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Let $ABCD$ be a square in which length of a side is $10$ meters. Suppose that we have $T$-shape brick which consists of $4$ smaller squares in which a side of each smaller square has length of $1$ meter. Can $ABCD$ be entirely covered by 25 $T$-shape bricks?

I have tried but can not figure out where to start. Please give me some hints, not the full solution!

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    $\begingroup$ Use colouring proofs. $\endgroup$ – Anik Bhowmick Aug 6 '18 at 7:12
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    $\begingroup$ If the grid is coloured like a checkerboard, $25$ T-shapes can never cover the same number of black squares as red. $\endgroup$ – Robert Israel Aug 6 '18 at 7:20
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    $\begingroup$ Closely related (but not a dupe!) $\endgroup$ – Jyrki Lahtonen Aug 6 '18 at 7:36
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Think to the given square as a $10\times 10$ chessboard with alternate black and white squares and assume that such covering with $T$ pieces exists. Then each of the $25$ $T$ pieces will have 1) $3$ black squares and $1$ white square or 2) $3$ white squares and $1$ black square. Let $b$ the number of $T$ pieces of the first category and $w$ the number of $T$ pieces of the second category. Then the integers $b$ and $w$ should satisfy the following equations: $$3b+w=\frac{10\cdot 10}{2},\quad b+3w=\frac{10\cdot 10}{2}.$$ What may we conclude?

Bonus question: What happens when the given square is $n\times n$?

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  • $\begingroup$ Do you know the answer for the $m$-by-$n$ rectangular case? It suffices that both $m$ and $n$ are divisible by $4$, but I cannot prove that this is necessary. I know that one of $m$ and $n$ is divisible by $4$ (or better, just as you showed, $8\mid mn$). $\endgroup$ – Batominovski Aug 6 '18 at 8:14
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    $\begingroup$ @Batominovski Yes, both $m$ and $n$ are divisible by 4 is also necessary. See jstor.org/stable/pdf/2313337.pdf?seq=1#page_scan_tab_contents $\endgroup$ – Robert Z Aug 6 '18 at 8:30

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