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Difference between revisions of "2023 AIME II Problems/Problem 5"
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==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>
 
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>
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==Solution==
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Denote <math>r = \frac{a}{b}</math>, where <math>\left( a, b \right) = 1</math>.
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We have <math>55 r = \frac{55a}{b}</math>.
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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>.
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This cannot be equal to the sum of the numerator and the denominator of <math>r</math>, <math>a + b</math>.
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Therefore, <math>\left( 55, b \right) \neq 1</math>.
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Case 1: <math>b</math> can be written as <math>5c</math> with <math>\left( c, 11 \right) = 1</math>.
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Thus, <math>55r = \frac{11a}{c}</math>.
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Because the sum of the numerator and the denominator of <math>r</math> and <math>55r</math> are the same,
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<cmath>
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\[
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a + 5c = 11a + c .
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\]
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</cmath>
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Hence, <math>2c = 5 a</math>.
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Because <math>\left( a, b \right) = 1</math>, <math>\left( a, c \right) = 1</math>.
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Thus, <math>a = 2</math> and <math>c = 5</math>.
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Therefore, <math>r = \frac{a}{5c} = \frac{2}{25}</math>.
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Case 2: <math>b</math> can be written as <math>11d</math> with <math>\left( d, 5 \right) = 1</math>.
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Thus, <math>55r = \frac{5a}{c}</math>.
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 +
Because the sum of the numerator and the denominator of <math>r</math> and <math>55r</math> are the same,
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<cmath>
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\[
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a + 11c = 5a + c .
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\]
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</cmath>
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Hence, <math>2a = 5 c</math>.
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Because <math>\left( a, b \right) = 1</math>, <math>\left( a, c \right) = 1</math>.
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Thus, <math>a = 5</math> and <math>c = 2</math>.
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Therefore, <math>r = \frac{a}{11c} = \frac{5}{22}</math>.
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Case 3: <math>b</math> can be written as <math>55 e</math>.
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Thus, <math>55r = \frac{a}{c}</math>.
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Because the sum of the numerator and the denominator of <math>r</math> and <math>55r</math> are the same,
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<cmath>
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\[
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a + 55c = a + c .
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\]
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</cmath>
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Hence, <math>c = 0</math>. This is infeasible.
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Thus, there is no solution in this case.
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Putting all cases together, <math>S = \left\{ \frac{2}{25}, \frac{5}{22} \right\}</math>.
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Therefore, the sum of all numbers in <math>S</math> is
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<cmath>
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\[
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\frac{2}{25} + \frac{5}{22} = \frac{169}{550} .
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\]
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</cmath>
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Therefore, the answer is <math>169 + 550 = \boxed{\textbf{(719) }}</math>.
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Video Solution by The Power of Logic==
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https://youtu.be/qUJtReB_9sU
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== See also ==
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{{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

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