Try to fill the total board (10x10-2) with bricks of size 2 (), so no overlaps, no gaps, and no bricks crossing the borders.

First Question: Is this possible? (Proof!)

Second Question: How many squares are present in the picture of the board?

First Answer: Imagine the board to be colored like a chess-board. Each brick will cover a white and a black square of the board, so the number of bricks equals the number of white squares equals the number of black squares. But by removing the two opposite corners of the chess-board (with an even numbers of rows and columns), two squares of the same color are removed, so there is an imbalance of white and black squares.

Conclusion: No, you can not fill the board.

Second Answer: Here is an overview of the number of squares of each size, as they are present in the figure:

9x9: 22 - 2 = 2

8x8: 32 - 2 = 7

7x7: 42 - 2 = 14

6x6: 52 - 2 = 23

5x5: 62 - 2 = 34

4x4: 72 - 2 = 47

3x3: 82 - 2 = 62

2x2: 92 - 2 = 79

1x1: 102 - 2 = 98

----------------- +

Total: 366 squares.

## Monday, July 28, 2008

### Placing Bricks

Posted by Covert Bay at 1:08 AM

Labels: Riddles/Puzzles

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