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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 corners, 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.

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Solution:
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 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 rectangles cannot cover these two squares.