I solved a puzzle about proving that if a rectangular board can be covered by L-Tetrominoes then the number of squares must be a multiple of 8.
I based the solution on a colored board (like a chessboard) and 2 types of L-Tetrominoes
Either:
_____
|BLACK|
|WHITE|_______
|BLACK| WHITE|
--------------
or
_____
|WHITE|
|BLACK|_______
|WHITE|BLACK |
--------------
The follow up puzzle is:
Prove that an 8x8 board can not be covered with 1 O-Tetromino and 15 L-Tetrominoes.
The O-Tetromino is:
____ ____
| | |
|----+----|
|____|____|
I thought of the following proof:
If I use 1 O-Tetromino then there remains a board of 8x8-4 = 60 squares.
60 is not a multiple of 8 and we have already shown that for a board to be covered with L-Tetrominoes must be a multiple of 8. Therefore the 8x8 board can not be covered with 1 O-Tetromino and 15 L-Tetrominoes
Is this proof valid? It seems to informal also in relation to the approach I followed to prove that the L-Tetrominoes cover a board of x8
If it is not, then how can I prove it?
If you have stated the earlier conclusion correctly, the proof does not work because the board less the O tetromino is not rectangular.
A simpler proof that you need an even number of tetrominos to cover a rectangle comes from coloring the board with alternate black and white stripes instead of the traditional checkerboard. With that coloring, each L tetromino covers three squares of one color and one square of the other. An odd number of L tetrominos cannot color an equal number of black and white squares. Now once you note that the O tetromino covers two squares of each coloring, you can just use the count of squares instead of the shape of the board to conclude that $15$ L tetrominos cannot cover what is left.