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2013 AIME II Problems/Problem 9

Revision as of 00:48, 5 April 2013 by Aux770 (talk | contribs)

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 $\boxed{106}$

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