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

m
m
Line 4: Line 4:
 
==Solution==
 
==Solution==
 
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.
 +
We ignore the cases of 1 or 2 pieces since we need at least one tile of each color.
  
# 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)
+
# 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
# 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)
+
# Four pieces: <math>\dbinom{6}{3}=20</math>
# 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>
# Four pieces:<math>\dbinom{6}{3}=20</math>
+
# Six pieces: <math>\dbinom{6}{5}=6</math>
# Five pieces:<math>\dbinom{6}{4}=15</math>
+
# Seven pieces: <math>\dbinom{6}{6}=1</math>
# Six pieces:<math>\dbinom{6}{5}=6</math>
 
# 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:
Line 21: Line 20:
 
# Seven pieces: <math>3^7-3\times 2^7+3=1806</math>
 
# 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 15:57, 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. We ignore the cases of 1 or 2 pieces since we need at least one tile of each color.

  1. Three pieces: $5+1+1$, $4+2+1$, $4+1+2$, etc, $\dbinom{6}{2}=15$ ways in total
  2. Four pieces: $\dbinom{6}{3}=20$
  3. Five pieces: $\dbinom{6}{4}=15$
  4. Six pieces: $\dbinom{6}{5}=6$
  5. 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 (ProblemsAnswer KeyResources)
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