2002 AMC 12A Problems/Problem 20

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 (no idea what it's called). We have $100(0.\overline{ab}) = ab.\overline{ab}$ $99(0.\overline{ab}) = 100(0.\overline{ab}) - 0.\overline{ab} = ab.\overline{ab} - 0.\overline{ab} = ab$ $0.\overline{ab} = \frac{ab}{99}$ Continuing in the same way, we have 5 different possibilities for the denomenator. $\mathrm

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2002 AMC 12A Problems/Problem 20

Contents

Problem

Solution

Solution 1

Solution 2

See Also