## Dominoes on a Checkerboard

Suppose we have a Checkerboard and we discard two diagonally opposed corners, leaving a total of 62 squares. If a domino covers two squares, is it possible to completely cover the surface of the trimmed board with 31 dominoes?

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 Suppose (1,1) and (8,8) are the discarded corners. Consider the column 1. Since there are an odd number of available squares (7) , there will be a odd number of horizontal dominoes from column 1 to column 2. Being so, there will be an odd number of available squares at column 2. Again, the same applies to the column 2, and so on, up to column 7. So, along 7 columns, an odd number of dominoes are necessary from each column to the next. So, the total number of horizontal dominoes is odd.S Now, consider the rows. Starting from row 1, with only 7 squares, the same argument applies, and there will be necessary an odd number of vertical dominoes. Then, the total number of dominoes (vertical + horizontal) is even although there are exactly 31 dominoes. So, it is not possible!
 Good work, also another proof: Since two diagonally opposite squares are of the same color, it leaves 30 squares of a color and 32 of another. Since a domino only covers two squares of opposite color, only 15 dominoes at most can be fitted on the board.

## Dominoes on a Checkerboard

 Quote by Werg22 Good work, also another proof: Since two diagonally opposite squares are of the same color, it leaves 30 squares of a color and 32 of another. Since a domino only covers two squares of opposite color, only 15 dominoes at most can be fitted on the board.
...considerably shorter!

 I said 15, meant to say 30. Wrote that late at night.