Competition · AMC preparation · step 4 of 4
AMC 10 · 2023A · #20
Grade 7 geometry-2d
Pick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The puzzle has a natural two-stage structure (Tool #7 — Identify Subproblems): (i) color the top-left 2 × 2 block in 4! = 24 ways, then (ii) count how many ways the third column and third row can be filled so that all four 2 × 2 sub-grids contain all four colors. By symmetry across the 24 initial choices, fix one (say R-W-B-G clockwise) and multiply at the end. For the completion step, the right column and bottom row each have 2 "local" choices, giving 4 combinations to check (Tool #2 — Systematic List). Tool #1 (Diagram) keeps the grid visible so the dependencies are easy to track.
Sketch the grid and blocks
Sketch the 3 × 3 grid and mark the four overlapping 2 × 2 blocks; the center cell lies inside all four, making it the most constrained.
Drawing the four 2 × 2 windows on a 3 × 3 grid makes the overlap pattern (and the strong constraints) visible.
3.G.A.2Draw A DiagramCount the top-left colorings
Every 2 × 2 block is a permutation of the four colors, so the top-left block can be filled in 4! = 24 ways; fix one and multiply back later.
Every 2 × 2 block uses all four colors exactly once, so the top-left block is a permutation of {R, W, B, G}.
7.SP.C.8Identify SubproblemsFix one representative coloring
Fix R, W, B, G in the top-left block; now the right column and bottom row each sit in a 2 × 2 block with two colors already set.
Pinning down the top-left block reduces the remaining work to filling 5 cells under heavy local constraints.
3.G.A.2Draw A DiagramFill the remaining cells
Top-right needs {R, B}, bottom-left needs {R, W}, each in two orders — 4 local options, filtered next by the bottom-right block.
Each adjacent 2 × 2 block needs the two colors missing from its known half — naturally two choices for each missing pair.
7.SP.C.8Identify SubproblemsList the four combinations
List all 4 combinations of the two binary choices and test each against the bottom-right block for a feasible C₃₃.
With only four cases, enumeration is faster than any clever argument — list, check, count.
4.OA.C.5Make A Systematic ListCheck each case
Three combinations leave the bottom-right block with three distinct colors so C₃₃ is forced; the fourth repeats R and dies — 3 valid.
Three of the four binary-choice combos pass the bottom-right 2 × 2 check; the fourth forces a color collision.
7.SP.C.8Make A Systematic ListMultiply the two stages
Multiply the stages: 24 top-left colorings × 3 valid completions each = 72, which is choice (D).
Multiplication principle: independent stages multiply, and the 3-completion count is the same for every top-left coloring by symmetry.
The two stages are independent, so their counts multiply into the total.
▸ Why?
Each stage is chosen without regard to the other, so every combination occurs exactly once.
▸ Why?
Relabelling the colours carries one starting block to any other, so each gives the same completion count.
The top-left 2 × 2 block can be colored in 4! = 24 ways. For each choice, the right column and bottom row each have 2 binary options, but only 3 of the 4 combined options keep the bottom-right 2 × 2 valid. Multiply: 24 × 3 = (D) 72.
- Sketch the grid and blocks
- Count the top-left colorings
- Fix one representative coloring
- Fill the remaining cells
- List the four combinations
- Check each case
- Multiply the two stages
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