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

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== Solution ==
 
== Solution ==
 +
 +
=== Solution 1 ===
  
 
The repeating decimal <math>0.\overline{ab}</math> is equal to  
 
The repeating decimal <math>0.\overline{ab}</math> is equal to  
 
<cmath>
 
<cmath>
−
\frac{ab}{100} + \frac{ab}{10000} + \cdots
+
\frac{10a+b}{100} + \frac{10a+b}{10000} + \cdots
 
=
 
=
−
ab\cdot\left(\frac 1{10^2} + \frac 1{10^4} + \cdots \right)
+
(10a+b)\cdot\left(\frac 1{10^2} + \frac 1{10^4} + \cdots \right)
 
=  
 
=  
−
ab \cdot \frac 1{99}
+
(10a+b) \cdot \frac 1{99}
 
=
 
=
−
\frac{ab}{99}
+
\frac{10a+b}{99}
 
</cmath>
 
</cmath>
  
−
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.
+
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.)
 +
 +
=== Solution 2 ===
 +
 +
Another way to convert the decimal into a fraction (simplifying, I guess?). We have <cmath>100(0.\overline{ab}) = ab.\overline{ab}</cmath> <cmath>99(0.\overline{ab}) = 100(0.\overline{ab}) - 0.\overline{ab} = ab.\overline{ab} - 0.\overline{ab} = ab</cmath> <cmath>0.\overline{ab} = \frac{ab}{99}</cmath>
 +
where <math>a, b</math> are digits. Continuing in the same way by looking at the factors of 99, we have 5 different possibilities for the denomenator. <math>\boxed{(C)}</math>
 +
 +
~ Nafer
 +
~ edit by SpeedCuber7
 +
 +
=== Solution 3 ===
 +
Since <math>\frac{1}{99}=0.\overline{01}</math>, we know that <math>0.\overline{ab} = \frac{ab}{99}</math>. From here, we wish to find the number of factors of <math>99</math>, which is <math>6</math>. However, notice that <math>1</math> is not a possible denominator, so our answer is <math>6-1=\boxed{5}</math>.
 +
<cmath></cmath>
 +
~AopsUser101
  
 
== 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}}
 +
{{MAA Notice}}

Revision as of 16:10, 6 September 2021

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

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 denomenator. $\boxed{(C)}$

~ Nafer ~ edit by SpeedCuber7

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

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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