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

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Firstly, we consider how many different ways possible to divide the <math>7\times 1</math> board.
 
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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# 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)
−
 
+
# 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)
−
(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)
+
# 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
−
 
+
# Four pieces:<math>\dbinom{6}{3}=20</math>
−
(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
+
# Five pieces:<math>\dbinom{6}{4}=15</math>
−
 
+
# Six pieces:<math>\dbinom{6}{5}=6</math>
−
(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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# Seven pieces:<math>\dbinom{6}{6}=1</math>
  
 
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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# Three pieces: <math>3^3-3\times 2^3+3=6</math>
−
 
+
# Four pieces: <math>3^4-3\times 2^4+3=36</math>
−
(2)Four pieces: <math>3^4-3\times 2^4+3=36</math>
+
# Five pieces: <math>3^5-3\times 2^5+3=150</math>
−
 
+
# Six pieces: <math>3^6-3\times 2^6+3=540</math>
−
(3)Five pieces: <math>3^5-3\times 2^5+3=150</math>
+
# Seven pieces: <math>3^7-3\times 2^7+3=1806</math>
−
 
 
−
(4)Six pieces: <math>3^6-3\times 2^6+3=540</math>
 
−
 
 
−
(5)Seven pieces: <math>3^7-3\times 2^7+3=1806</math>
 
  
−
Finally, we combine then together:<math>15\times 6+20\times 36+15\times 150+6\times 540+1\times 1806= 8106</math>
+
Finally, we combine then together: <math>15\times 6+20\times 36+15\times 150+6\times 540+1\times 1806= 8106</math>
  
 
So the answer is <math>\boxed{106}</math>.
 
So the answer is <math>\boxed{106}</math>.

Revision as of 15:55, 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. Three 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 $\boxed{106}$.

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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