AMC 8 · 2025 · #11

Easy mode Grade 3

Problem

A $\textit{tetromino}$ is a shape made of four little squares joined along their edges. Picture five different tetrominoes, called $I$ , $O$ , $L$ , $T$ , and $S$ (the $I$ is a straight bar, the $O$ is a square, the $L$ looks like the letter L, the $T$ looks like the letter T, and the $S$ has a zig-zag step). You may rotate them or flip them over.

Now imagine a flat rectangle with $3$ rows and $4$ columns of small squares. We cover this rectangle completely using exactly three tetrominoes, with no gaps and no overlaps. At least one of the three pieces we used is an $S$ .

What are the other two pieces?

Pick an answer.

(A)

I and L

(B)

I and T

(C)

L and L

(D)

L and S

(E)

O and T

View mode:

Toolkit + CCSS Solution

Understand

Restated: A $3 \times 4$ rectangle is completely covered by three tetrominoes chosen from the five shapes $I, O, L, T, S$ (rotations and reflections allowed). One of the three tiles must be an $S$ tile. Which pair of tiles makes up the other two?

Givens: Board is a $3 \times 4$ rectangle (area $= 12$ unit squares); Three tetrominoes are used, each made of $4$ unit squares; Tile types allowed: $I, O, L, T, S$ (rotations and reflections allowed); Exactly one of the three tiles is an $S$ tile; Answer choices: (A) I and L, (B) I and T, (C) L and L, (D) L and S, (E) O and T

Unknowns: The pair of tetromino shapes that join the mandatory $S$ tile to fully cover the $3 \times 4$ board

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #3 Eliminate Possibilities, #10 Create a Physical Representation

Tiling problems live on a picture, so Tool #1 (Draw a Diagram) is the starting move: color the $3 \times 4$ board like a chessboard and count how many black/white squares each tetromino must cover. That single picture turns a hard geometry question into easy parity arithmetic. Tool #3 (Eliminate Possibilities) then knocks out the choices that violate the black/white balance — this is the classic AMC multiple-choice move. Finally, Tool #10 (Physical Representation) finishes the job: with only three options left, cut out paper tetrominoes (or shade cells on graph paper) and actually try to place an $S$ together with the candidates. The combination that fits is the answer.

Execute — Answer: C

#1 Draw a Diagram 3.MD.C.6 Step 1

Color the $3 \times 4$ board like a chessboard.
Because $3 \times 4 = 12$ and the colors alternate, there are exactly $6$ black squares and $6$ white squares on the board.

$$3 \times 4 = 12, \quad 12 \div 2 = 6 \text{ black} + 6 \text{ white}$$

💡 Counting unit squares of each color on a grid is exactly the Grade 3 "measure area by counting unit squares" skill.

#1 Draw a Diagram 1.G.A.2 Step 2

Check the black/white footprint of each tetromino on a chessboard.
Place each shape mentally and count: an $I$, $O$, $L$, or $S$ tile always covers $2$ black and $2$ white squares no matter how it is rotated or flipped.
A $T$ tile is the odd one out — its three-in-a-row body and one foot in the middle always cover $3$ of one color and $1$ of the other.

$$\text{I, O, L, S} \to 2B+2W \quad ; \quad \text{T} \to 3B+1W \text{ or } 1B+3W$$

💡 Composing tetrominoes from $4$ unit squares and reading off their color counts is the Grade 1 "compose 2D shapes" idea.

#3 Eliminate Possibilities 2.OA.C.3 Step 3

Use the color totals to rule out choices.
The mandatory $S$ tile uses $2$ black and $2$ white, leaving $4$ black and $4$ white for the other two tiles.
If one of them is a $T$, that $T$ uses $3$ of one color and $1$ of the other, so the third tile would need to use $1$ of the first color and $3$ of the second — which forces the third tile to also be a $T$.
So a $T$ can only appear paired with another $T$.
Choices (B) I and T and (E) O and T mix a $T$ with a non-$T$, so they break the color balance and are eliminated.

$$\text{Remaining after } S: 6-2=4 \text{ B}, \; 6-2=4 \text{ W}$$

💡 Checking that 'black count = white count' is even-vs-odd / balance reasoning — a Grade 2 odd-or-even style argument.

#10 Create a Physical Representation 1.G.A.2 Step 4

Three choices remain: (A) I and L, (C) L and L, (D) L and S.
Try each by cutting paper tetrominoes (or shading squares on graph paper).
For (A), the $I$ tile is $1 \times 4$, so in a $3 \times 4$ board it must occupy a full row; the leftover is a $2 \times 4$ strip that has to be tiled by one $L$ and one $S$, and no placement of an $L$ in a $2 \times 4$ strip leaves an $S$-shaped hole — every leftover is itself another $L$, never an $S$.
For (D) with $S, S, L$, every placement of the first $S$ either creates an isolated square or leaves a region that cannot be split into a second $S$ plus an $L$.
Both (A) and (D) fail.

$$\text{(A) fails: } I \text{ forces a full row} \Rightarrow 2 \times 4 \text{ strip cannot be } L + S$$

💡 Physically arranging the small shapes to compose the rectangle is hands-on Grade 1 shape composition.

#10 Create a Physical Representation 1.G.A.2 Step 5

Now place choice (C): one $S$ tile plus two $L$ tiles.
Put the $S$ (in its reflected form) so it covers the cells $(1,2), (2,2), (2,1), (3,1)$ in (column, row) coordinates from the bottom-left.
The eight uncovered cells split cleanly into two $L$-shaped pieces: the left/top region $(1,1), (1,3), (2,3), (3,3)$ is one rotated $L$, and the right region $(3,2), (4,1), (4,2), (4,3)$ is another rotated $L$.
A valid tiling exists, so the partners of the $S$ tile are two $L$ tiles, giving choice $\textbf{(C)}$.

$$S \text{ at } \{(1,2),(2,2),(2,1),(3,1)\} \;+\; L \text{ at } \{(1,1),(1,3),(2,3),(3,3)\} \;+\; L \text{ at } \{(3,2),(4,1),(4,2),(4,3)\} \;\Rightarrow\; \textbf{(C)}$$

💡 Building the $3 \times 4$ rectangle out of three pre-cut tetrominoes is exactly the kindergarten/Grade 1 "compose a larger shape from smaller shapes" idea.

[1] #1 3.MD.C.6 Color the $3 \times 4$ board like a chessboard. Because $3 \times 4 = 12$ and th

[2] #1 1.G.A.2 Check the black/white footprint of each tetromino on a chessboard. Place each sh

[3] #3 2.OA.C.3 Use the color totals to rule out choices. The mandatory $S$ tile uses $2$ black

[4] #10 1.G.A.2 Three choices remain: (A) I and L, (C) L and L, (D) L and S. Try each by cutting

[5] #10 1.G.A.2 Now place choice (C): one $S$ tile plus two $L$ tiles. Put the $S$ (in its refle

Review

Reasonableness: Three tiles of $4$ squares cover $12$ squares, which matches the $3 \times 4 = 12$ area of the board — the area arithmetic checks out. The chessboard parity ($6$ black, $6$ white) is preserved by the $S$ tile ($2+2$) and by the two $L$ tiles ($2+2$ each), summing to $6+6$. The explicit placement of $S$, $L$, $L$ given above tiles the rectangle with no overlaps and no holes (one can verify by shading the cells on graph paper), so the answer (C) L and L is both internally consistent and constructively demonstrated.

Alternative: Tool #2 (Make a Systematic List): list every essentially different placement of the $S$ tile inside the $3 \times 4$ board (there are only a handful up to symmetry). For each placement, look at the $8$ uncovered cells and ask 'can this leftover be split into two tetrominoes?' This brute-force enumeration is slower but doesn't require any clever coloring — it directly proves that the only successful pairing is two $L$ tiles.

CCSS standards used (min grade 3)

3.MD.C.6 Measure areas by counting unit squares (Counting that the $3 \times 4$ board has $12$ unit squares which split into $6$ black and $6$ white squares under a chessboard coloring.)
1.G.A.2 Compose two-dimensional shapes or three-dimensional shapes (Composing each tetromino from $4$ unit squares to read its black/white footprint, and composing the $3 \times 4$ rectangle from one $S$ plus two $L$ tetrominoes to demonstrate the tiling.)
2.OA.C.3 Determine whether a group of objects has an odd or even number (Using even-vs-odd balance ($4$ black $=$ $4$ white remaining after the $S$ tile) to eliminate choices that pair a $T$ tile with a non-$T$ tile.)

⭐ This AMC 8 problem only needs Grade 3 unit-square counting (and a clever chessboard picture) that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 3)

More like this