2023 AMC 10A Problems/Problem 20
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Video Solution by Little Fermat
- 8 Video Solution by Math-X (First fully understand the problem!!!)
- 9 Video Solution by MegaMath
- 10 Video Solution by Power Solve (easy to digest!)
- 11 Video Solution 1
- 12 Video Solution by OmegaLearn
- 13 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 14 Video Solution by Solve It (Simple)
- 15 Video Solution
- 16 Video Solution by TheBeautyofMath
- 17 See Also
Problem
Each square in a grid of squares is colored red, white, blue, or green so that every square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
Solution 1
Let a "tile" denote a \(1 \times 1\) square, and a "square" refer to a \(2 \times 2\) square.
We have \(4! = 24\) possible ways to fill out the top-left square. Next, we fill out the bottom-right tile. In the bottom-right square, one corner is already filled (from our initial coloring), so we have 3 options remaining.
For the right-middle tile, it is part of two squares: the top-right and the top-left. Among these squares, 3 colors have already been used, leaving us with only 1 remaining option. Similarly, every other square has only one available option for coloring.
Thus, the total number of ways is \(3 \times 4! = \boxed{\textbf{(D) }72}\).
~ stevehan
~ Technodoggo
Solution 2
We can split this problem into cases as shown above. We can swap a set of equal colors for another set of equal colors to create a new square.
Square 1: The first square can be rotated to create another square so we have to multiply the number of arrangements by . We have arrangements without rotating and arrangements in total for the first square.
Square 2: There are ways to arrange the colors.
In total, we have arrangements.
~South (LaTeX and Solution) Edit by: Mismatchedcubing/Andrew_Lu
Solution 3
Let’s call the top-right corner color A, the top-middle color B, the top-right color C, and so on, with color D being the middle row, and right corner square, and color G being the bottom-left square’s color. WLOG A=Red, B=White, D=Blue, and E=Green. We will now consider squares C and F’s colors. Case 1 : C=Red and F=Blue In this case, we get that G and H have to be Red and White in some order, and the same for H and I. We can color this in 2 ways. Case 2 : C=Blue and F=Red In this case, one of G and H needs to be White and Red, and H and I needs to be White and Blue. There is 1 way to color this. In total, we get 24*(2+1)=72 ways to color the grid. .
-paixiao
Solution 4
We will choose colors step-by-step:
1. There are ways to choose a color in the center.
2. Then we select any corner and there would be ways to choose a color as we can't use the same color as the one in the center.
3. Consider the square that contains the center and the corner we have selected. For the other squares, there are ways to choose colors.
4. Now, consider how many configurations it makes sense to construct the square opposite to the corner we have selected using the other squares, and we get configurations.
Finally, the answer is
~jjaamm
Solution 5
Note that we could fill the 3 by 3 square with numbers of choices, rather than letters or color names. The top-left corner receives a 3 because there are 3 total choices to choose from: R, G and B. The squares bordering them has values of 2 and 1, regardless of order. 2 indicates that the small square can have any color excluding the one existing color, 1 indicates the remaining color of the 2 by 2 square. Finally, the middle square receives 3, since the first 2 by 2 square has RGB already, and it does not matter what color it has. Moving onto the next 2 by 2 square, we see that there are already 2 decided colors, so the other squares must be 2 and 1, again, regardless of the order. The same applys to the third 2 by 2 square, and finally the last square has the value of one, as it is the only square left. Multiplying the numbers, = (Similar to Solution 1)
-MEZE_RUN
Video Solution by Little Fermat
https://youtu.be/h2Pf2hvF1wE?si=TCQJ2UEm-sB8rVjZ&t=4487 ~little-fermat
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/GP-DYudh5qU?si=YXUPq1RZv92E0d8H&t=7113
~Math-X
Video Solution by MegaMath
https://www.youtube.com/watch?v=AhqOj8502Dc&t=65s
~megahertz13
Video Solution by Power Solve (easy to digest!)
Video Solution 1
https://www.youtube.com/watch?v=VWZBpT9lt0Q&t=6s
-paixiao
Video Solution by OmegaLearn
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=vlaZ5P8UZls
Video Solution by Solve It (Simple)
https://www.youtube.com/watch?v=ke4ZOV4KA6Y
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by TheBeautyofMath
~IceMatrix Solution 5.
Let's name the cells A,B,C,D,E,F,G,H,I from the top left to the bottom right.
Case 1. Cell B and cell H have the same color. The middle one cell E has 4 choices, cell B has 3 choices, then cell E has 2 choices and cell F has 2 choices, this gives ways.
Case 2. Cell B and cell H have different colors. The middle one cell E has 4 choices, cell B has 3 choices, cell H has 2 choices, then cell D and F each can only have one choice(different from B,E,H). This gives ways.
The answer= 48+24=72.
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.