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.