# Mock AIME 1 2006-2007 Problems/Problem 15

## Problem

Let be the set of integers . An element (in) is chosen at random. Let denote the sum of the digits of . The probability that is divisible by 11 is where and are relatively prime positive integers. Compute the last 3 digits of

## Solution

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Counting all , , and digit combinations and then permuting only those up to , we find that there are 186 numbers whose sums are either or . We need not account for the sum 33, as it is not achievable with a as the lowest digit. Since there are a total of numbers and that work, we get or . Our sum is then . The last three digits are .