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

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=== Solution 4 (Alcumus) ===
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Since <math>0.\overline{ab} = \frac{ab}{99}</math>, the denominator must be a factor of <math>99 = 3^2 \cdot 11</math>. The factors of <math>99</math> are <math>1,</math> <math>3,</math> <math>9,</math> <math>11,</math> <math>33,</math> and <math>99</math>. Since <math>a</math> and <math>b</math> are not both nine, the denominator cannot be <math>1</math>. By choosing <math>a</math> and <math>b</math> appropriately, we can make fractions with each of the other denominators.
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Thus, the answer is <math>\boxed{5}</math>.
  
 
== Video Solution ==
 
== Video Solution ==
  
 
https://youtu.be/0_0XpawpVVs
 
https://youtu.be/0_0XpawpVVs
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=x086uFh-i00
  
 
== See Also ==
 
== See Also ==
 
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[[Category:Introductory Number Theory Problems]]
 
{{AMC12 box|year=2002|ab=A|num-b=19|num-a=21}}
 
{{AMC12 box|year=2002|ab=A|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:23, 16 June 2024

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 the 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 $\boxed{\mathrm{(C) }5}$ 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 (simplifying, I guess?). We have \[100(0.\overline{ab}) = ab.\overline{ab}\] \[99(0.\overline{ab}) = 100(0.\overline{ab}) - 0.\overline{ab} = ab.\overline{ab} - 0.\overline{ab} = ab\] \[0.\overline{ab} = \frac{ab}{99}\] where $a, b$ are digits. Continuing in the same way by looking at the factors of 99, we have 5 different possibilities for the denominator. $\boxed{(C)}$

~ Nafer ~ edit by SpeedCuber7 ~ edit by PojoDotCom

Solution 3

Since $\frac{1}{99}=0.\overline{01}$, we know that $0.\overline{ab} = \frac{ab}{99}$. From here, we wish to find the number of factors of $99$, which is $6$. However, notice that $1$ is not a possible denominator, so our answer is $6-1=\boxed{5}$. \[\] ~AopsUser101

Solution 4 (Alcumus)

Since $0.\overline{ab} = \frac{ab}{99}$, the denominator must be a factor of $99 = 3^2 \cdot 11$. The factors of $99$ are $1,$ $3,$ $9,$ $11,$ $33,$ and $99$. Since $a$ and $b$ are not both nine, the denominator cannot be $1$. By choosing $a$ and $b$ appropriately, we can make fractions with each of the other denominators.

Thus, the answer is $\boxed{5}$.

Video Solution

https://youtu.be/0_0XpawpVVs

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=x086uFh-i00

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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