Difference between revisions of "2013 AMC 12A Problems/Problem 10"

Revision as of 19:42, 22 February 2013

Problem

Let $S$ be the set of positive integers $n$ for which $\tfrac{1}{n}$ has the repeating decimal representation $0.\overline{ab} = 0.ababab\cdots,$ with $a$ and $b$ different digits. What is the sum of the elements of $S$ ?

$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad$

Solution

Solution 1

Note that $\frac{1}{11} = 0.\overline{09}$ .

Dividing by 3 gives $\frac{1}{33} = 0.\overline{03}$ , and dividing by 9 gives $\frac{1}{99} = 0.\overline{01}$ .

$S = \{11, 33, 99\}$

$11 + 33 + 99 = 143$

The answer must be at least $143$ , but cannot be $155$ since no $n \le 12$ other than $11$ satisfies the conditions, so the answer is $143$ .

Solution 2

Let us begin by working with the condition $0.\overline{ab} = 0.ababab\cdots,$ . Let $x = 0.ababab\cdots$ . So, $100x-x = ab \Rightarrow x = \frac{ab}{99}$ . In order for this fraction $x$ to be in the form $\frac{1}{n}$ , $99$ must be a multiple of $ab$ . Hence the possibilities of $ab$ are $1,3,9,11,33,99$ . Checking each of these, $\frac{1}{99} = 0.\overline{01}, \frac{3}{99}=\frac{1}{33} = 0.\overline{03}, \frac{9}{99}=\frac{1}{11} = 0.\overline{09}, \frac{11}{99}=\frac{1}{9} = 0.\overline{1}, \frac{33}{99} =\frac{1}{3}= 0.\overline{3},$ and $\frac{99}{99} = 1$ . So the only values of $n$ that have distinct $a$ and $b$ are $11,33,$ and $99$ . So, $11+33+99=143 \textbf{D}$

@@ Line 18: / Line 18: @@
 ==Solution 2==
-Let us begin by working with the condition <math>0.\overline{ab} = 0.ababab\cdots,</math>. Let <math>x = 0.ababab\cdots</math>. So, <math>100x-x = ab \Rightarrow x = \frac{ab}{99}</math>. In order for this fraction <math>x</math> to be in the form <math>\frac{1}{n}</math>, <math>99</math> must be a multiple of <math>ab</math> less than <math>99</math>. Hence the possibilities of <math>ab</math> are <math>1,3,9,11,33</math>. Checking each of these, <math>\frac{1}{99} = 0.\overline{01}, \frac{3}{99}=\frac{1}{33} = 0.\overline{03}, \frac{9}{99}=\frac{1}{11} = 0.\overline{09}, \frac{11}{99}=\frac{1}{9} = 0.\overline{1}, \frac{33}{99} =\frac{1}{3}= 0.\overline{3}</math>. So the only values of <math>n</math> that have distinct <math>a</math> and <math>b</math> are <math>11,33,</math> and <math>99</math>. So, <math>11+33+99=143 \textbf{D}</math>
+Let us begin by working with the condition <math>0.\overline{ab} = 0.ababab\cdots,</math>. Let <math>x = 0.ababab\cdots</math>. So, <math>100x-x = ab \Rightarrow x = \frac{ab}{99}</math>. In order for this fraction <math>x</math> to be in the form <math>\frac{1}{n}</math>, <math>99</math> must be a multiple of <math>ab</math>. Hence the possibilities of <math>ab</math> are <math>1,3,9,11,33,99</math>. Checking each of these, <math>\frac{1}{99} = 0.\overline{01}, \frac{3}{99}=\frac{1}{33} = 0.\overline{03}, \frac{9}{99}=\frac{1}{11} = 0.\overline{09}, \frac{11}{99}=\frac{1}{9} = 0.\overline{1}, \frac{33}{99} =\frac{1}{3}= 0.\overline{3},</math> and <math>\frac{99}{99} = 1</math>. So the only values of <math>n</math> that have distinct <math>a</math> and <math>b</math> are <math>11,33,</math> and <math>99</math>. So, <math>11+33+99=143 \textbf{D}</math>
 == See also ==
 {{AMC12 box|year=2013|ab=A|num-b=9|num-a=11}}

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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

Page

Toolbox

Search