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

Revision as of 19:59, 29 May 2023

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 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

@@ Line 30: / Line 30: @@
 </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 denominator <math>1</math> can not be achieved, leaving us with <math>\boxed{(C)5}</math> possible denominators.
+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.)

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
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