2006 AIME I Problems/Problem 11
Problem
A sequence is defined as follows and, for all positive integers
Given that
and
find the remainder when
is divided by 1000.
Solution
Define the sum as . Notice that
, so the sum will be:
The first two groupings almost completely cancel. The third resembles .
and
are both given; the last four digits of the sum is
, and half of that is
. Therefore, the answer is
.
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AIME Problems and Solutions |