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Math challange!
Asked by: aanatoliev
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You have a chess board. It has 64 squares. Now, you also have 32 dominoes. Each domino covers two squares. You can place them horizontally or vertically. Now, the question. I will take away one domino and cut two opposite corners from chess board (let's say top right corner and bottom left). You will end up with 31 dominoes and 62 squares. Is it possible to cover all squares without breaking dominoes? If yes, explain how, if not - tell me why.
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The mathematical insight is simply that a domino covers one white square and one black square on the checkerboard, and if we remove two blacks, as you did, (or two whites for that matter) we cannot put the 31 dominoes down to cover the remaining 62 squares.
Answer Date: 08:42pm 07/10/08
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The 64 squares can be covered with 32 dominos, but if we remove two of the squares we might or might not be able to cover the remaining 62 squares with 31 dominos. There are 64 X 63 / 2 = 2016 different ways we can remove two squares. With half of those ways we can cover them with the 31 dominos, and with the other half we cannot.
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Answer Date: 05:53pm 07/10/08
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