2006 AIME II 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 . Since
, the sum will be:
The first two groups telescope. The third resembles .
and
are both given; the last four digits of their sum is
, and half of that is
. Therefore, the answer is
.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |