AMC 10 · 2025 · #21
Grade 8 countingEach of the 9 squares in a 3×3 grid is to be colored red, blue, or yellow in such a way that each red square shares an edge with at least one blue square, each blue square shares an edge with at least one yellow square, and each yellow square shares an edge with at least one red square. Colorings that can be obtained from one another by rotations and/or reflections are considered the same. How many different colorings are possible?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Color each of the $9$ cells of a $3\times 3$ grid red, blue, or yellow so that every red cell touches a blue cell along an edge, every blue cell touches a yellow cell, and every yellow cell touches a red cell. Two colorings are the same if a rotation or reflection turns one into the other. Count the different colorings.
Givens: A $3\times 3$ grid; each cell gets one of three colors: red, blue, yellow; Every red cell must share an edge with at least one blue cell; Every blue cell must share an edge with at least one yellow cell; Every yellow cell must share an edge with at least one red cell; Colorings related by any rotation or reflection of the square count as one; Answer choices: (A) $3$, (B) $9$, (C) $12$, (D) $18$, (E) $27$
Unknowns: The number of colorings that are different once rotations and reflections are treated as the same
Understand
Restated: Color each of the $9$ cells of a $3\times 3$ grid red, blue, or yellow so that every red cell touches a blue cell along an edge, every blue cell touches a yellow cell, and every yellow cell touches a red cell. Two colorings are the same if a rotation or reflection turns one into the other. Count the different colorings.
Givens: A $3\times 3$ grid; each cell gets one of three colors: red, blue, yellow; Every red cell must share an edge with at least one blue cell; Every blue cell must share an edge with at least one yellow cell; Every yellow cell must share an edge with at least one red cell; Colorings related by any rotation or reflection of the square count as one; Answer choices: (A) $3$, (B) $9$, (C) $12$, (D) $18$, (E) $27$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement, #3 Eliminate Possibilities
The question is a "how many colorings" count, so Tool #2 (Make a Systematic List) is the backbone: once the grid is mostly pinned down, we list every way to finish it. But listing all $3^9$ raw colorings is hopeless, so two ideas cut it down first. Tool #16 (Change Focus) exploits symmetry twice: rotations and reflections merge equal colorings, and relabeling colors along the loop red$\to$blue$\to$yellow$\to$red is itself allowed, which lets us fix the center cell to red and multiply by $3$ at the end. Tool #1 (Draw a Diagram) keeps the grid in front of us so each rule is visible, and Tool #3 (Eliminate Possibilities) forces most cells: once the center and one neighbor are set, the loop rule leaves only one legal color for cell after cell.
Execute — Answer: C
6.EE.B.5 Step 1 Read the rules as a color loop
- Draw the $3\times 3$ grid and restate the three rules as a single loop: red must touch blue, blue must touch yellow, yellow must touch red.
- So each color "points to" the next one in the cycle red$\to$blue$\to$yellow$\to$red.
- Every cell, whatever its color, must have an edge neighbor holding the next color in this loop.
💡 Turning three separate rules into one repeating loop means you only have to remember a single arrow.
8.G.A.2 Step 2 Use symmetry to fix the center as red
- Two kinds of symmetry shrink the job.
- First, rotations and reflections make matching colorings count as one.
- Second, swapping every color forward along the loop (red becomes blue, blue becomes yellow, yellow becomes red) turns any legal coloring into another legal coloring, because the loop of rules is unchanged.
- That swap changes the center cell's color, so legal colorings split into three equal groups by the center's color.
- Count the ones with a red center, then multiply by $3$.
💡 Because relabeling colors keeps every rule true, the center is red exactly one-third of the time.
8.G.A.1 Step 3 Place the forced first neighbors
- Fix the center red.
- It must touch a blue cell; the four edge cells around it are interchangeable by rotation, so put that blue on the left-middle edge.
- That blue must touch a yellow cell.
- Its free neighbors are the two left corners, and a reflection across the middle row swaps them, so put the yellow at the top-left corner.
- Using symmetry to lock these two placements costs nothing and pins down a starting corner.
💡 Spending the rotations and reflections now fixes a corner, so later choices are real choices, not repeats.
6.EE.B.5 Step 4 Chase the loop around the top
- Now the loop forces cell after cell.
- The top-left yellow needs a red neighbor; its only open neighbor is the top-middle, so that is red.
- This red needs a blue neighbor; its only open neighbor is the top-right corner, so that is blue.
- This blue needs a yellow neighbor; its only open neighbor is the right-middle edge, so that is yellow.
- The top two rows are now completely forced, and every rule for those six cells is already satisfied.
💡 With a neighbor's color known, the loop leaves only one legal color, so each cell falls like a domino.
7.SP.C.8 Step 5 List every legal bottom row
- Only the three bottom cells are still open.
- Each must obey its own rule, using neighbors already known: the bottom-left sees the left blue, the bottom-middle sees the center red, the bottom-right sees the right yellow, and the three share edges with each other.
- Checking all combinations by the loop rule leaves exactly four legal bottom rows: $(\text{R},\text{R},\text{B})$, $(\text{Y},\text{R},\text{B})$, $(\text{R},\text{Y},\text{B})$, and $(\text{B},\text{Y},\text{B})$.
💡 With everything above fixed, the whole puzzle shrinks to filling three cells you can just list out.
8.G.A.2 Step 6 Multiply by the three center colors
- There are $4$ legal colorings with a red center.
- By the color-loop symmetry from earlier, blue-center and yellow-center each also give $4$, and these three groups never overlap because the center color is different.
- So the total is $4\times 3 = 12$ different colorings.
- The answer is $(\text{C})\ 12$.
💡 Each of the four red-center pictures has a twin in blue and in yellow, tripling the count.
6.EE.B.5 Draw the $3\times 3$ grid and restate the three rules as a single loop: red must 8.G.A.2 Two kinds of symmetry shrink the job. First, rotations and reflections make matc 8.G.A.1 Fix the center red. It must touch a blue cell; the four edge cells around it are 6.EE.B.5 Now the loop forces cell after cell. The top-left yellow needs a red neighbor; i 7.SP.C.8 Only the three bottom cells are still open. Each must obey its own rule, using n 8.G.A.2 There are $4$ legal colorings with a red center. By the color-loop symmetry from Review
Reasonableness: The four red-center grids are all legal and none is a rotation or reflection of another, so they are genuinely different, and multiplying by the three center colors gives $12$, matching (C). A direct check by computer of all $3^9$ colorings finds $84$ legal ones; grouping them by the square's $8$ symmetries collapses them to $12$ classes, confirming the count. The plausible wrong answers line up too: (E) $27$ ignores symmetry, and (A) $3$ or (B) $9$ would come from undercounting the four bottom-row options.
Alternative: Skip the color-loop shortcut and instead fix the center to each color in turn, doing the same forced-chain and bottom-row listing three separate times. Each center color yields $4$, so $4+4+4=12$. This is more work but avoids the relabeling argument, and it lands on the same (C).
CCSS standards used (min grade 8)
8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Using rotations and reflections of the square to fix the blue neighbor on the left edge and the yellow at the top-left corner without losing generality.)8.G.A.2Understand that a two-dimensional figure is congruent to another using transformations (Treating colorings that match under a symmetry as the same, and using the color-relabeling symmetry to split colorings into three equal center-color groups and multiply by 3.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Reading each adjacency rule as a condition a cell must satisfy and deducing the single color that keeps it true.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Making an organized list of all colorings of the three free bottom cells and keeping only the four that obey every rule.)
⭐ Pin down whatever the rules force, use the shape's symmetry to avoid counting the same picture twice, and then just list the few choices that are left.
⭐ Pin down whatever the rules force, use the shape's symmetry to avoid counting the same picture twice, and then just list the few choices that are left.
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