Imagine an 8x8 chessboard has two opposite corners taken out, with 62 squares left over. Is it possible to lay 31 dominoes with the size 2x1 so they cover all of the empty, left over, squares?
This is the mutilated chess board problem - I apologise for ruining this cool puzzle which a quick answer, but I'm afraid it's well-known:
The puzzle is impossible to complete. A domino placed on the chessboard will always cover one white square and one black square. Therefore a collection of dominoes placed on the board will cover an equal numbers of squares of each colour. If the two white corners are removed from the board then 30 white squares and 32 black squares remain to be covered by dominoes, so this is impossible. If the two black corners are removed instead, then 32 white squares and 30 black squares remain, so it is again impossible.