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>. | ||
Revision as of 14:50, 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
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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 
.
