2006 AIME I Problems/Problem 6

Revision as of 10:48, 13 March 2007 by Thechancellor (talk | contribs) (Solution)

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, .abc there will be exactly one other number, such that when they are added together, the sum will equal $0.\overline{.999}$, or, more precisely, 1.

Ex. $.123+.876=.999$ -> 1

Thus, the solution can be determined by dividing the total number of permutations by 2.

$\frac{720}{2}=360$

See also

2006 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions