2013 AMC 12A Problems/Problem 10

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, $11+33+99= \boxed{\textbf{(D)} 143}$

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

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2013 AMC 12A Problems/Problem 10

Contents

Problem

Solution

Solution 1

Solution 2

See also