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

(New page: == Problem == Suppose that <math>a</math> and <math>b</math> are digits, not both nine and not both zero, and the repeating decimal <math>0.\overline{ab}</math> is expressed as a fraction...)
 
 
(33 intermediate revisions by 17 users not shown)
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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 denumerator 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.
+
When expressed in the 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{\mathrm{(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 denominator. <math>\boxed{(C)}</math>
 +
 +
~ Nafer
 +
~ edit by SpeedCuber7
 +
~ edit by PojoDotCom
 +
 +
=== 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
 +
=== Solution 4 (Alcumus) ===
 +
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.
 +
 +
Thus, the answer is <math>\boxed{5}</math>.
 +
 +
== Video Solution ==
 +
 +
https://youtu.be/0_0XpawpVVs
 +
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=x086uFh-i00
  
 
== See Also ==
 
== See Also ==
−
 
+
[[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}}

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 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AMC 12 Problems and Solutions

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