You have 15 3 × 1 rectangular tiles, plus one “L” shaped tile (made up of three 1 × 1 squares). You would like to use these tiles to cover the floor of your 7 × 7 dorm room. Of course, doing so will leave a 1×1 square missing, but you figure you will put a plant there to cover up the hole.
Prove that such a tiling is possible, but the missing square cannot be adjacent to the wall (i.e., on any side of the square).
First part, show a tiling is possible:
AAABCCC
DXXBEEE
DXXBFFF
DGGGHHH
IJKLMNO
IJKLMNO
IJKLMNO
A-O represent the 15 straight triominoes, and X a 2x2 square which the L-triomino can be put in any of its four orientations. Rotating the entire 7x7 layout shows 16 possible places for the uncovered square.
The second part is then to show these are the only possible places, which then implies the uncovered square cannot be along the perimeter.
Apply a stripe parity of three colors to the full 7x7 square:
ABCABCA
ABCABCA
ABCABCA
ABCABCA
ABCABCA
ABCABCA
ABCABCA
When a straight triomino is placed it either must cover one of each parity or all three of the same parity. When the L-triomino it must cover two tiles of one parity type and one tile of a second.
Now look at the difference between B and C. The differential created by placing the L-triomino cannot be compensated for by any number of straight triominoes.
If the single uncovered square is on an A tile then it is not possible to end up with B and C being the same. So the uncovered tile must be on B or C
Rotate the parity stripe 90 degrees so that it runs horizontally, then the same conclusion applies in that orientation as well. The union of the sets of A squares is then
AAAAAAA
ABBABBA
ABBABBA
AAAAAAA
ABBABBA
ABBABBA
AAAAAAA
This leaves the 16 placed marked as B to be the only places the uncovered square can occur and none of them are the perimeter squares along the walls of the dorm room.
Solution
