# Tiling $41$ unit squares with $L$ tetrominoes and $L$ trominoes

In the given figure, there are $41$ unit squares. We want tile to the figure with $L$-tetrominoes and $L$-trominoes. Determine all possible numbers of usable $L$-tetrominoes? Please, prove your answer wiht supporter figures.

Notes:

1. We can rotate and reflect some of $L$-tetrominoes and $L$-trominoes for the tiling operations. Also, number of $L$-tetrominoes and number of $L$-trominoes can be different.

2. Problem is mine and I have its solution. I have sent for sharing. I hope that you like it.

• Have you thought about how many ways there are to add up $3$s and $4$s to make $41$? I strongly suspect all of those are possible, so would just try to find a solution. Where are you stuck? – Ross Millikan Apr 19 '17 at 0:14
• It's own writing problem and I sent it for sharing. (I didn't stuck). – scarface Apr 19 '17 at 0:18
• Are you saying the number of tetrominoes is equal to the number of trominoes (both equal to $L$)? That would be impossible because 41 is not a multiple of 7. – Joel Reyes Noche Apr 19 '17 at 2:35
• No, I didn't say anywhere that number of tetrominoes is equal to the number of trominoes. They must be different. – scarface Apr 19 '17 at 11:39
• If you are not stuck then perhaps you should clarify what your Question is about. Have you found some numbers $m$ of $L$-tetrominoes that work? Perhaps you intend the Question not to help you learn (e.g. about the techniques for solving such tiling problems) but rather as a challenge to Readers. The Question may be off-topic if offered in that spirit here, but look at some previous posts at Puzzling.SE for comparison. – hardmath Apr 19 '17 at 16:23

If there are $p$ trominoes and $q$ tetrominoes, then $3p + 4q = 41$, which has $3$ solutions over the positive integers: $(p,q) \in \{(3,8), (7,5), (11, 2)\}$. Each of these is possible, as shown below: