# Checkerboard: Use proof by contradiction

• Feb 14th 2007, 09:14 PM
chillerbros17
ee
• Feb 15th 2007, 12:01 AM
CaptainBlack
Quote:

Originally Posted by chillerbros17
Attachment 1730

The problem is in the attachment.

Suppose it can be made from the given tiles, then there will be 15 of them.
This is because there are 60 small squares on the board to be covered, 30
light and 30 dark.

So there must be an even number of one type of tiles and an odd number of
the other.

Now the tile type that is used an even number of times is covering an even
number of light squares and an even number of dark squares.

Also the tile type that is used an odd number of times is covering an odd
number of light squares and an odd number of dark squares (as the two
tile types both have an odd number of light and dark squares in them).

So the tiling covers an odd number of light squares and an odd number of
dark squares, which as the board actualy has an even number of both
is a contradiction. Hence no such tiling exists.

RonL
• Feb 15th 2007, 09:18 AM
Soroban
Hello, chillerbros17!

There are 30 black squares and 30 white squares to be covered.

Each type-A tile covers 1 black and 3 white squares.
Each type-B tile covers 3 black and 1 white squares.

A pair (AB) covers 4 black and 4 white squares.

Using seven pairs of tiles, we can cover 28 black and 28 white squares.

But there will be 2 black and 2 white squares remaining to be covered.
It is impossible to cover them with either a type-A or a type-B tiles.