AMC 8 · 2024 · #17

Grade 3 geometry-2d

Archetype Casework by Position / Vertex Type └ Sub-archetype Grid / Solid Cell Position

systematic-enumerationcombinations-basic caseworksystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticmental-arithmetic

📏 Long solution 💡 4 insights 📊 Diagram

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Problem

Place a distinct white king and black king on different squares of a 3 × 3 grid so that they do not attack each other (no horizontal, vertical, or diagonal neighbor). Count the number of such ordered placements.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Everything happens on a tiny 3 × 3 picture, so first use Tool #1 — Draw the grid and look directly at where the white king can sit. The diagram makes clear that the 9 squares naturally split into three types: corner, edge, and center, and the king attacks a different number of squares from each type. So Tool #7 — Identify subproblems — splits the big count into three smaller cases (white king on a corner / on an edge / in the center). Inside each case, Tool #2 — a systematic listing via the multiplication rule — gives the number of legal black-king squares. Finally Tool #3 — eliminate possibilities — checks our total against the answer choices. Since this is multiple choice on a tiny grid, simple visual + counting tools suffice; no algebra is needed.

1STEP 1

Split the 9 squares by position into 4 corners, 4 edges, and 1 center — 4 + 4 + 1 = 9, every square classified once.

corners=4, edges=4, center=1, total=9

2STEP 2

A king attacks 3 squares from a corner, 5 from an edge, and 8 from the center — the key numbers 3, 5, 8.

corner attacks=3, edge attacks=5, center attacks=8

3STEP 3

Split by where the white king sits — corner, edge, or center; each case counts (white squares) × (9 - 1 - attacks).

count per case = (white-king squares) × (9 - 1 - attacks)

4STEP 4

Case ①, white king on a corner: 4 corners each leave 9 - 4 = 5 squares, so 4 × 5 = 20.

4 × (9 - 1 - 3) = 4 × 5 = 20

5STEP 5

Case ②, white king on an edge: it blocks 6 squares, leaving 9 - 6 = 3, so 4 × 3 = 12.

4 × (9 - 1 - 5) = 4 × 3 = 12

6STEP 6

Case ③, white king in the center blocks all 9 squares, leaving 9 - 9 = 0, so 1 × 0 = 0 — no room for the black king.

1 × (9 - 1 - 8) = 1 × 0 = 0

7STEP 7

The three positions partition the grid with no overlap, so add the cases: 20 + 12 + 0 = 32 total placements.

20 + 12 + 0 = 32

8STEP 8

Match 32 to the choices: (A) 20 is only case ①, (D) 28 falls 4 short — our value is exactly (E).

32 → (E)

Answer

Sanity-check with the complement. Total ordered placements on different squares: 9 × 8 = 72. Adjacent pairs on a 3 × 3 grid: 3 · 2 + 2 · 3 = 12 horizontal/vertical and 2 · 2 · 2 = 8 diagonal, 20 pairs in all. Since the two kings are distinguishable, attacking placements count as 20 × 2 = 40, and non-attacking ones are 72 - 40 = 32 — matching our casework answer exactly. It also fits intuition that the center contributes 0 and the corner case is the largest at 20.

💡Key takeaway

This AMC 8 problem only needs Grade 3 multiplication and the "split into cases, then add" idea you already know!

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