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


  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.

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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:


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.