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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.

enter image description here

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.

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    $\begingroup$ 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? $\endgroup$ – Ross Millikan Apr 19 '17 at 0:14
  • $\begingroup$ It's own writing problem and I sent it for sharing. (I didn't stuck). $\endgroup$ – scarface Apr 19 '17 at 0:18
  • $\begingroup$ 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. $\endgroup$ – Joel Reyes Noche Apr 19 '17 at 2:35
  • $\begingroup$ No, I didn't say anywhere that number of tetrominoes is equal to the number of trominoes. They must be different. $\endgroup$ – scarface Apr 19 '17 at 11:39
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    $\begingroup$ 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. $\endgroup$ – hardmath Apr 19 '17 at 16:23
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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:

tilings

Both in the tilings and in the positive integer equations, the idea is that we can replace three tetrominoes by four trominoes while still covering the same total area.

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  • $\begingroup$ Nice explantion! +1 $\endgroup$ – scarface Apr 19 '17 at 13:39

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