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

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

@@ Line 37: / Line 37: @@
 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>
+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 ===
@@ Line 46: / Line 47: @@
 <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 ==
+[[Category:Introductory Number Theory Problems]]
 {{AMC12 box|year=2002|ab=A|num-b=19|num-a=21}}
 {{MAA Notice}}

2002 AMC 12A (Problems • Answer Key • Resources)
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