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2013 AMC 12A Problems/Problem 10

Revision as of 22:05, 6 February 2013 by Epicwisdom (talk | contribs) (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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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 $143$.

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