# 2002 AMC 12A Problems/Problem 20

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## Problem

Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?

$\text{(A) }3 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }8 \qquad \text{(E) }9$

## Solution

### Solution 1

The repeating decimal $0.\overline{ab}$ is equal to $$\frac{10a+b}{100} + \frac{10a+b}{10000} + \cdots = (10a+b)\cdot\left(\frac 1{10^2} + \frac 1{10^4} + \cdots \right) = (10a+b) \cdot \frac 1{99} = \frac{10a+b}{99}$$

When expressed in lowest terms, the denominator of this fraction will always be a divisor of the number $99 = 3\cdot 3\cdot 11$. This gives us the possibilities $\{1,3,9,11,33,99\}$. As $a$ and $b$ are not both nine and not both zero, the denominator $1$ can not be achieved, leaving us with $\boxed{(C)5}$ possible denominators.

(The other ones are achieved e.g. for $ab$ equal to $33$, $11$, $9$, $3$, and $1$, respectively.)

### Solution 2

Another way to convert the decimal into a fraction (simplifying, I guess?). We have $$100(0.\overline{ab}) = ab.\overline{ab}$$ $$99(0.\overline{ab}) = 100(0.\overline{ab}) - 0.\overline{ab} = ab.\overline{ab} - 0.\overline{ab} = ab$$ $$0.\overline{ab} = \frac{ab}{99}$$ where $a, b$ are digits. Continuing in the same way by looking at the factors of 99, we have 5 different possibilities for the denomenator. $\boxed{(C)}$

~ Nafer ~ edit by SpeedCuber7

### Solution 3

Since $\frac{1}{99}=0.\overline{01}$, we know that $0.\overline{ab} = \frac{ab}{99}$. From here, we wish to find the number of factors of $99$, which is $6$. However, notice that $1$ is not a possible denominator, so our answer is $6-1=\boxed{5}$. $$ ~AopsUser101

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