2006 AIME A Problems/Problem 11

Problem

A collection of 8 cubes consists of one cube with edge-length $k$ for each integer $k, 1 \le k \le 8.$ 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 $k$ must have edge-length at most $k+2.$

Let $T$ be the number of different towers than can be constructed. What is the remainder when $T$ is divided by 1000?

Solution

Define the sum as $x$ . Notice that $a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$ , so the sum will be:

$x = (a_4 - a_3 - a_2) + (a_5 - a_4 - a_3) + \ldots (a_{30} - a_{29} - a_{28}) + a_{28}$

$x = (a_4+ a_5 \ldots a_{30}) - (a_3 + a_4 + \ldots a_{29}) - (a_2 + a_3 + \ldots a_{28}) + a_{28} + (a_1 - a_1)$

The first two groupings almost completely cancel. The third resembles $x$ .

$x\ = a_1 - a_3 + a_{28} + a_{30} - x$

$2x\ = a_{28} + a_{30}$

$x\ = \frac{a_{28} + a_{30}}{2}$

$a_{28}$ and $a_{30}$ are both given; the last four digits of the sum is $3668$ , and half of that is $1834$ . Therefore, the answer is $834$ .

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
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2006 AIME A Problems/Problem 11

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