2006 AIME I Problems/Problem 6
Problem
Let be the set of real numbers that can be represented as repeating decimals of the form where are distinct digits. Find the sum of the elements of
Solution 1
Numbers of the form can be written as . There are such numbers. Each digit will appear in each place value times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is .
Solution 2
Alternatively, for every number, , there will be exactly one other number, such that when they are added together, the sum is , or, more precisely, 1. As an example, .
Thus, the solution can be determined by dividing the total number of permutations by 2. The answer is .
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 . Then the total sum becomes which reduces to
Solution 3
By symmetry, the average over all numbers is . Then, their sum is .
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.