Prove that for a m$\times$n rectangle, if this rectangle can be covered completely by trominoes of the shape indicated in the picture, then mn is divisible by 3.
I came up with a tentative way to prove the statement above. There are two problems that are bothering me, first, I am not sure if I've covered all the cases in my proof; second, I am guessing there will be a more elegant way to prove this. I'd appreciate if you can help!
Assume that 3 divides m. Suppose that n is even, we can partition the rectangle into 3$\times$2 blocks(the following graph indicates how this would be done.) Suppose n is odd, clearly we can't have n=1. If m is even, the first three columns of the rectangle may be partitioned into 2$\times$3 blocks while the remaining can be partitioned into 3$\times$2 blocks. Supppose m is odd, obviously, we can't have m=3, or n =3, since we can only pack a 3$\times$n rectangle by partitioning it into 3$\times$2 blocks. But that requires n to be even. If m$\geq$9 and n$\geq$5, the first five columns of the first nine rows consist of a 9$\times$5 block(indicated below).
The remaining columns of these rows are partitioned into 3$\times$2 blocks. The remaining rows can be covered if and only if 3 divides mn. Provided that both m and b are greater than 1, and neither is equal to 3 when the other is odd.
The tromino's area is a multiple of 3, so if there's a covering by them then the total area is a multiple of the tromino's area and hence a multiple of 3. Read the problem carefully.
This is an example of what's called a necessary but not sufficient condition. If the rectangle is to be tilable then we have to have 3 divides $mn$, but that fact alone won't guarantee the rectangle is tilable.