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
numbers whose sums are either
or
. We need not account for the sum
, 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
.