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

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==Problem 9==
 
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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Secondly, we consider how many ways to color them:
 
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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(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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(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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(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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(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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(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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Finally, we combine then together:<math>15\times 6+20\times 36+15\times 150+6\times 540+1\times 1806= \boxed{8106}</math>
  
 
So the answer is <math>\boxed{106}</math>.
 
So the answer is <math>\boxed{106}</math>.

Revision as of 16:52, 6 April 2013

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:$15\times 6+20\times 36+15\times 150+6\times 540+1\times 1806= \boxed{8106}$

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

See Also

2013 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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