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You have a chess board with the top left and bottom right corner removed. You have to place dominoes on the borad in such a way so that the board is fully covered with dominoes. Each domino covers 2 squares on the chess board. I also think u can not place the domino diagonnaly. However if u think u can place them diagonally explain how, otherwise explain your answer if u have another way. THNX!!

2007-03-11 03:30:38 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

Impossible - count the black and white squares on the dismembered board.

Each domino, no matter wher it is placed, must cover 1 white and one black square.

i.e. the number of black squares MUST equal the number of white squares.

As this does not hold once you have removed the corners, it is no longer possible to cover the board with non-overlapping dominoes.

2007-03-11 03:33:39 · answer #1 · answered by sumzrfun 3 · 2 0

This cannot be done because 2 corners are of same color so we are left with 30 white and 32 black or vice versa. 31 dominos have 31 white and 31 black as each shall cover one white and one black so we shall be left with 2 of the same color after 30 dominos which cannot be filled

2007-03-11 03:37:09 · answer #2 · answered by Mein Hoon Na 7 · 1 0

Each domino has to cover a white square and a black square. But you have removed two white squares so you're left with 30 white and 32 black squares. Therefore it's impossible.

2007-03-11 03:35:35 · answer #3 · answered by Anonymous · 2 0

Well since the chess board has 8 * 8 squares on it

That means there are 64 Squares right?

If you take one off the top row that leaves 7 so you'd have to go 6 across ways and one down ways, and sort of go left to right until you got to the bottom and then the last line you'd do the same as the top line.

2007-03-11 03:39:25 · answer #4 · answered by hey mickey you're so fine 3 · 0 2

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