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Difference between revisions of "2013 AIME II Problems/Problem 9"

(Problem 9)
 
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A <math>7\times 1</math> board is completely covered by <math>m\times 1</math> 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 <math>N</math> be the number of tilings of the <math>7\times 1</math> board in which all three colors are used at least once. For example, a <math>1\times 1</math> red tile followed by a <math>2\times 1</math> green tile, a <math>1\times 1</math> green tile, a <math>2\times 1</math> blue tile, and a <math>1\times 1</math> green tile is a valid tiling. Note that if the <math>2\times 1</math> blue tile is replaced by two <math>1\times 1</math> blue tiles, this results in a different tiling. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 
A <math>7\times 1</math> board is completely covered by <math>m\times 1</math> 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 <math>N</math> be the number of tilings of the <math>7\times 1</math> board in which all three colors are used at least once. For example, a <math>1\times 1</math> red tile followed by a <math>2\times 1</math> green tile, a <math>1\times 1</math> green tile, a <math>2\times 1</math> blue tile, and a <math>1\times 1</math> green tile is a valid tiling. Note that if the <math>2\times 1</math> blue tile is replaced by two <math>1\times 1</math> blue tiles, this results in a different tiling. Find the remainder when <math>N</math> is divided by <math>1000</math>.
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==Solution==
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Firstly, we consider how many different ways possible to divide the <math>7\times 1</math> board.
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(1)One piece: <math>7</math>,<math>\dbinom{6}{0}=1</math>way in total(since the 3 colors are used at least once, we reject this case)
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(2)Two pieces:<math>6+1</math>,<math>5+2</math>,etc,<math>\dbinom{6}{1}=6</math>ways in total(rejected for the same reason)
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(3)Three pieces:<math>5+1+1</math>,<math>4+2+1</math>,<math>4+1+2</math>,etc,<math>\dbinom{6}{2}=15</math>ways in total
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(4)Four pieces:<math>\dbinom{6}{3}=20</math>  (5)Five pieces:<math>\dbinom{6}{4}=15</math> (6)Six pieces:<math>\dbinom{6}{5}=6</math> (7)Seven pieces:<math>\dbinom{6}{6}=1</math>
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Secondly, we consider how many ways to color them:
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(1)There pieces:<math>3^3-3\times 2^3+3=6</math>
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(2)Four pieces:<math>3^4-3\times 2^4+3=36</math>
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(3)Five pieces:<math>3^5-3\times 2^5+3=150</math>
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(4)Six pieces:<math>3^6-3\times 2^6+3=540</math>
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(5)Seven pieces:<math>3^7-3\times 2^7+3=1806</math>
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Finally, we combine then together:<math>15\times 6+20\times 36+15\times 150+6\times 540+1\times 1806= 8106</math>
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So the answer is <math>\boxed{106}</math>

Revision as of 00:48, 5 April 2013

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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