2002 AMC 12A Problems/Problem 20

Revision as of 14:45, 17 June 2009 by Mathaops (talk | contribs) (Solution)

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

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

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 denumerator $1$ can not be achieved, leaving us with $\boxed{5}$ possible denumerators.

(The other ones are achieved e.g. for $ab$ equal to $99$, $33$, $11$, $9$, $3$, and $1$, respectively.)

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 12 Problems and Solutions