The Mutilated Chessboard
Consider a regular chessboard and 32 dominoes. The size of the chessboard is such that one domino can cover 2 adjacent squares on the board. Therefore the 32 dominoes can cover all 64 of the chessboard squares.
Now, imagine that 2 diagonally opposite corners of the board are cut from the board and discarded. Is is possible to place 31 of the dominoes on the mutilated chessboard in a manner such that all of the remaining 62 squares are covered.
If so, explain; it not, prove it impossible.
Now, imagine that 2 diagonally opposite corners of the board are cut from the board and discarded. Is is possible to place 31 of the dominoes on the mutilated chessboard in a manner such that all of the remaining 62 squares are covered.
If so, explain; it not, prove it impossible.
posted by Zaux at
4:26 AM
1 Comments:
It's impossible
A chess board has 64 squares. 32 black 32 white. Opposite diagonals have the same color i.e. both black or both white
A domino has a black side and a white side. So for n dominos, n black and n white are covered. We can see that the #white covered must = the # black covered
So if we eliminate the 2 opposite diagonals we have:
30 black squares+32 white squares
OR
32 black squares+30 white squares
black and white are unequal thus the dominos cannot cover them
Cam
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