AMC 8 · 2007 · #11

Grade 4 logic

Archetype Small-Domain Constrained Search

logical-deductionsystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: logical-deduction

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

Tiles $I, II, III$ and $IV$ are translated so one tile coincides with each of the rectangles $A, B, C$ and $D$ . In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle $C$ ?

$\mathrm{(A)}\ I \qquad \mathrm{(B)}\ II \qquad \mathrm{(C)}\ III \qquad \mathrm{(D)}\ IV \qquad \mathrm{(E)}$ cannot be determined

Pick an answer.

(A)

(B)

(C)

III

(D)

(E)

cannot be determined

View mode:

Toolkit + CCSS Solution

Understand

Restated: Four square tiles I, II, III, IV each have a number on each of their four sides. They are slid (without rotation or reflection) into a $2 \times 2$ grid of rectangles A (top-left), B (top-right), C (bottom-left), D (bottom-right). Wherever two tiles touch, the two numbers on the shared edge must match. Which tile lands in rectangle C?

Givens: Tile I — top $8$, right $9$, bottom $7$, left $3$; Tile II — top $6$, right $3$, bottom $2$, left $4$; Tile III — top $7$, right $5$, bottom $0$, left $1$; Tile IV — top $2$, right $1$, bottom $6$, left $9$; Grid layout: A top-left, B top-right, C bottom-left, D bottom-right; Tiles are translated only (no rotation), so each tile's top/right/bottom/left stays fixed; Answer choices: (A) I, (B) II, (C) III, (D) IV, (E) cannot be determined

Unknowns: Which of the four tiles ends up in rectangle C

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #13 Count Systematically

Tool #2 (Make a Systematic List) lines up all $16$ tile-edge numbers in a table so nothing is missed. Tool #13 (Count Systematically) then tallies how often each value $0$-$9$ appears. The key observation: any number on an interior edge of the $2 \times 2$ grid is shared by two tiles, so its value must appear at least twice across the four tiles. A value that appears only once cannot lie on an interior edge — it must be on the outer boundary. That uniqueness forces one tile into a specific corner, and the remaining three are pinned down by matching the shared edges.

Execute — Answer: D

#2 Make a Systematic List 3.MD.B.3 Step 1

List every edge value for every tile in a table.

$$\begin{array}{c|cccc} & \text{top} & \text{right} & \text{bottom} & \text{left} \\ \hline \text{I} & 8 & 9 & 7 & 3 \\ \text{II} & 6 & 3 & 2 & 4 \\ \text{III} & 7 & 5 & 0 & 1 \\ \text{IV} & 2 & 1 & 6 & 9 \end{array}$$

💡 A clean table makes it easy to scan for repeats and one-offs without losing track.

#13 Count Systematically 3.MD.B.3 Step 2

Count how often each digit appears across all $16$ edges.
The digits $0$ and $5$ each appear exactly once — both on tile III ($0$ on its bottom, $5$ on its right).

Frequency: $0{:}1,\; 1{:}2,\; 2{:}2,\; 3{:}2,\; 4{:}1,\; 5{:}1,\; 6{:}2,\; 7{:}2,\; 8{:}1,\; 9{:}2$

💡 An interior edge needs the same number on two touching sides, so any value used on an interior edge must appear at least twice.

#13 Count Systematically 4.OA.C.5 Step 3

Place tile III.
The grid has four interior edge-sides: the bottom side of A, top of C, bottom of B, top of D (sharing the horizontal middle line) and the right side of A, left of B, right of C, left of D (sharing the vertical middle line).
Tile III's bottom ($0$) and right ($5$) are both unique, so neither can lie on an interior edge — both must face the outer boundary.
The only spot where a tile's bottom and right both face outward is the bottom-right corner.

$$\text{III} \to \text{D}$$

💡 Bottom-right is the only corner whose bottom and right sides are both on the outside of the $2 \times 2$ grid.

#2 Make a Systematic List 4.OA.C.5 Step 4

Match tile III's interior sides.
Its top is $7$ and its left is $1$.
The tile in B (above D) must have bottom $= 7$, and the tile in C (left of D) must have right $= 1$.
Scanning the table: tile I has bottom $7$, so I goes in B; tile IV has right $1$, so IV goes in C.

$$\text{I} \to \text{B}, \quad \text{IV} \to \text{C}$$

💡 Only one tile carries each needed value, so each placement is forced.

#2 Make a Systematic List 4.OA.C.5 Step 5

Check the remaining edges.
Tile II is the only tile left, so it goes in A.
The remaining shared edges must also match: A-B share I's left ($3$) with II's right ($3$); A-C share II's bottom ($2$) with IV's top ($2$).
Both match.

$$\text{II} \to \text{A};\quad 3 = 3,\; 2 = 2 \;\checkmark$$

💡 Once three tiles are placed and every shared edge checks out, the last tile fits automatically.

#2 Make a Systematic List 4.OA.C.5 Step 6

Read off the answer.

$$\text{C contains tile IV} \;\Rightarrow\; \textbf{(D)}$$

💡 Tile IV is the one with right $= 1$, the value that matches tile III's left in column 2.

[1] #2 3.MD.B.3 List every edge value for every tile in a table.

[2] #13 3.MD.B.3 Count how often each digit appears across all $16$ edges. The digits $0$ and $5$

[3] #13 4.OA.C.5 Place tile III. The grid has four interior edge-sides: the bottom side of A, top

[4] #2 4.OA.C.5 Match tile III's interior sides. Its top is $7$ and its left is $1$. The tile in

[5] #2 4.OA.C.5 Check the remaining edges. Tile II is the only tile left, so it goes in A. The r

[6] #2 4.OA.C.5 Read off the answer.

Review

Reasonableness: Double-check every shared edge in the final layout $\begin{smallmatrix} \text{II} & \text{I} \\ \text{IV} & \text{III} \end{smallmatrix}$: A-B (II's right $3$ = I's left $3$) $\checkmark$; C-D (IV's right $1$ = III's left $1$) $\checkmark$; A-C (II's bottom $2$ = IV's top $2$) $\checkmark$; B-D (I's bottom $7$ = III's top $7$) $\checkmark$. All four interior edges match. The unique values $0, 4, 5, 8$ all end up on the outer boundary, exactly as the uniqueness argument predicted.

Alternative: Tool #6 (Guess and Check): try placing tile I in C. Then C's right ($9$) must equal D's left, so D's tile needs left $= 9$ — only tile IV qualifies. But then C's top ($8$) must equal A's bottom, and no remaining tile has bottom $= 8$. Dead end. Trying tile II in C: II's right is $3$, so D needs left $= 3$ — no tile has left $3$. Trying tile III in C: III's right is $5$, so D needs left $= 5$ — no tile has left $5$. The only survivor is tile IV in C, confirming (D).

CCSS standards used (min grade 4)

3.MD.B.3 Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories (Organizing each tile's four edge values into a table and tallying how many times each digit appears — Grade 3 data-organization work.)
4.OA.C.5 Generate a number or shape pattern that follows a given rule (Following the matching-edge rule to chain placements: III $\to$ D forces I $\to$ B and IV $\to$ C, and then II $\to$ A is the only leftover.)

⭐ Tally the digits on every tile edge. The ones that show up only once cannot match anything, so they must face the outside — and that single observation pins the puzzle down.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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