2013 AIME II Problems/Problem 9

A $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$ .

Solution

Firstly, we consider how many different ways possible to divide the $7\times 1$ board.

(1)One piece: $7$ , $\dbinom{6}{0}=1$ way in total(since the 3 colors are used at least once, we reject this case)

(2)Two pieces: $6+1$ , $5+2$ ,etc, $\dbinom{6}{1}=6$ ways in total(rejected for the same reason)

(3)Three pieces: $5+1+1$ , $4+2+1$ , $4+1+2$ ,etc, $\dbinom{6}{2}=15$ ways in total

(4)Four pieces: $\dbinom{6}{3}=20$ (5)Five pieces: $\dbinom{6}{4}=15$ (6)Six pieces: $\dbinom{6}{5}=6$ (7)Seven pieces: $\dbinom{6}{6}=1$

Secondly, we consider how many ways to color them:

(1)There pieces: $3^3-3\times 2^3+3=6$

(2)Four pieces: $3^4-3\times 2^4+3=36$

(3)Five pieces: $3^5-3\times 2^5+3=150$

(4)Six pieces: $3^6-3\times 2^6+3=540$

(5)Seven pieces: $3^7-3\times 2^7+3=1806$

Finally, we combine then together: $15\times 6+20\times 36+15\times 150+6\times 540+1\times 1806= 8106$

So the answer is $\boxed{106}$ .

2013 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2013 AIME II Problems/Problem 9

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