A $60\times 60$ square $S$ is divided into $2\times 5$ rectangles. Prove that no matter how this is done there is a way to divide $S$ into $1\times 3$ rectangles so that each of the above $2\times 5$ rectangles contains one of the $1\times 3$ rectangles.
A bashy idea (which I cannot actually complete) is to show that we can always remove one-by-one connected figures of $30$ squares, composed of three $2\times 5$ rectangles, so that the resulting figure remains connected; and then check that each such figure of $30$ squares can be tiled with a $1\times 3$ rectangles.
Any idea of something smarter? (Or if not, at least a fast way to complete the details of the above strategy?) Any help appreciated!