Competition · AMC preparation · step 4 of 4

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

📘 View easy version →

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)

(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

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

Count the squares each tile covers

Area splits the job: the board is 3 × 7 = 21 squares and big tiles cover 4 each, so the 1 × 1 count c is 21 minus a multiple of 4.

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

2STEP 2

Divide 21 by 4 to fix the remainder

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 one 1×1 tile

Try c=1 (guess & check): 5 big tiles would cover 20 squares, but no layout leaves just one hole — the next step’s colouring shows why.

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

4STEP 4

Color the columns to rule it out

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

Coloring the diagram turns the impossibility argument into a Grade 4 multi-step word problem about how many more red squares than blue squares are left over.

The 3 × 7 board cannot be covered using only a single 1 × 1 tile.

▸ Why?

Paint the 7 columns alternately red and blue starting with red; this gives 4 red columns and 3 blue columns, so the board holds exactly 3 more red squares than blue squares.

▸ Why?

Each column contains 3 squares, so the 4 red columns hold 4 × 3 = 12 red squares and the 3 blue columns hold 3 × 3 = 9 blue squares, and 12 - 9 = 3.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

Every 2 × 2 tile and every 1 × 4 tile covers equally many red and blue squares, so laying big tiles never changes the surplus of 3 red squares over blue squares.

▸ Why?

A 2 × 2 tile spans 2 adjacent columns and a horizontal 1 × 4 tile spans 4 adjacent columns, and any block covering an even run of neighboring columns takes equal numbers of red and blue columns.

▸ Why?

Because the columns run red, blue, red, blue, an even number of neighboring columns splits exactly into red-blue pairs — one red beside one blue in each pair — so the reds and the blues come out equal in number.

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?

Since each big tile shifts the red-minus-blue count by 0, adding any number of big tiles leaves that difference exactly at 3.

Additive identity (zero)Adding zero adds nothing, so the number stays exactly the same.

▸ Why?

The 1 × 1 tiles alone must therefore supply the entire surplus of 3, but a single 1 × 1 tile can raise red over blue by at most 1, which is short of 3.

▸ Why?

One tile covers one square of one color, so from a single 1 × 1 tile the count placed on red minus the count placed on blue can be at most the one tile itself, never 3.

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.A.3Draw A Diagram

5STEP 5

Build a layout with five

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 the rest.

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

Take the smallest that works

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

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!

Count the squares each tile covers
Divide 21 by 4 to fix the remainder
Try one 1×1 tile
Color the columns to rule it out
Build a layout with five
Take the smallest that works

A parent dashboard for the family lives at sensimlab.com.

Count the squares each tile covers

Divide 21 by 4 to fix the remainder

Try one 1×1 tile

Color the columns to rule it out

Build a layout with five

Take the smallest that works

More like this