Competition · AMC preparation · step 4 of 4

AMC 8 · 2007 · #11

Grade 4 logic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

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

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

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 × 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?

Pick an answer.

(A)

(B)

(C)

III

(D)

(E)

cannot be determined

AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

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 × 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.

1STEP 1

Table every tile edge

List every edge value for every tile in a table.

& top & right & bottom & left ; I & 8 & 9 & 7 & 3 ; II & 6 & 3 & 2 & 4 ; III & 7 & 5 & 0 & 1 ; IV & 2 & 1 & 6 & 9

2STEP 2

Count how often each digit appears

Tally each digit over all 16 edges: only 0 and 5 appear exactly once, both on tile III (0 bottom, 5 right).

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

3STEP 3

Place tile III

Unique digits can't sit on an interior edge, so III's 0 and 5 face out — only the bottom-right corner does that, so III goes in D.

III → D

Bottom-right is the only corner whose bottom and right sides are both on the outside of the 2 × 2 grid.

The tile carrying two numbers that appear only once in the whole set — a 0 on the bottom and a 5 on the right — is forced into the bottom-right corner of the grid.

▸ Why?

Every interior edge is shared by two tiles that must show the same number where they meet, so any number on an interior edge is written on two different tiles; but 0 and 5 each appear on just one tile, so they cannot lie on any interior edge and must face the outer boundary instead.

▸ Why?

Each interior edge writes its number onto both touching tiles at once, so every interior appearance of a value is bought in a pair and the count of a value's interior appearances is always even; but 0 and 5 each show up an odd number of times in the whole set — exactly once — and an odd count cannot be spent entirely on even pairs, so at least one appearance is forced off the interior and onto the outer boundary.

Even and odd (parity)A whole number is even when it splits into two equal whole groups (a factor of 2), else odd; odd×odd stays odd and even×anything is even.

▸ Why?

Both of those once-only numbers keep pointing the way they started — 0 downward and 5 to the right — because the tiles are slid into place without being turned.

▸ Why?

Sliding a tile without rotating it sets it straight onto its own copy, so each side still faces the same direction it faced before the move.

Rigid motion preserves length and angleSliding, turning, or flipping a shape lays it exactly onto a copy, so every length and angle stays the same (folding is the flip case).

▸ Why?

Among the four cells of the 2 × 2 grid, only the bottom-right cell has both its bottom side and its right side on the outer boundary, so the tile whose bottom and right must both face outward can only go there.

▸ Why?

Sorting every cell's four sides into those that touch a neighbor and those that face outward accounts for all the sides with none left over, and this shows the bottom-right cell is the only one with both its bottom and right facing out.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

4.OA.C.5Convert To Algebra

4STEP 4

Match tile III's neighbors

III's top 7 needs a tile with bottom 7 above it (tile I → B), and its left 1 needs a tile with right 1 beside it (tile IV → C).

I → B, IV → C

5STEP 5

Check the remaining edges

Only tile II remains, so II → A; its shared edges match too (3 = 3 with B, 2 = 2 with C).

II → A; 3 = 3, 2 = 2 ✓

6STEP 6

Read off the answer

Read off the answer.

C contains tile IV → (D)

Answer

Double-check every shared edge in the final layout smallmatrix II & I ; IV & III smallmatrix: A-B (II's right 3 = I's left 3) ✓; C-D (IV's right 1 = III's left 1) ✓; A-C (II's bottom 2 = IV's top 2) ✓; B-D (I's bottom 7 = III's top 7) ✓. 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.

💡Key takeaway

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.

Table every tile edge
Count how often each digit appears
Place tile III
Match tile III's neighbors
Check the remaining edges
Read off the answer

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Table every tile edge

Count how often each digit appears

Place tile III

Match tile III's neighbors

Check the remaining edges

Read off the answer

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