2023 AMC 10A Problems/Problem 20

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 $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?

$[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("B", (19.5,7.5)); label("R", (22.5,7.5)); label("B", (25.5,7.5)); [/asy]$

$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$

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

$[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); label("R", (1.5,1.5)); label("B", (4.5,1.5)); label("R", (7.5,1.5)); label("G", (1.5,4.5)); label("W", (4.5,4.5)); label("G", (7.5,4.5)); label("B", (1.5,7.5)); label("R", (4.5,7.5)); label("B", (7.5,7.5)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("R", (19.5,7.5)); label("B", (22.5,7.5)); label("R", (25.5,7.5)); [/asy]$ We can split this problem into $2$ 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 $2$ . We have $4! = 24$ arrangements without rotating and $24\cdot 2 = 48$ arrangements in total for the first square.

Square 2: There are $4! = 24$ ways to arrange the colors.

In total, we have $48 + 24 = \boxed{\textbf{(D) }72}$ 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. $\boxed{\textbf{(D) }72}$ .

-paixiao

Solution 4

We will choose colors step-by-step:

1. There are $4$ ways to choose a color in the center.

2. Then we select any corner and there would be $3$ ways to choose a color as we can't use the same color as the one in the center.

3. Consider the $2\times 2$ square that contains the center and the corner we have selected. For the other $2$ squares, there are $2$ ways to choose colors.

4. Now, consider how many configurations it makes sense to construct the $2\times 2$ square opposite to the corner we have selected using the $2$ other $2\times 2$ squares, and we get $3$ configurations.

Finally, the answer is $4 \cdot 3 \cdot 2 \cdot 3 = \boxed{\textbf{(D) }72}$

~jjaamm

Solution 5

$[asy] unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); label("2", (1.5,1.5)); label("1", (4.5,1.5)); label("1", (7.5,1.5)); label("2", (1.5,4.5)); label("3", (4.5,4.5)); label("1", (7.5,4.5)); label("3", (1.5,7.5)); label("1", (4.5,7.5)); label("2", (7.5,7.5)); [/asy]$

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, $2\times2\times2\times3\time3\times3$ = $\boxed{\textbf{(D) }72}$ (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!)

https://youtu.be/0_eRTysDEsI

Video Solution 1

https://www.youtube.com/watch?v=VWZBpT9lt0Q&t=6s

-paixiao

Video Solution by OmegaLearn

https://youtu.be/Zrqy5yGYlvQ

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

https://youtu.be/8ki0D3m_mEo

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by TheBeautyofMath

https://youtu.be/BDNsMEIzHF0

~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 $4\cdot 3\cdot 2\cdot 2=48$ 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 $4\cdot 3\cdot 2=24$ ways.

The answer= 48+24=72. $(D)$

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

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