Proof that given 8x8 Mboard and 21 trominoes there will no overlap

Question

Given an 8x8 Mboard ( top left piece missing), as given below in the image, and 21 trominoes, how can we proof that there exists a configuration where there will be no overlap after all trominoes are placed.

What I tried doing was - a 8x8 Mboard will have 63 squares, now $63 / 21 = 3$, ie. it is divisble. But recall a tromino is L-shaped, so just show that the total number of squares is divisble with the total number of trominoes really isnt telling much. I mean how do you account for the shape of the tromino in that case.

I actually manually plotted out a configuration but could only get up 20 trominoes to fit. I have no idea how 21 fits in, and how we can say that there will be no overlap.

The red squares are the squares that were blank. There are 3 of them, but how can we rearrange the configuration to included 21 trominoes. And more than that how can we say that there will be no overlap?

Answer 1

     

Answer 2

Instead of filling in row-by-row, go layer-by-layer. First tackle the $2\times 2$ with one corner missing. This is trivial. Now tackle the $4\times 4$, by first covering the top $2\times 2$ the way you did.

Here is the kicker: The way you placed your trominoes in this last step makes them one large tromino! (They cover the $4\times4$ board except the top $2\times2$.) Now you can tackle the $8\times8$ by first taking the top $2\times2$, then the top $4\times4$, then take the rest of the board using your large tromino piece. This makes this last step identical to the $4\times 4$ covering, only scaled up.

Answer 3

This picture shows how you can build up the solution, exactly as Arthur describes in his answer.