Question 24

For tiling to be possible, the total area of the region must be a multiple of the area of the tile. If the tile is a domino, we also check the number of black and white squares.

Step 1 — Analyze Tiling 1

Let us find the area of the region to be tiled. The region is a $3 \times 3$ square with one square removed. The area of a $3 \times 3$ square is $3 \times 3$ squares.

$3 \times 3 = 9 \text{ squares}$

One square is removed from this $3 \times 3$ square. So, the area of the region is $9 - 1$ squares.

$9 - 1 = 8 \text{ squares}$

Let us find the area of the tile. The tile is an L-shaped piece made of 3 squares. So, the area of the tile is 3 squares.

For tiling to be possible, the area of the region must be perfectly divisible by the area of the tile. We divide the area of the region by the area of the tile.

$8 \div 3$

$= 2 \text{ with a remainder of } 2$

Since 8 is not perfectly divisible by 3, the tiling is not possible.

$\boxed{\text{Tiling 1 is not possible}}$

Step 2 — Analyze Tiling 2: Area Check

Let us find the area of the region to be tiled. The region is a $7 \times 8$ rectangle with two squares removed. The area of a $7 \times 8$ rectangle is $7 \times 8$ squares.

$7 \times 8 = 56 \text{ squares}$

Two squares are removed from this $7 \times 8$ rectangle. So, the area of the region is $56 - 2$ squares.

$56 - 2 = 54 \text{ squares}$

Let us find the area of the tile. The tile is a $1 \times 2$ rectangle, also called a domino. So, the area of the tile is 2 squares.

For tiling to be possible, the area of the region must be perfectly divisible by the area of the tile. We divide the area of the region by the area of the tile.

$54 \div 2$

$= 27$

Since 54 is perfectly divisible by 2, the area condition is met. This means tiling might be possible.

Step 3 — Analyze Tiling 2: Chessboard Coloring

Let us imagine coloring the region like a chessboard. We color each square alternately black and white. Each $1 \times 2$ tile (domino) will always cover exactly one black square and one white square. So, for tiling to be possible, the region must have an equal number of black and white squares.

Let us count the number of black and white squares in the region. A full $7 \times 8$ chessboard has $56$ squares. Since $56$ is an even number, a full $7 \times 8$ chessboard has an equal number of black and white squares. It has $56 \div 2 = 28$ white squares and $28$ black squares.

The region has two squares removed. Let us assume the top-left square is white. The removed squares are the top-rightmost square and the bottom-leftmost square. The top-rightmost square is at position (row 1, column 8). Its color is black (since $1+8=9$, which is odd). The bottom-leftmost square is at position (row 7, column 1). Its color is white (since $7+1=8$, which is even).

So, from the $28$ white and $28$ black squares of the full board, we remove one white square and one black square. Number of white squares remaining: $28 - 1 = 27$. Number of black squares remaining: $28 - 1 = 27$.

The region has 27 white squares and 27 black squares. Since the number of white squares equals the number of black squares, the chessboard coloring condition is met. Both conditions (area and coloring) are met. This means the tiling is possible.

$\boxed{\text{Tiling 2 is possible}}$

Answer

(i) No (ii) Yes