Due Friday, April 19, 2024 at 3 pm submitted to this Google form. You must be logged into your NAU gmail to submit via this form.

This problem is related to the problem from last week. Last week, we considered an 8 by 8 grid where two diagonally opposite corner squares are removed. It is impossible to cover the 62 remaining squares with rectangles size 2 by 1?

There is an elegant way to prove this. If we think of the grid as a checkerboard, then the two corners will be the same color. Every rectangle will cover a black and a white square. When you have covered 60 squares with 30 rectangles, you will be left with two squares of the same color, which cannot be adjacent.

This week: Take an 8 by 8 checkerboard and remove two squares of opposite color. It is always possible to cover the remaining 62 squares with 2 by 1 rectangles. Make the argument why this is always possible.