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Wall Street Quant Β· Puzzles & Problems Β· question 46 of 155

A standard 8x8 chessboard has two diagonally opposite corners removed. Is it possible to cover the remaining 62 squares with 31 dominos, each covering exactly two adjacent squares?

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The problem you present is not within the scope of mathematical finance but rather belongs to combinatorics. However, as an experienced mathematical professional, I would be happy to help you understand the solution to this problem.

Let us color the chessboard like an actual chessboard; using black and white squares. Without loss of generality, assume that the top-left corner of the chessboard is colored white and the bottom-right corner is also white. Now, let’s observe the parity of the colors of the chessboard considering that we removed the two diagonally opposite corners, i.e., we start at a square, and after moving one square to the right, we switch the color of the square.

For an 8x8 chessboard, we have 64 squares with two diagonally opposite corners removed, so we have 64β€…βˆ’β€…2 = 62 remaining squares in total. Out of these 62 squares, 32 are black and 30 are white.

Observe that when we place a domino on the chessboard, it will always cover one white and one black square. This is due to the positioning of the colors: moving to an adjacent square horizontally or vertically will result in a change in color.

We have 31 dominos to cover the 62 remaining squares. Let W be the number of white squares covered by the dominos and B be the number of black squares covered by the dominos. Since there are 31 dominos and each covers 2 squares, we know that Wβ€…+β€…B = 62. We also know that B = Wβ€…+β€…2 since we have 32 black and 30 white squares on the board.

However, even though the total number of squares is 62, we cannot lay 31 dominos in such a way that we cover exactly 30 white and 32 black squares, as each domino will cover equal numbers of black and white squares. In other words, each domino will cover one black and one white square, so the number of white squares they cover should be equal to the number of black squares they cover.

In this case, we have

W = Bβ€…βˆ’β€…2

But each domino covers equal numbers of black and white squares, so we should have

W = B

These two equalities contradict each other, thus it is impossible to cover the remaining 62 squares of the chessboard with 31 dominos.

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