Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 3)
(→‎Solution 2 (Recursive): more latex)
 
(5 intermediate revisions by 3 users not shown)
Line 6: Line 6:
 
We ignore the cases of 1 or 2 pieces since we need at least one tile of each color.
 
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>
 
* Four pieces: <math>\dbinom{6}{3}=20</math>
 
* Five pieces: <math>\dbinom{6}{4}=15</math>
 
* Five pieces: <math>\dbinom{6}{4}=15</math>
Line 29: Line 29:
 
<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)
 
<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 just stars and bars. The colorings are PIE giving <math>3^i-3\cdot2^i+3</math>. Plus, you don't have to worry about the cases where i=1 or i=2 since they both give no solutions. So our equation becomes:
+
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} (\dbinom{6}{i})\cdot(3^i-3\cdot2^i+3)</math>
+
<math>\sum_{i=3}^{7} \left(\dbinom{6}{I}\right)\cdot\left(3^i-3\cdot2^i+3\right)</math>
  
−
Write it all out, keep the numbers small with mod 1000, and you'll eventually arrive at your answer of <math>\boxed{106}</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
 
~Rowechen Zhong
  
 
==Solution 2 (Recursive)==
 
==Solution 2 (Recursive)==
−
3 colors is a lot. How many ways can we tile an n x 1 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>
+
3 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 (n+1) x 1 based off a n x 1. There are 2 cases we should consider:
+
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:
  
−
1) The n x 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 <math>2f_1(n)</math>
+
1) 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>
  
−
2) The n x 1 was a two-color tiling so now we've got 3 cases to form the (n+1) x 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 x 1.  
+
2) 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>
 
This gives us <math> f_2(n+1)=2f_1(n)+3f_2(n)</math>
Line 50: Line 50:
 
Time to tackle the 3-color tiling. Again, we split into 2 cases:
 
Time to tackle the 3-color tiling. Again, we split into 2 cases:
  
−
1) The n x 1 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 3C2 because we have to pick 2 different colors for the two-color tiling.  
+
1) 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.  
  
−
2) The n x 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 x 1. This gives <math>4f_3(n)</math>
+
2) 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>.  
 
So in total we have <math>f_3(n+1)=3f_2(n)+4f_3(n)</math>.  
Line 70: Line 70:
 
In the given problem, we have <math>n=7</math>. By PIE, we have
 
In the given problem, we have <math>n=7</math>. By PIE, we have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
&\quad\chi(7, 3) - {3 \choose 2} \, \chi(7, 2) + {3 \choose 1} \, \chi(7, 1) \\
+
&\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) \\
 
&= 3 \cdot 4^6 - 3(2 \cdot 3^6) + 3(1 \cdot 2^6) \\
 
&= 8106 \rightarrow \boxed{106}.
 
&= 8106 \rightarrow \boxed{106}.
 
\end{align*}</cmath>
 
\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==
 
==See Also==

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 $n-1$ squares, we have $x+1$ 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).

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_9&oldid=205091"