A conjecture on tiling Here's the problem:
An $L-tile$ is the one which looks like this, and covers $3$ square units:
A A
A
Now we will define a term called an $L_k-good$ rectangles where $k$ is a nonnegative integer. First of all, for two positive integers $m,n$ let $p_{mn}$ be the remainder when $mn$ is divided by $3$. Now, we call an $m*n$ rectangle $L_k-good$ if the following statement is true:
For any possible choice of $p_{mn}+3k$ mutually non-adjacent unit squares(non-adjacent means that it doesn't even share a vertex) from the $mn$ squares of the rectangle, if we remove those $p_{mn}+3k$ squares, the remaining grid becomes completely tilable with only using $L-tiles$. This property holds true for all choices satisfying the "mutual non-adjacency" condition, only then the $m*n$ rectangle is $L_k-good$. So, here are my conjectures:
1)For any $k$, there exist finite positive integers $A,B$ such that for any integers $m,n$ such that $m>A,n>B$, the rectangle $m*n$ is $L_k-good$. As an example, for $k=0$, we can show $A=7,B=7$ to work.
2)For any $k$, $A=B=8+12k$ is a pair for which the first part holds.
I tried this for quite a while, got no progress. This seems quite hard. Can anyone help me with this conjecture?
Edit: I got some progress last year.
Parts $1)$ and $2)$ are equivalent (and they imply a more general version of the conjecture). For any specific value of $k$, the conjecture is equivalent to a finite computation, namely checking whether the conjecture holds for all such $m,n$ with $12k+8<m,n \leq 24k+8$. As mentioned earlier in a comment, $k=0$ case works. Some weaker versions of the conjecture are true, e.g.:
$1)$ Allowing tiling using two kinds of tiles: $L-tile$ and the $3*1$ rectangle instead of just the $L-tile$.
$2)$ Using only the $L-tile$ as in the original conjecture, but instead of the unit squares being non-adjacent, we require that no $4*4$ square contains two or more of the unit squares which are removed.
.
 A: I'm pretty sure the first part of your conjecture is more or less true.  
I'm approaching the problem slightly differently. Let's consider rectangles $R(m, n)$ with 0, 1, or 2 holes, such that the area is always divisible by three. The question is, for such potentially holy rectangles, whether they are tileable by right trominoes.  


*

*For rectangles with 0 holes, they are all tileable by trominoes except for $R(3, k)$ for odd $k$. For rectangles of the form $R(3m, 2n)$, you can split the rectangle into $3\times 2$ rectangles, that can be tiled by trominoes. For rectangles of the form $R(6m, 2n + 1)$, you can split the rectangle into $R(6, 2n + 1)$, which can in turn be split into $R(6, 3)$ and $R(6, 2n - 2)$, both tileable by right trominoes. 

*For 1 hole, it is a bit more complicated, but the reason is similar. The idea is to break a rectangle up into some primitives which are tileable. $R(m, n)$ with one hole is tileable for $m, n \geq 6$, and the hole in any position. This can be derived from the "DEFICIENT RECTANGLE THEOREM" in Tiling Deficient Rectangles with Trominoes (PDF).

*For 2 holes, it seems the problem is still open. There are some partial results in Tromino tilings of domino-deficient rectangles. (Your requirement for the holes to be non-adjacent rule out many impossible situations, but at least for small rectangles, not all.) I'm pretty sure if you use the same type of reasoning you can show for large enough rectangles with two holes in any permitted position, you can always find a tiling. 
For the second part, the papers give better limits than your guess:


*

*For 0 holes, tilings always exist for $R(m, n)$ if $m, n > 3$.

*For 1 holes, tilings always exist for $R(m, n)$ if $m, n > 5$.

*For 2 holes, the problem seems to be open. 

