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2006 AIME I Problems/Problem 6

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Problem

Let \mathcal{S} be the set of real numbers that can be represented as repeating decimals of the form 0.\overline{abc} where a, b, c are distinct digits. Find the sum of the elements of \mathcal{S}.

Solution

Numbers of the form 0.\overline{abc} can be written as \frac{abc}{999}. There are 10\times9\times8=720 such numbers. Each digit will appear in each place value \frac{720}{10}=72 times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is \frac{45\times72\times111}{999}=360.

Alternatively, for every number, 0.\overline{abc}, there will be exactly one other number, such that when they are added together, the sum is 0.\overline{999}, or, more precisely, 1. As an example, .\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1.

Thus, the solution can be determined by dividing the total number of permutations by 2. The answer is \frac{10 \cdot 9 \cdot 8}{2} = \frac{720}{2}= 360.

See also

2006 AIME I (ProblemsResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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