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{4}</math> possible denumerators.
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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{5}</math> possible denumerators.
  
(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.)
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(The other ones are achieved e.g. for <math>ab</math> equal to <math>99</math>, <math>33</math>, <math>11</math>, <math>9</math>, <math>3</math>, and <math>1</math>, respectively.)
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2002|ab=A|num-b=19|num-a=21}}
 
{{AMC12 box|year=2002|ab=A|num-b=19|num-a=21}}

Revision as of 14:45, 17 June 2009

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
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