An Olympiad Problem (tiling a rectangle with the L-tetromino)

Question

An L block that is 3 unit blocks high and 2 unit blocks wide . It is true that if an n by m rectangle can be covered by such L blocks with out overlap that we would require an even amount of L blocks, why?

Answer 1

This can be solved in one fell swoop, with just one coloring.

The total area of the rectangle is even (and even a multiple of 4). Then we can assume without loss of generality that the width is even. Now let us paint all cells in the first column black, in the second column - white, in the third - black, and so on. Clearly, exactly half of the cells will be black, and the other half white.

Notice that each L-block either covers 3 black cells and 1 white one, or 3 white cells and 1 black one. In the former case let's call this L-block mostly black, in the latter - mostly white. It is easy to see that the number of mostly black L-blocks must be the same as the number of mostly white L-blocks, because the total amounts of black and white cells are equal to each other. It follows that the total amount of L-blocks is even.

Answer 2

Apply a black-white chessboard colouring to the $n\times m$ rectangle, that is the colour of the square in the $i$th row and $j$th column is determined by $i+j\bmod 2$. Each of the $t$ tetrominoes covers two white and two black squares, hence if there are $t$ terominoes, there are $2t$ white and $2t$ black squares. Let $w$ be the number of tetrominoes having a white corner square and $b=t-w$ the number of thise with a black corner square.

If we assume that $t$ is odd, exactly one of $w,b$ must be odd. Assume wlog. that $b$ is odd. Repaint the white squares: If the square in the $i$th row and $j$th column is white then paint the squares red or blue according to th eparity of $i$ (or equivalently: according to the parity of $j$).

Now each of the $b$ black-corner tetrominoes covers two black, one red, and one blue square, whereas each of the $w$ originally-white-corner tetrominoes covers two black and either two red or two blue squares. Since $b$ is odd, the total number of red squares (and also of blue squares) is odd.

How many red and blue squares are there in an $n\times m$ rectangle? If $n=2k+1$ is odd (and accordingly $m=4l$ is a multiple of $4$) then there are $2lk$ squares of one of these two colours and $2lk+2l$ of the other - both numbers are even, contradiction. Similarly, $m$ cannot be odd. So $n=2k$ and $m=2l$ and hence there are exactly $kl$ red and blue squares each.

Undo the red-blue painting and now repaint the black squares instead: green or purple according to $i\bmod 2$. Then each of the $b$ tetrominoes covers zero or two green and zero or two purple squares whereas each of the $w$ tetrominoes covers one green and one purple square. Thus the number of green (or purple) squares is even just as $w$ is. But as alreeady seen, $n$ and $m$ are even. As the red, blue, purple, green colours repeat per $2\times 2$ block, the number of red, blue, purple, or green squares are all the same - contradiction.