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Ok, so define a tromino as a $1$x$3$ tile. If a corner is removed from an $8$x$8$ board, is it possible to tile it with these trominoes? So far, I have concluded that you would need $21$ trominoes to tile the remaining $63$ squares. If you color the squares three different colors, it is possible to have an equal number of each. I alternated the diagonals for each color. Theoretically, a tiling should be possible right? I have been trying to find such a tiling, but each time I end up with three squares that cannot fit the shape of the given tromino. Are there any possible tilings? Thanks.

RXY15
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  • Try coloring with the diagonals in the other direction. – arkeet Oct 16 '14 at 00:53
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    If you color the board by diagonals, you get $22$ of the color used for the squares adjacent to the missing corner, $21$ of the next color, and $20$ of the third. – Brian M. Scott Oct 16 '14 at 00:56
  • I colored it in such a way that you have $21$,$21$,$22$ of each color and then removing one square from the color with $22$ squares gives $21$ squares to each color. – RXY15 Oct 16 '14 at 00:57
  • Ok so according to the link given from @MateusSampaio, there does not exist a tiling. Thanks. – RXY15 Oct 16 '14 at 01:00

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