Let the colors be 0, 1, and 2 if a block is exclusively in column k, then color it k (mod 3). If a block is in column k and column k+1, then count how many blocks above it are also in column k and k+1. If this number is even, then color it k (mod 3). If it's odd, color it (k+1) (mod 3). This coloring satisfies all properties in the problem.

No, this doesn't work (it already fails on a $ 6\times 6$ board). And if you claim that this coloring solves the problem, you have to prove that. Further note: the number of tiles that are in columns $ k$ and $ k + 1$ is always even (more generally on any board with even height).

Imagine the board as a chessboard. Each domino occupies a white and a black cell. If its white cell is (i,j) then color the domino with color $ (i - j) \mod3$. And that's all. This coloring will form a sort of diagonal strips. The only neighbors of a domino which have the same color are on the same diagonal strip, so there are at most two of them.

You can have three neighbours in the same diagonal strip: XxXxX xXbAx XxBaX cCsSx XxXxX Here the dominos are aA, bB, cC and sS (where capital letters are white fields), you see that the first three will be coloured the same but all touch sS.

ZetaX wrote: You can have three neighbours in the same diagonal strip: XxXxX xXbAx XxBaX cCsSx XxXxX Here the dominos are aA, bB, cC and sS (where capital letters are white fields), you see that the first three will be coloured the same but all touch sS. But Ss has a different color. the diagonal ABC is different mod 3 from the parallel one that contains S. Note that it says the same color as $ d$.

color the table with 1,2,3 diagonally. 1,2,3,1,2,3 2,3,1,2,3,1 3,1,2,3,1,2 1,2,3,1,2,3 now for each domino write the some of two numbers in the cells of it mod 3 and we are done

Color the table with red, green and blue like this, and color each domino with the same colored cell it covers. Done ?: ROGOBOROGOBO... OGOBOROGOBOR... GOBOROGOBORO... OBOROGOBOROG... BOROGOBOROGO... OROGOBOROGOB... ...............

Here is a simple solution:Color the dominos with one part in the first column blue the remaining ones with one part in the second column red the remaining ones with one part in third column green and continue this alnternatingly.It is easy to see this works.

wow amazing

Label each cell with $x+y \pmod{3}$. Color each domino with its missing label.