AMC 8 · 2007 · #11

Easy mode Grade 4

Problem

Picture four rectangular tiles labeled $I, II, III,$ and $IV$ . Each tile has a number written on each of its four sides.

We are going to slide each tile into one of four spots, labeled $A, B, C,$ and $D$ . The spots sit next to each other in a $2 \times 2$ arrangement, so some tiles will share a side with another tile.

The rule: wherever two tiles touch along a side, the numbers written on that shared side must match.

(The figure below shows the four tiles on the left and the four spots on the right.)

Which tile ends up in spot $C$ ?

$\mathrm{(A)}\ I \qquad \mathrm{(B)}\ II \qquad \mathrm{(C)}\ III \qquad \mathrm{(D)}\ IV \qquad \mathrm{(E)}$ 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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