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Difference between revisions of "2023 AIME II Problems/Problem 5"

m (→‎Solution)
m (→‎Solution)
 
Line 10: Line 10:
 
Therefore, <math>gcf\left( 55, b \right) \neq 1</math>.
 
Therefore, <math>gcf\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>.
+
Case 1: <math>b</math> can be written as <math>5c</math> with <math>gcf\left( c, 11 \right) = 1</math>.
  
 
Thus, <math>55r = \frac{11a}{c}</math>.
 
Thus, <math>55r = \frac{11a}{c}</math>.
Line 23: Line 23:
 
Hence, <math>2c = 5 a</math>.
 
Hence, <math>2c = 5 a</math>.
  
−
Because <math>\left( a, b \right) = 1</math>, <math>\left( a, c \right) = 1</math>.
+
Because <math>gcf\left( a, b \right) = 1</math>, <math>gcf\left( a, c \right) = 1</math>.
 
Thus, <math>a = 2</math> and <math>c = 5</math>.
 
Thus, <math>a = 2</math> and <math>c = 5</math>.
 
Therefore, <math>r = \frac{a}{5c} = \frac{2}{25}</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>.
+
Case 2: <math>b</math> can be written as <math>11d</math> with <math>gcf\left( d, 5 \right) = 1</math>.
  
 
Thus, <math>55r = \frac{5a}{c}</math>.
 
Thus, <math>55r = \frac{5a}{c}</math>.
Line 40: Line 40:
 
Hence, <math>2a = 5 c</math>.
 
Hence, <math>2a = 5 c</math>.
  
−
Because <math>\left( a, b \right) = 1</math>, <math>\left( a, c \right) = 1</math>.
+
Because <math>gcf\left( a, b \right) = 1</math>, <math>gcf\left( a, c \right) = 1</math>.
 
Thus, <math>a = 5</math> and <math>c = 2</math>.
 
Thus, <math>a = 5</math> and <math>c = 2</math>.
 
Therefore, <math>r = \frac{a}{11c} = \frac{5}{22}</math>.
 
Therefore, <math>r = \frac{a}{11c} = \frac{5}{22}</math>.

Latest revision as of 13:30, 14 August 2025

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 $gcf\left( a, b \right) = 1$. We have $55 r = \frac{55a}{b}$. Suppose $gcf\left( 55, b \right) = 1$, 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, $gcf\left( 55, b \right) \neq 1$.

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

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

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

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

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

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)

Minor latex edits by T3CHN0B14D3

Note

This problem mainly comes down to noticing that $55r$ has to be simplifiable such that the numerator and denominator both change, so they potentially equal their original sum. Then you proceed with casework just as Solution 1.

~BigBrain_2009

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

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