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!
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 $n\times n$?