If we remove the square in the upper right corner of a $8\times 8$ chessboard.
The question is: Is it possible to cover the entire remaining area, with $1\times 3$ chocolate bars? (they can be laid on the chessboard vertically and horizontally aswell)
For the answer: We know that there will be $63$ squares left on the chessboard, of which $32$ are white and $31$ are black, since the one in the upper right corner was black. $63$ is divisible by $3$, so it could be possible to cover the area, but that doesn't prove that an arrangement exists to do it.
I'm not sure how to continue and am seeking help.
Thanks.
Best Answer
[Copied from When chessboards meet dominoes .]
This problem appears in Solomon Golomb's book Polyominoes. The solution involves coloring the chessboard in three colors in alternating diagonal stripes:
This colors the $64$ squares of the chessboard with $21$ green squares, $21$ yellow squares, and $22$ blue squares. Each $1\times $3 rectangle must cover exactly one square of each color. The deleted square therefore cannot be any of the green or yellow ones, nor any of the squares equivalent to one of these under a rotation or reflection of the chessboard:
(This is four copies of the first diagram, superimposed, with suitable rotations.)
This eliminates all but $4$ squares from consideration, namely the four bright blue ones in the previous diagram. So the only solutions involve deleting one of these four blue squares. In particular, the corner square won't do.
Showing that the chessboard can be tiled if one of the four special squares is deleted is then an easy exercise.