# Is it possible to cover with dominos a chess-board with opposite corners removed?

This question was in my course book but without an answer.

From a chessboard, two boxes each located on an opposite corner, get cut away. So for example the most top-left and bottom-right box get cut away. Is it possible to cover the whole chess-board with Domino-stone's, when an half of the Domino equals the size of one box on the chess-board (1 domino-stone covers two boxes).

My solution (math): Possible

Amount of boxes on this chessboard? 64 - 2 = 62 (Two cut away)

Amount of Domino's needed to cover the whole board: 31 (62 / 2)

Amount of horizontal boxes: 7 + 7 + (8 * 6) - 14

=> #horizontalFirstRow + #horizontalLastRow + #allOtherBoxes - #verticalBoxes = 48

How many Domino's fit on 48 horizontal boxes? 48/2 = 24

31 - 24 = 7 (7 Domino's left)

We have 14 vertical boxes, so 14/2 = 7 (Used the 7 left domino's), so 0 domino's left and all boxes covered.

My solution (graphical): Impossible You will have to zig-zag the domino-stones, but in the 7th row you won't be able to do the zig-zag because the bottom-right corner is cut-away.

Question: Is it possible or not? And how to prove that mathematically?

• It's always worth trying it with a smaller example - take a $4\times 4$ board instead, and remove the top left and bottom right, then try to cover it. Nov 1, 2017 at 17:29
• This is a classing problem. Search for covering chessboard dominoes Nov 1, 2017 at 17:30