Mock AIME 1 2006-2007 Problems/Problem 15
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
This problem needs a solution. If you have a solution for it, please help us out by.
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 .