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

Latest revision as of 13:14, 21 November 2023

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 1

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.

Three pieces: $5+1+1$ , $4+2+1$ , $4+1+2$ , etc, $\dbinom{6}{2}=15$ ways in total (just apply stars and bars here)
Four pieces: $\dbinom{6}{3}=20$
Five pieces: $\dbinom{6}{4}=15$
Six pieces: $\dbinom{6}{5}=6$
Seven pieces: $\dbinom{6}{6}=1$

Secondly, we use Principle of Inclusion-Exclusion to consider how many ways to color them:

Three pieces: $3^3-3\times 2^3+3=6$
Four pieces: $3^4-3\times 2^4+3=36$
Five pieces: $3^5-3\times 2^5+3=150$
Six pieces: $3^6-3\times 2^6+3=540$
Seven pieces: $3^7-3\times 2^7+3=1806$

Finally, we combine them together: $15\times 6+20\times 36+15\times 150+6\times 540+1\times 1806= 8106$ .

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

Solution 1.1 (Faster/Streamlined 1)

This solution is basically solution 1 with more things done at once. The game plan:

$\sum_{i=0}^{7} ($ the amount of ways to divide the board into $i$ pieces $) \cdot ($ the amount of ways to color the respective divisions)

The amount of ways to divide the board is determined using stars and bars. The colorings are found using PIE giving $3^i-3\cdot2^i+3$ . Plus, we don't have to worry about the cases where $i=1$ or $I=2$ since they both give no solutions. So our equation becomes:

$\sum_{i=3}^{7} \left(\dbinom{6}{I}\right)\cdot\left(3^i-3\cdot2^i+3\right)$

Writing it all out and keeping the numbers small with mod 1000, we will eventually arrive at the answer of $\boxed{106}$ .

~Rowechen Zhong

Solution 2 (Recursive)

3 colors is a lot. How many ways can we tile an $n \times 1$ board with one color? It's going to be $2^{n-1}$ because if you draw out the $n$ spaces, you can decide whether each of the borders between the tiles are either there or not there. There are $n-1$ borders so there are $2^{n-1}$ tilings. Define a one-tiling of an mx1 as $f_1(m)$

Now let's look at two colors. Let's see if we can get a two-tiling of an $(n+1) \times 1$ based off a $n \times 1$ . There are 2 cases we should consider:

1) The $n \times 1$ was a one-coloring and the block we are going to add consists of the second color. The number of ways we can do this is $2f_1(n)$

2) The $n \times 1$ was a two-color tiling so now we've got 3 cases to form the $(n+1) \times 1$ : We can either add a block of the first color, the second color, or we can adjoin a block to the last block in the $n \times 1$ .

This gives us $f_2(n+1)=2f_1(n)+3f_2(n)$

Time to tackle the 3-color tiling. Again, we split into 2 cases:

1) The $n \times 1$ was a two-color tiling, and the block we are adding is of the 3rd color. This gives $f_2(n)$ ways but we have to multiply by $3C2 = 3$ because we have to pick 2 different colors for the two-color tiling.

2) The $n \times 1$ was a 3-color tiling, and we have to consider what we can do with the block that we are adding. It can be any of the 3 colors, or we can adjoin it to whatever was the last block in the $n \times 1$ . This gives $4f_3(n)$

So in total we have $f_3(n+1)=3f_2(n)+4f_3(n)$ .

Finally, we just sorta bash through the computation to get $f_3(7)=8\boxed{106}$

Solution 3

Let $n$ be the length of the board and $x$ be the number of colors. We will find the number of ways to tile the $n \times 1$ board with no color restrictions (some colors may be unused) and then use PIE.

By stars and bars, the number of ways to divide the board into $k$ pieces is ${n-1 \choose k-1}$ . There are $x^k$ ways to color each of these divisions. Therefore, the total number of ways to divide and color the board is $\begin{align*} \chi(n, x) &:= \sum_{k=1}^n {n-1 \choose k-1} x^k \\ &= x\sum_{k=0}^{n-1} {n-1 \choose k} x^k \\ &= x(x+1)^{n-1}. \end{align*}$

In the given problem, we have $n=7$ . By PIE, we have $\begin{align*} &\quad {3 \choose 3} \chi(7, 3) - {3 \choose 2} \, \chi(7, 2) + {3 \choose 1} \, \chi(7, 1) \\ &= 3 \cdot 4^6 - 3(2 \cdot 3^6) + 3(1 \cdot 2^6) \\ &= 8106 \rightarrow \boxed{106}. \end{align*}$

The above formula for $\chi(n, x)$ can also be derived directly as follows:

The first square can be colored in one of $x$ ways.
Proceeding left to right, at each of the remaining squares, we have options:
- $x$ ways to begin a new tile, selecting an arbitrary color. Note that this includes the case where we begin a new tile of the same color.
- $1$ way to expand the current tile by attaching a square of the current color.

This results in $x(x+1)^{n-1}$ overall ways to divide and color the board (without the color use restriction).

@@ Line 2: / Line 2: @@
 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>.
-==Solution==
+==Solution 1==
 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.
-* 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
+* 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 (just apply stars and bars here)
 * Four pieces: <math>\dbinom{6}{3}=20</math>
 * Five pieces: <math>\dbinom{6}{4}=15</math>
@@ Line 12: / Line 12: @@
 * Seven pieces: <math>\dbinom{6}{6}=1</math>
-Secondly, we consider how many ways to color them:
+Secondly, we use Principle of Inclusion-Exclusion to consider how many ways to color them:
 * Three pieces: <math>3^3-3\times 2^3+3=6</math>
@@ Line 20: / Line 20: @@
 * 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 them 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>.
+==Solution 1.1 (Faster/Streamlined 1)==
+This solution is basically solution 1 with more things done at once. The game plan:
+<math>\sum_{i=0}^{7} (</math>the amount of ways to divide the board into <math>i</math> pieces<math>) \cdot (</math>the amount of ways to color the respective divisions)
+The amount of ways to divide the board is determined using stars and bars. The colorings are found using PIE giving <math>3^i-3\cdot2^i+3</math>. Plus, we don't have to worry about the cases where <math>i=1</math> or <math>I=2</math> since they both give no solutions. So our equation becomes:
+<math>\sum_{i=3}^{7} \left(\dbinom{6}{I}\right)\cdot\left(3^i-3\cdot2^i+3\right)</math>
+Writing it all out and keeping the numbers small with mod 1000, we will eventually arrive at the answer of <math>\boxed{106}</math>.
+~Rowechen Zhong
+==Solution 2 (Recursive)==
+colors is a lot. How many ways can we tile an <math>n \times 1</math> board with one color? It's going to be <math>2^{n-1}</math> because if you draw out the <math>n</math> spaces, you can decide whether each of the borders between the tiles are either there or not there. There are <math>n-1</math> borders so there are <math>2^{n-1}</math> tilings. Define a one-tiling of an mx1 as <math>f_1(m)</math>
+Now let's look at two colors. Let's see if we can get a two-tiling of an <math>(n+1) \times 1</math> based off a <math>n \times 1</math>. There are 2 cases we should consider:
+) The <math>n \times 1</math> was a one-coloring and the block we are going to add consists of the second color. The number of ways we can do this is <math>2f_1(n)</math>
+) The <math>n \times 1</math> was a two-color tiling so now we've got 3 cases to form the <math>(n+1) \times 1</math>: We can either add a block of the first color, the second color, or we can adjoin a block to the last block in the <math>n \times 1</math>.
+This gives us <math> f_2(n+1)=2f_1(n)+3f_2(n)</math>
+Time to tackle the 3-color tiling. Again, we split into 2 cases:
+) The <math>n \times 1</math> was a two-color tiling, and the block we are adding is of the 3rd color. This gives <math>f_2(n)</math> ways but we have to multiply by <math>3C2 = 3</math> because we have to pick 2 different colors for the two-color tiling.
+) The <math>n \times 1</math> was a 3-color tiling, and we have to consider what we can do with the block that we are adding. It can be any of the 3 colors, or we can adjoin it to whatever was the last block in the <math>n \times 1</math>. This gives <math>4f_3(n)</math>
+So in total we have <math>f_3(n+1)=3f_2(n)+4f_3(n)</math>.
+Finally, we just sorta bash through the computation to get <math>f_3(7)=8\boxed{106}</math>
+==Solution 3==
+Let <math>n</math> be the length of the board and <math>x</math> be the number of colors. We will find the number of ways to tile the <math>n \times 1</math> board with no color restrictions (some colors may be unused) and then use PIE.
+By stars and bars, the number of ways to divide the board into <math>k</math> pieces is <math>{n-1 \choose k-1}</math>. There are <math>x^k</math> ways to color each of these divisions. Therefore, the total number of ways to divide and color the board is
+<cmath>\begin{align*}
+\chi(n, x) &:= \sum_{k=1}^n {n-1 \choose k-1} x^k \\
+&= x\sum_{k=0}^{n-1} {n-1 \choose k} x^k \\
+&= x(x+1)^{n-1}.
+\end{align*}</cmath>
+In the given problem, we have <math>n=7</math>. By PIE, we have
+<cmath>\begin{align*}
+&\quad {3 \choose 3} \chi(7, 3) - {3 \choose 2} \, \chi(7, 2) + {3 \choose 1} \, \chi(7, 1) \\
+&= 3 \cdot 4^6 - 3(2 \cdot 3^6) + 3(1 \cdot 2^6) \\
+&= 8106 \rightarrow \boxed{106}.
+\end{align*}</cmath>
+The above formula for <math>\chi(n, x)</math> can also be derived directly as follows:
+* The first square can be colored in one of <math>x</math> ways.
+* Proceeding left to right, at each of the remaining <math>n-1</math> squares, we have <math>x+1</math> options:
+** <math>x</math> ways to begin a new tile, selecting an arbitrary color. Note that this includes the case where we begin a new tile of the same color.
+** <math>1</math> way to expand the current tile by attaching a square of the current color.
+This results in <math>x(x+1)^{n-1}</math> overall ways to divide and color the board (without the color use restriction).
 ==See Also==

2013 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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