2021 AMC 10 A Problem 25

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To help you master 2021 AMC 10 A Problem 25, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

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2021 AMC10 A Problem 25

Solution

Solution 1

Checkerboard Parity Method

By coloring the grid like a checkerboard, we see that the center square touches all four edge-centers. This completely bans the center’s color from the edge-centers, forcing a rigid placement that simplifies the rest of the grid.

When a grid has a rule against adjacent elements, divide the squares into a checkerboard pattern. Squares of the same color (black or white) never share an edge.

Step 1: Identify the independence of squares using a checkerboard

Imagine the $3\times 3$ grid colored like a checkerboard. The center square and the 4 corners are Black (5 squares). The 4 edge-centers are White (4 squares). The adjacency rule means no two chips of the same color can occupy adjacent squares.

Step 2: Determine the placement of the center’s color

Notice that all 4 White squares (the edge-centers) share an edge with the center square. Therefore, whatever color is placed in the center CANNOT be placed in any White square. Since each color has exactly 3 chips, the color in the center must place all 3 of its chips on the Black squares. One chip is in the center, so the remaining 2 chips of that color must go into the 4 corners.

Step 3: Analyze the remaining corners when the center is Red (Case 1: Opposite)

Assume the center is Red. There are two geometric ways to place the remaining 2 Reds in the corners. First, the Reds can be in opposite corners (e.g., Top-Left and Bottom-Right). This can be done in 2 ways. The other 2 corners must be Blue and Green. If both were Blue, then all 4 White squares would touch a Blue corner, leaving nowhere for the third Blue chip. Thus, they must be different colors (2 ways to assign Blue and Green). The remaining 2 Blue and 2 Green chips must go into the 4 White squares. To avoid touching their own color in the corners, their positions are uniquely forced (1 way). This gives $2\times 2\times 1=4$ valid grids.

$$2\times 2\times 1=4$$

Step 4: Analyze the remaining corners when the center is Red (Case 2: Adjacent)

Second, the Reds can be in adjacent corners (e.g., Top-Left and Top-Right). This can be chosen in 4 ways. The other 2 corners (Bottom-Left and Bottom-Right) must be the same color. If one were Blue and one Green, both colors would need two White squares, but they both share the Bottom-Center square as a neighbor, meaning neither could go there. This would leave only 3 valid White squares for 4 chips, which is impossible. So, the two corners must both be Blue or both be Green (2 ways). This uniquely forces the placement of the remaining chips (1 way). This gives $4\times 2\times 1=8$ valid grids.

$$4\times 2\times 1=8$$

Step 5: Calculate the total number of valid grids

Adding the cases together, there are $4+8=12$ valid arrangements if Red is in the center. Because the center can be Red, Blue, or Green, we multiply the result by 3.

$$12\times 3=36$$

Correct answer: E. $36$
The total number of valid placements is 36, which corresponds to answer choice E. We can verify the logic by noticing that every placement exactly mirrors the strict bipartite nature of the grid without overlaps.

Solution 2

Row and Column Pattern Classification

Every valid row must either contain 3 different colors ($A-B-C$) or alternate 2 colors ($A-B-A$). By proving that these row types cannot be mixed, we can exhaustively count the grids as Latin squares or striped patterns.

A Latin square of order $n$ is an $n\times n$ grid filled with $n$ different symbols, where each symbol appears exactly once in each row and column.

Step 1: Classify the possible row types

Because adjacent squares cannot be the same color, a row cannot contain adjacent duplicates. A row must either have 3 distinct colors (e.g., Red-Blue-Green, which we call an $A-B-C$ row) or 2 distinct colors alternating (e.g., Red-Blue-Red, which we call an $A-B-A$ row).

Step 2: Prove that row types cannot be mixed

An $A-B-C$ row contains exactly 1 Green. If a grid had exactly two $A-B-C$ rows, those rows would use at most 2 Greens. The third row, being an $A-B-A$ row, can hold at most 0 or 2 Greens, but no combination properly accounts for all 3 chips of all 3 colors without violating adjacency. Thus, the grid must consist entirely of $A-B-C$ rows, or entirely of $A-B-A$ rows.

Step 3: Count grids with entirely A-B-C rows (Latin Squares)

If all rows are $A-B-C$, the grid must be a $3\times 3$ Latin square. There are $3!=6$ ways to color the first row. For each choice, there are exactly 2 valid ways to complete the remaining two rows without vertical clashes. This yields 12 grids.

$$6\times 2=12$$

Step 4: Count grids with entirely A-B-A rows (Row-Striped)

In an $A-B-A$ row, one color appears twice (the majority) and another appears once (the minority). To use exactly 3 chips of each color across the whole grid, the 3 rows must have different majority colors (say $M_1,M_2,M_3$) and different minority colors ($m_1,m_2,m_3$). There are $3!=6$ ways to assign the sequence of majorities to the 3 rows. For the rows to be valid, the minority color cannot equal the majority color ($m_i\ne M_i$). This means the sequence of minorities must be a derangement of the majorities. There are exactly 2 derangements of 3 items. Because the $M_i$ and $m_i$ sequences are permutations, vertical adjacencies are automatically avoided (e.g., $M_1\ne M_2$). This yields 12 grids.

$$6\times 2=12$$

Step 5: Count grids with A-B-A columns (Column-Striped) and sum

By transposing the row-striped grids, we get grids where the columns are $A-B-A$ and the rows are $A-B-C$. By symmetry, there are 12 of these grids. These three cases (Latin squares, row-striped, column-striped) are mutually exclusive and collectively exhaustive.

$$12+12+12=36$$

Correct answer: E. $36$
Summing the mutually exclusive structural cases gives $12+12+12=36$. Both distinct mathematical models (checkerboard spatial logic vs. row-parity combinatorics) confidently arrive at 36. This matches option E.

Similar Problems in Real Exams

The following five AMC 10 A problems are closely related to AMC 10 A 2021 Question 25. These real AMC 10 A problems test the same key concept—Combinatorics. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2023 AMC 10 A Problem 20

2022 AMC 10 A Problem 9

2025 AMC 10 A Problem 8

2025 AMC 10 A Problem 12

2025 AMC 10 A Problem 21

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Checkerboard Bipartite Coloring:

Applies to square grids where moves or restrictions happen strictly horizontally and vertically. By noticing that the center square is black and its 4 adjacent neighbors are all white, we instantly proved that the center color could not be placed on any of those 4 white squares.

• Derangements:

Useful when pairing items from two identical sets such that no item pairs with its exact counterpart. In Method 2, matching a minority color to a majority color in a row required that they not be the same color, meaning the minority sequence had to be a derangement of the majority sequence.

Key Takeaway: Remember This Strategy for Next Time

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