AMC 8 · 2024 · #7

Grade 4 geometry-2d
Archetype Grid Tiling / Packing with Divisibility └ Sub-archetype Rectangular Tiling — Minimize Filler Count
area-rectanglesparitymodular-arithmetic modular-arithmeticparity-coloring ↑ Prerequisites: area-rectanglesdivisibility-rules
📏 Long solution 💡 4 insights 📊 Diagram
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Problem
A 3 × 7 rectangle (21 unit squares) must be covered, with no overlap and no gaps, using only three tile shapes: 2× 2, 1× 4, and 1× 1. Choose the smallest possible number of 1× 1 tiles among (A) 1, (B) 2, (C) 3, (D) 4, (E) 5.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5

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

How to solve
Strategy Draw a Diagram

Both the rectangle and the tiles live on a grid, so Tool #1 (draw a diagram) of the 3× 7 board and physically placing tiles into it is the most natural attack. Split the task into two subproblems (Tool #7): (i) For which counts c of 1× 1 tiles do the areas even add up to 21? (ii) For those candidate counts, can a tiling actually be built? Part (i) is a Grade 4 'divide 21 by 4 and look at the remainder' question. Part (ii) is solved by Tool #6 (guess & check) — try placing tiles directly on the diagram. Since the candidates are only c=1 and c=5, Tool #2 (make a systematic list) just checks those two cases. No algebra (#13) needed.

1STEP 1

Grade-4 area splits the job: the board is 3× 7 = 21 squares, big tiles cover multiples of 4, so the 1× 1 count is 21 − 4×(big tiles).

3 × 7 = 21 = 4 × (number of big tiles) + c
2STEP 2

Divide: 21 ÷ 4 = 5 remainder 1, so c must be 1, 5, 9, …. Of choices A–E only 1 and 5 qualify, so the answer is 1 or 5.

21 ÷ 4 = 5 remainder 1 → c ∈ {1, 5, 9, 13, …}
3STEP 3

Try c=1 on the grid (guess & check): 5 big tiles must cover 20 squares, but a 3-tall board always strands a 2× 1 strip, not a single hole.

c = 1 → 4× 5 + 1 = 21, need 5 big tiles to cover 20 squares
4STEP 4

Color columns R,B,R,…: 12 red, 9 blue. Every big tile covers 2 red + 2 blue, so it never touches the gap of 3; c=1 can't supply it.

Red = 12, Blue = 9, Red - Blue = 3; big tiles do not change this difference
5STEP 5

Now try c=5: three 2× 2 tiles fill the top 2× 6 block, one 1× 4 the bottom-left, and five 1× 1s fill column 7 and the last two bottom cells.

A & A & C & C & D & D & s ; A & A & C & C & D & D & s ; B & B & B & B & s & s & s → c = 5
6STEP 6

So c is 1, 5, 9, …; c=1 is killed by the color gap and c=5 is built above, so the minimum is 5 — choice (E).

c_min = 5 → (E) 5
Answer
5
c=5 being the answer requires two facts to fit together: (1) 21 - 5 = 16 = 4× 4, so 4 big tiles can cover the other 16 squares; and (2) the smaller candidate c=1 cannot work because the red-minus-blue gap of 3 in our coloring can never be closed by big tiles alone. Fact (1) was shown by an explicit drawing (three 2× 2 tiles filling the top 2× 6 area, one 1× 4 filling the bottom-left 1× 4 area). Fact (2) was shown by the column-coloring on the same diagram. Both diagram facts agree, so the answer (E) 5 is consistent.
💡Key takeaway

This AMC 8 problem only needs Grade 4 rectangle area and multiples-with-remainders thinking you already know!

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