2015 AMC 10A Problem 15

P11-15~5 minPrint

Diophantine Equations Fractions & Decimals GCD & LCM Bound the quantity above/below or estimate to pin it down Organized listing / direct enumeration

Consider the set of all fractions $\tfrac{x}{y},$ where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1$ , the value of the fraction is increased by $10\%$ ?

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Key insight

Cross-multiply and rearrange to get y times (10 minus x) equals 11x. So x is at most 9; checking x from 1 to 9 leaves only 5/11 in lowest terms.

Solution

Write the condition as an equation, rearrange it so it limits $x$ , and then check every possible $x$ .

1. "Increased by $10\%$ " means multiplied by $1.1 = \frac{11}{10}$ . So
$\frac{x + 1}{y + 1} = \frac{11}{10} \cdot \frac{x}{y}.$
2. Cross-multiply:
$10y(x + 1) = 11x(y + 1), \quad\text{so}\quad 10xy + 10y = 11xy + 11x.$
3. Subtract $10xy$ from both sides: $10y = xy + 11x$ . Then move $xy$ to the left and factor out $y$ :
$y(10 - x) = 11x.$
4. The right side $11x$ is positive, and $y$ is positive, so $10 - x$ must be positive. That means $x$ is one of $1, 2, \ldots, 9$ . There are only nine cases.
5. For each case, $y = \dfrac{11x}{10 - x}$ must be a whole number:

$x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$\frac{11x}{10-x}$	$\frac{11}{9}$	$\frac{22}{8}$	$\frac{33}{7}$	$\frac{44}{6}$	$11$	$\frac{66}{4}$	$\frac{77}{3}$	$44$	$99$

Whole-number $y$ only for $x = 5, 8, 9$ , giving $\frac{5}{11}$ , $\frac{8}{44}$ and $\frac{9}{99}$ .
The problem wants $x$ and $y$ with no common factor. $\frac{8}{44}$ shares a factor $4$ , and $\frac{9}{99}$ shares a factor $9$ , so they are out. Only $\frac{5}{11}$ is left.
Check: $\frac{6}{12} = \frac{1}{2}$ , and $\frac{11}{10} \cdot \frac{5}{11} = \frac{5}{10} = \frac{1}{2}$ . It works.

The answer is $\boxed{\textbf{(B)}\ 1}$ .

Why this works

After cross-multiplying, get one letter by itself. Here that shows $10 - x$ has to be positive, which cuts the problem down to nine cases you can simply check. The "no common factor" condition is doing real work: without it the answer would be $3$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The condition is
$\frac{x + 1}{y + 1} = \frac{11}{10} \cdot \frac{x}{y}.$
Cross-multiplying: $10y(x + 1) = 11x(y + 1)$ , so $10xy + 10y = 11xy + 11x$ , which simplifies to
$10y = xy + 11x \quad\Longrightarrow\quad y(10 - x) = 11x.$
Since $x, y > 0$ , the right side is positive, so $10 - x > 0$ : $x$ is one of $1, 2, \ldots, 9$ , and $y = \dfrac{11x}{10 - x}$ .

Now test each $x$ . For $x = 1, 2, 3, 4, 6, 7$ the denominators $9, 8, 7, 6, 4, 3$ do not divide $11x$ (they share no factor with $11$ and do not divide $x$ ). The survivors:
- $x = 5$ : $y = 55/5 = 11$ , and $\gcd(5, 11) = 1$ . Valid.
- $x = 8$ : $y = 88/2 = 44$ , but $\gcd(8, 44) = 4$ . Not in lowest terms.
- $x = 9$ : $y = 99/1 = 99$ , but $\gcd(9, 99) = 9$ . Not in lowest terms.

Only $\frac{5}{11}$ works: indeed $\frac{6}{12} = \frac{1}{2}$ and $\frac{11}{10} \cdot \frac{5}{11} = \frac{1}{2}$ .

The answer is $\boxed{\textbf{(B)}\ 1}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Number Theory, Chapter 11: Decimals and Fractions
Introduction to Number Theory, Chapter 3: Multiples and Divisors (§3.3, §3.4, §3.7)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)