# Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 15"

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+ | Counting all <math>2</math>, <math>3</math>, and <math>4</math> digit combinations and then permuting only those up to <math>2045</math>, we find that there are 186 numbers whose sums are either <math>11</math> or <math>22</math>. We need not account for the sum 33, as it is not achievable with a <math>2</math> as the lowest digit. Since there are a total of <math>2048</math> numbers and <math>186</math> that work, we get <math>186/2048</math> or <math>93/1024</math>. Our sum is then <math>93 + 1024 = 1117</math>. The last three digits are <math>117</math>. | ||

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## Revision as of 14:51, 25 April 2009

## 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

*This problem needs a solution. If you have a solution for it, please help us out by adding it.*

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 .