In the above picture, there are 24 squares. Can you only use L trominos to fill the figure? If yes, give an example. Otherwise, please explain why.
An L tromino is like this:
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No, it's not possible.
This is not a popular way of approaching this, and I personally love @Magma's proof, yet I have to disagree.
The problem as stated is perfectly solveable, since nowhere in the rules it states, that you are limited by the borders of the 5x5 quare. So if you are to leave some hangover beyond the bounding box of the original shape, you are easily able to cover the original square, but you will have these ugly hangover parts left. I do know that this is not a very elagant solution, but it is a valid one. But I do agree that the problem as it is (probably) meant to be solved is impossible.
I was going to post a different answer, only to realize it was the same as Magma's. So I had to find a new one:
If one of the squares in a tromino is at a corner, it being the tromino's middle square is the better option, because otherwise it would always force another tromino to close the gap, forming a 3x2 rectangle combined with the first (1s and 2s together):
With the corner being in the middle, this option is also available, but not necessarily the only one. Hovever, even this advantage isn't enough to reach our goal: