Can a 4x11 rectangular grid of square blocks be covered (having no overlaps), with multiple copies of the tetromino (including rotations, but not reflections) as shown below?
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If so, give an example. If not, provide a proof that this cannot be achieved.
Note: Each of the 44 square blocks of the 4x11 grid has the same shape and size as each of the 4 square blocks of the tetromino.
(In reply to
Divisibility by Brian Smith)
So the idea behind this problem popped back into my mind, and I was able to recreate a proof I thought of years ago.
Let a rectangle be tiled by L-tetrominos. At least one of the dimensions must be even, so we can say the rectangle has dimensions 2X by Y.
Now apply a stripe parity with the stripes running parallel to the Y side. Call the stripes black and white. As an example the 4x11 rectangle in the original problem colored according to this stripe parity (X=2 and Y=11):
BBBBBBBBBBB
WWWWWWWWWWW
BBBBBBBBBBB
WWWWWWWWWWW
Then there are X black stripes and X white stripes, so the rectangle has an equal number of black and white cells. Any placement of an L-tetromino must have an unequal covering of either 3 black + 1 white cells or 1 black + 3 white cells.
But then for the two totals of cells to be equal there must be an equal number of each coloring of L-tetrominos, so then the number of L-tetrominos is even. But then the area of the rectangle must be divisibly by 8 to be covered by an even number of L-tetrominos, which each have an area of 4.
So then for the 4x11 rectangle in this problem the area is 44, which is not divisible by 8 so cannot be tiled by L-tetrominos.
re: Divisibility - Proof
