
 +
{{AIME box|year=2023|num-b=4|num-a=6|n=II}}
 +
 
 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:13, 18 June 2023

Problem

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 \[ \frac{2}{25} + \frac{5}{22} = \frac{169}{550} . \]

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

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

Video Solution by The Power of Logic

https://youtu.be/qUJtReB_9sU

See also

2023 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_II_Problems/Problem_5&oldid=194599"