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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 6]] |
| − | Let <math> \mathcal{S} </math> be the set of real numbers that can be represented as repeating decimals of the form <math> 0.\overline{abc} </math> where <math> a, b, c </math> are distinct digits. Find the sum of the elements of <math> \mathcal{S}. </math>
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| − | == Solution ==
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| − | Numbers of the form <math>0.\overline{abc}</math> can be written as <math>\frac{abc}{999}</math>. There are <math>10\times9\times8=720</math> such numbers. Each digit will appear in each place value <math>\frac{720}{10}=72</math> times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is <math>\frac{45\times72\times111}{999}=360</math>.
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| − | Alternatively, for every number, <math>0.\overline{abc}</math>, there will be exactly one other number, such that when they are added together, the sum is <math>0.\overline{999}</math>, or, more precisely, 1. As an example, <math>.\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1</math>.
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| − | Thus, the solution can be determined by dividing the total number of [[permutation]]s by 2. The answer is <math>\frac{10 \cdot 9 \cdot 8}{2} = \frac{720}{2}= 360</math>.
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| − | == See also ==
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| − | {{AIME box|year=2006|n=II|num-b=5|num-a=7}}
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| − | [[Category:Intermediate Geometry Problems]]
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