You have a chessboard and 32 dominoes. Each domino is of such size that it exactly covers two different adjacent squares on the board. The 32 dominoes therefore can cover all 64 squares of the chessboard.

But now suppose we cut off two squares at diagonally opposite corners of the chessboard and discard one of the dominoes. Is it possible to place the 31 dominoes on the board so that all remaining 62 squares are covered? If so, show how it can be done. If not, prove it impossible.

## Question

## Guest

You have a chessboard and 32 dominoes. Each domino is of such size that it exactly covers two different adjacent squares on the board. The 32 dominoes therefore can cover all 64 squares of the chessboard.

But now suppose we cut off two squares at diagonally opposite corners of the chessboard and discard one of the dominoes. Is it possible to place the 31 dominoes on the board so that all remaining 62 squares are covered? If so, show how it can be done. If not, prove it impossible.

## Link to post

## Share on other sites

## 1 answer to this question

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.