I know!
[tab]No
By my calculations, no it’s not. I’ve seen a similar problem before, and the logic works like this:
First of all, let’s call these ‘2x1 tiles’ dominos, and let’s assume that the dimensions of the dominos are exactly 2x1 relative to the size of each equally sized tile on the chessboard.
Also, we’re assuming that the only way to place them is for each to fully cover 2 tiles, of course, and the only way to do that is to align them with the chess grid (no partial covers, no diagonal pieces, etc).
Now, notice this: whichever way you place the domino, such that it fully covers two squares and is perfectly aligned with the grid, it covers 1 black and 1 white square. You cannot place it anywhere where it covers 2 squares fully, it’s not diagonal, and the squares are of the same color.
So if you think about it, every domino must be covering 1 black and 1 white square, so the number of blacks must be equal to the number of whites covered.
Now think back to the problem statement, in which opposite corners of the chess board were removed. You start out with a chessboard of 64 squares, 32 black and 32 white. Equal black and white. When you remove 2 corners that are opposite each other, you remove 2 corners of the same color. This means that you’re left with 30black and 32white, or 32black and 30 white. In either case, you’re left with an unequal number of black and white squares.
Since each domino, to be placed correctly, must be covering 1 black and 1 white square, you can only fully cover 60 of the squares with even the best strategy, as the remaining 2 will be of the same color.[/tab]