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

(Created page with "Note that <math>\frac{1}{11} = 0.\overline{09}</math>. Dividing by 3 gives <math>\frac{1}{33} = 0.\overline{03}</math>, and dividing by 9 gives <math>\frac{1}{99} = 0.\overline{...")
 
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== Problem==
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Let <math>S</math> be the set of positive integers <math>n</math> for which <math>\tfrac{1}{n}</math> has the repeating decimal representation <math>0.\overline{ab} = 0.ababab\cdots,</math> with <math>a</math> and <math>b</math> different digits.  What is the sum of the elements of <math>S</math>?
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<math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad </math>
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==Solution==
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Note that <math>\frac{1}{11} = 0.\overline{09}</math>.
 
Note that <math>\frac{1}{11} = 0.\overline{09}</math>.
  
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The answer must be at least <math>143</math>, but cannot be <math>155</math> since no <math>n \le 12</math> other than <math>11</math> satisfies the conditions, so the answer is <math>143</math>.
 
The answer must be at least <math>143</math>, but cannot be <math>155</math> since no <math>n \le 12</math> other than <math>11</math> satisfies the conditions, so the answer is <math>143</math>.
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=9|num-a=11}}

Revision as of 17:37, 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

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 $143$.

See also

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
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