2006 AIME I Problems/Problem 6

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 $\frac{45\times72\times111}{999}= \boxed{360}$ .

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 $\frac{10 \cdot 9 \cdot 8}{2} = \frac{720}{2}= \boxed{360}$ .

Another method, albeit a little risky, that can be used is to note that the numbers between 1 and 999 with distinct digits average out to $\frac{999}{2}$ . Then the total sum becomes $\frac{\frac{999}{2}\times720}{999}$ which reduces to $\boxed{360}$

2006 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2006 AIME I Problems/Problem 6

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