Question

Prove that a 15 x 8 board cannot be covered by 2 L-tetrominoes and 28 skew tetrominoes using coloring.

Answer 1

First colour the 15×8 board with vertical stripes of black starting from the second column, just like shown below in the given pic..

Then there are 7×8=56 black squares in total.

Now if we place skew-tetrominoes in any way, one skew-tetrominoes will cover 2 black and 2 white squares. Since there are 28 skew-tetrominoes, so there will be 28×2=56 black squares covered.

But we still have 2 L-tetrominoes. Each of which, placed in any way, covers 3 or 1 black squares. Which can't be done because we've seen all 56 black squares are already covered before.

Hence a 15×8 board can't be covered by 2 L-tetrominoes & 28 skew-tetrominoes.