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Difference between revisions of "2002 AMC 12A Problems/Problem 20"
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When expressed in lowest terms, the denominator of this fraction will always be a divisor of the number <math>99 = 3\cdot 3\cdot 11</math>. This gives us the possibilities <math>\{1,3,9,11,33,99\}</math>. As <math>a</math> and <math>b</math> are not both nine and not both zero, the denumerator <math>1</math> can not be achieved, leaving us with <math>\boxed{(C)5}</math> possible denominators.
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When expressed in lowest terms, the denominator of this fraction will always be a divisor of the number <math>99 = 3\cdot 3\cdot 11</math>. This gives us the possibilities <math>\{1,3,9,11,33,99\}</math>. As <math>a</math> and <math>b</math> are not both nine and not both zero, the denominator <math>1</math> can not be achieved, leaving us with <math>\boxed{(C)5}</math> possible denominators.
  
 
(The other ones are achieved e.g. for <math>ab</math> equal to <math>33</math>, <math>11</math>, <math>9</math>, <math>3</math>, and <math>1</math>, respectively.)
 
(The other ones are achieved e.g. for <math>ab</math> equal to <math>33</math>, <math>11</math>, <math>9</math>, <math>3</math>, and <math>1</math>, respectively.)

Revision as of 17:47, 7 April 2014

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 \[\frac{10a+b}{100} + \frac{10a+b}{10000} + \cdots = (10a+b)\cdot\left(\frac 1{10^2} + \frac 1{10^4} + \cdots \right) = (10a+b) \cdot \frac 1{99} = \frac{10a+b}{99}\]

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.)

See Also

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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