Difference between revisions of "2002 AMC 12A Problems/Problem 20"

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== Solution ==
 
== Solution ==
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=== Solution 1 ===
  
 
The repeating decimal <math>0.\overline{ab}</math> is equal to  
 
The repeating decimal <math>0.\overline{ab}</math> is equal to  
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(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.)
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=== Solution 2 ===
  
 
== See Also ==
 
== See Also ==

Revision as of 21:10, 1 July 2019

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

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

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