Suppose you have two types of tetrominoes made of four squares each; the two on the left are called “S-type”, and the two on the right are called “Z-type”.
Consider a white chess board of infinite size, on which each square is the same size as a square of the tetrominoes. We pick some squares on the board and paint them black. Assume that we can use only S-type tetrominoes to cover all black squares without covering any white one. Prove that, if we use S and (or) Z-type tetrominoes to cover all black squares instead, without covering any white one, then we must use an even number of Z-type tetrominoes.
, 2015
Solution
Let be the set of black squares. Color the infinite chess board in red and green as follows: notice that an S-type tetromino must cover an even number of red squares, while a Z-type tetromino must cover an odd number of red squares.
Since can be formed by S-type tetrominoes, must contain an even number of red squares. Therefore, if we instead use S and (or) Z-type tetrominoes, the number of Z-type tetrominoes must be even. Q.E.D.
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