A Surprising Chessboard Puzzle: Can You Solve It?

Mathematics is full of puzzles that challenge our intuition and reveal beautiful truths. Today, we explore a classic problem involving a chessboard and dominoes. At first glance, it seems straightforward, but there's a clever twist that makes it fascinating. This problem not only tests your problem-solving skills but also introduces a powerful mathematical concept in a visual and intuitive way.

The Problem

Imagine a standard 8x8 chessboard, which has 64 squares in total. Now, suppose we remove two squares from the board: the top-left corner and the bottom-right corner. These are diagonally opposite corners. After removing these two squares, we are left with 62 squares.

Next, consider dominoes, each of which can cover exactly two adjacent squares on the chessboard (either horizontally or vertically adjacent, not diagonally).

The question is: Can you cover the entire remaining chessboard (62 squares) with 31 dominoes, without any overlaps or gaps?

Take a moment to think about it. It might seem possible since 62 is an even number, and each domino covers two squares. But is there a way to arrange the dominoes to cover all 62 squares after removing those two corners?

Hint

To solve this, think about the colors of the squares on the chessboard. In a standard chessboard, squares are colored in an alternating pattern of black and white. Consider how the removed squares affect the balance of colors on the board.

Answer

Surprisingly, it is impossible to cover the remaining 62 squares with 31 dominoes after removing the two opposite corners.

Here's why:

First, recall that a standard chessboard has an 8x8 grid with squares colored alternately black and white. Specifically, the top-left corner (position (1,1)) is black, and the bottom-right corner (position (8,8)) is also black because both are positioned such that the sum of their row and column indices is even.

In general, a square at position (i,j) is black if i + j is even, and white if i + j is odd. Since both the top-left (1,1) and bottom-right (8,8) squares have even sums (1+1=2 and 8+8=16), they are both black.

Originally, the chessboard has 32 black squares and 32 white squares. By removing two black squares, we are left with 30 black squares and 32 white squares.

Now, each domino covers exactly two adjacent squares, which are always one black and one white because adjacent squares have different colors.

Therefore, each domino covers one black and one white square. To cover the entire board with dominoes, we would need an equal number of black and white squares because each domino pairs one black with one white.

However, after removing the two black squares, we have 30 black squares and 32 white squares. Since 30 ≠ 32, there are not enough black squares to pair with all the white squares.

Thus, it is impossible to cover the remaining 62 squares with 31 dominoes.

This problem illustrates the power of invariant-based reasoning in mathematics. The coloring of the chessboard provides an invariant that reveals the impossibility of the tiling, despite the total number of squares being even.

If you enjoyed this puzzle, try thinking about variations. For example, what if you remove two adjacent squares instead? Or what if the chessboard is of a different size? Exploring these questions can deepen your understanding of mathematical invariants and tiling problems.