2006 AIME A Problems/Problem 11
Problem
A collection of 8 cubes consists of one cube with edge-length for each integer A tower is to be built using all 8 cubes according to the rules:
- Any cube may be the bottom cube in the tower.
- The cube immediately on top of a cube with edge-length must have edge-length at most
Let be the number of different towers than can be constructed. What is 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 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 |