AIME 2023 II · 第 5 题

Solution 1

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 $11c$ with $gcf\left( c, 5 \right) = 1$.

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

Solution 2 (equation with GCDs)

When we simplify a fraction $\frac{a}{b}$ over the integers, we divide the numerator and denominator by $\gcd(a,b).$ So we can clearly see that the fraction $r$ in lowest terms will be $\frac{(\frac{a}{\gcd(a,b)})}{(\frac{b}{\gcd(a,b)})}$ and similarly, the fraction $55r$ in lowest terms will be $\frac{(\frac{55a}{\gcd(55a,b)})}{(\frac{b}{\gcd(55a,b)})}.$

Now, from the problem's condition, we obtain the following equation:

$$\frac{a + b}{\gcd(a,b)} = \frac{55a + b}{\gcd(55a,b)}.$$ Let $\gcd(a,b) = c,$ where $c$ is some positive integer. Then we can write $a$ as $cx$ and $b$ as $cy$ for positive integers $x$ and $y$ such that $\gcd(x,y) = 1.$ Substituting in $cx$ for $a$ and $cy$ for $b$ we get

$$\frac{cx + cy}{c} = \frac{55cx + cy}{\gcd(55cx, cy)}$$ which simplifies to

$$x + y = \frac{55x + y}{\gcd(55x,y)}$$ Now, note that since $x$ and $y$ share no common factors, any common factors shared between $55x$ and $y$ must be factors of $55,$ so from here we do casework on the possible values of $\gcd(55,y).$ We know that $\gcd(55,y) \in \{1, 5, 11, 55\},$ so we only have to look at four different cases.

Case 1: $\gcd(55,y) = 1:$ This implies

$$x + y = 55x + y$$ which would make one of the variables $0.$ This case doesn't work.

Case 2: $\gcd(55,y) = 5:$

$$5x + 5y = 55x + y \implies 50x = 4y \implies x = \frac{2}{25}y.$$ Since $a = cx$ and $b = cy,$ the value of $r$ is just $\frac{a}{b} = \frac{x}{y},$ so this case gives $\frac{2}{25}$ as a possible value for $r$.

Case 3: $\gcd(55,y) = 11:$

$$11x + 11y = 55x + y \implies 44x = 10y \implies 22x = 5y \implies r = \frac{5}{22}.$$ Case 4: $\gcd(55,y) = 55:$

$$55x + 55y = 55x + y$$ This also makes one of the variables $0$ which we can't have.

So the answer is $\frac{2}{25} + \frac{5}{22} = \frac{169}{550}.$

$169 + 550 = \boxed{719}.$

~grogg007

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