You are given 21 3x1 rectangular pieces to cover an 8x8 chessboard. Since the board has 64 squares, which square on the chessboard must you cut out so that the 21 given pieces exactly cover the remaining 63 squares? Or it is impossible, no matter which square you remove?
Each row must have at least two tiles that are placed vertically; since horizontally placed tiles consume 3 squares at a time. So, satisfy this requirement by placing 2 vertical tiles in the upper right corner, two immediately below that, and 2 in the lower left corner, like so:
0 0 0 0 0 0 x x
0 0 0 0 0 0 x x
0 0 0 0 0 0 x x
0 0 0 0 0 0 x x
0 0 0 0 0 0 x x
x x 0 0 0 0 x x
x x 0 0 0 0 0 0
x x 0 0 0 0 0 0
Now it's easy to fill the remaining spaces with horizontal tiles covering every square but either the 3rd or 6th square in in the 6th row. So, cut out either of those two squares and you have a solution.
The square to be removed is c3 (or equivalent square, by rotation or reflection). Charlie, in his post, has an elegant proof that this is the only solution
A solution
