Prove that a 4 x 11 rectangle cannot be tiled by Lshaped tetrominoes.
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I have no idea how to start the proof. – Tyler Lewandowski Jul 25 '19 at 04:29

I can't resist linking to [this (mildly embarrassing for me) old thread](https://math.stackexchange.com/q/80246/11619). – Jyrki Lahtonen Jul 25 '19 at 04:48

But the question is in need of a bit of context as outlined in [our guide fore new askers](https://math.meta.stackexchange.com/q/9959/11619) :( – Jyrki Lahtonen Jul 25 '19 at 04:50
1 Answers
I am considering a grid with $4$ column and $11$ rows. Now, color the first and the third row with black color and the second and fourth with white. Clearly, whenever you place a $L$Shaped domino there, it will cover either $(\hbox{3 black position and 1 white})$ or $(\hbox{1 black and 3 white})$. However, the number of black and white position is same in the hole grid. Thus, $L$Shaped domino must exist in pairs. To be more precise, for every domino covering $3$ black and $1$ white space, there must be another domino covering $1$ black and $3$ white positions. Thus, total number of $L$Shaped domino must be even. But each $L$Shaped domino has $4$ blocks in it. Thus total number of blocks that an even number of dominoes can cover will be a multiple of $4\times 2=8$. But total number of blocks in the grid is $44$ which is not a multiple of $8$. Hence, a $4\times 11$ board cannot be covered with $L$shaped dominoes.
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