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 | |
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All AIME Problems and Solutions |