Suppose I have a 6x6 board that I wish to tile with 1x3 vertical or horizontal trominoes. I want to prove the number of horizontal trominoes in any tiling must be divisible by 3. Intuitively, it makes sense since as soon as you place a horizontal 1x3 tromino, that takes up 1 space in that row and a vertical 1x3 tromino will always take 3 spaces so the remaining row spaces available will always be # horizontal trominoes mod 3. So if you place 1 horizontal 1x3 tromino, then you must place 2 others since the height of the board is 0 mod 3.
This all sounds kinda hand wavy. I'm wondering if I can turn this into a colouring proof or if there is a more systematic way of proving this fact.
Colour every third column. The total area of these coloured columns is a multiple of 3 since their height is a multiple of 3.
Every vertical tromino covers either 0 or 3 or the coloured squares, so the vertical trominoes cover a coloured area that is a multiple of 3.
The remaining coloured area is therefore also a multiple of 3. Each horizontal tromino covers exactly one coloured square, wherever it is located, so the number of horizontal trominoes is also a multiple of 3.