True Color

let z = no. of Z-tile used, v = no. of V-tile used.

obviously, we have: $4z+3v=9^2 \implies v=27-\frac{4}{3}z \quad —— \ (1)$ next, we color the 9 by 9 board as shown: notice that Z-tiles always cover exactly 1 dark tile, V-tiles cover either 0 or 1 dark tile. since there are 25 dark tiles, we will need at least 25 Z or V tiles. hence $z+v\geq 25 \quad —— \ (2)$ subs (1) into (2), simplify, we have: $z+27-\frac{4}{3}z\geq 25 \implies z\leq 6$ we prove that more than 6 is impossible, but that doesnt mean 6 is the maximum (yet).

to complete the proof, we have to construct an example by trial and error. fortunately we can use the dark tiles as a guide to improve our chance.

from (1), if $z=6, \ v=19, \ z+v=25$ , this means every Z and V-tile must cover exactly one dark tile. within reasonable time, we can fit them together. $\therefore z=6$

5x9 tiles have maximum 0 (Z type) tetraminoes and 2x9 tiles have maximum 3 (Z type) tetraminoes.

Both rectangular tiles are considered in combination of L type triominoes with Z type tetraminoes.

Combining together {5x9+2x9+2x9}we get 9x9 tiling with 19 (L type) triominoes and 6 (Z type) tetraminoes as the answer.