We all know that a chess board has 64 squares. This can
be completely covered by 32 cardboard rectangles, each cardboard covering just
2 squares.
Supposing we remove 2 squares of the chess board at
diagonally opposite comers, can we cover the modified board with 31 rectangles?
If it can be done, how can we do it? And if it cannot be done, prove it
impossible.
Answer: No it cannot be done.
Each rectangle covers one white square and one black square,
because on a chess board the white and black squares are always adjacent.
The two squares which we remove from the chess board are of
the same colour, and so the remaining board has two more boxes of one colour
than the other. And after the rectangles have covered 60 boxes, there will be
left two squares of the same colour. Obviously the remaining rectangle cannot
cover these two squares.
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