SOLUTIONS FOR HOMEWORK 3
2.40. (a)
Suppose, for sake of contradiction, that the chessboard with corners removed (CWCR, for
short) is covered by dominoes. The number of dominoes required equals 62
/
2 = 31 (each domino
covers 2 squares, out of the 62 available).
Moreover, each domino covers one black square, and
one white square, hence the CWCR must contain 31 squares of each color. However, the original
chessboard had 8
×
8
/
2 = 32 squares of each type, and the two squares removed are of the same
color (white on the picture in your textbook).
Thus, CWCR has 32 black and 30 white squares
left. This is a contradiction. Therefore, our assumption (that dominoes can cover CWCR) is false.
(b)
Now we consider a “crippled chessboard”, with four squares removed (two in each of the two
opposite corners). This new board contains 60 squares – 30 white and 30 black. Suppose, for the
sake of contradiction, that this “crippled chessboard” is covered by “Tshapes”.
As each of the
Tshapes covers four squares, the total number of them equals 60
/
4 = 15.
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 Fall '09
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 Math, Logic, Logical connective, white squares

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