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© Copyright 2000, Jim Loy
Let's say that a domino covers two adjacent squares of
a checker (or chess) board. On the left, we see such a domino, in black, on my
board. There are 64 squares on the board, and 32 dominoes can cover the entire
board. I have colored two opposite corner squares red. Can 31 dominoes cover
the rest of the board, leaving the two red squares uncovered? Think about this
before you look at the answer.
Answer: It is fairly well-known that it is impossible for the dominoes to cover the non-red squares. The proof is that each domino covers one white (buff colored) square and one black (green) square. So 31 dominoes would cover 31 white and 31 black squares, leaving 1 white and 1 black square uncovered. But the two red squares were originally both white. So the task is impossible.
This puzzle is interesting and educational. The solution has to do with mathematical parity. If we succeed in numbering the squares, so that 1 is next to 2, 2 is next to 3, 3 is next to 4, etc., then no even number is next to an even number. No odd number is next to an odd number. Such a situation concerns even and odd parity. The first square has odd parity, the second has even parity, etc. Do you think the sum of these numbers: 22, 16, 30, 17, 24, 6, 18 is even or odd? You may notice that they are all even numbers, except for 17. So the sum is odd. You don't have to add them up. If I replace 17 with 18, then the sum is even. In addition, an even number of odds becomes even, and an odd number of odds becomes odd. In our checker board puzzle, the sum of all of the numbers on the squares (1 through 64) is even (32 evens + 32 odds). Each domino covers one even and one odd number. 31 dominoes cover 31 evens and 31 odds, which sum to an odd number. So the two remaining red squares must sum to an odd number. But they are both odd (or even, if you numbered starting with a black square), and they sum to an even number. So it cannot be done.