[Guest]

You have a regular 8x8 chess board. Two squares on the board are "neighbors" if they share a vertex or an edge. So, the four corner squares have 3 neighbors, any other square on the boundary edge has five, and all other squares have eight neighbors.

Your task is to fill the squares with numbers (any real) such that, the number on each square is the average of the numbers in its neighbors.

a. (easy) Give any one solution that will fit the requirement

b. (little hard) Find all possible solutions that will fit the requirement

[Guest]

entering the same number into each will work

[Guest]

entering the same number into each will work

Agreed.. that is the first part. Are there any other possible solutions?

[bushindo]

Agreed.. that is the first part. Are there any other possible solutions?

Well, I tried writing a matrix, but logic is so much easier.

Imagine the cell values for the chess board for such a configuration that satisfies the original post. There must be a maximum value for the entire 64 cells. Select the cell with this maximum value, call it M. By definition of maximum, since its neighbors' values can not exceed M's value, all the neighbors must be equal to M in order for M to be the average of its neighbors. Extend this argument for the entire chessboard and we'll see that all cells have to be equal.

[Guest]

Well, I tried writing a matrix, but logic is so much easier.

Imagine the cell values for the chess board for such a configuration that satisfies the original post. There must be a maximum value for the entire 64 cells. Select the cell with this maximum value, call it M. By definition of maximum, since its neighbors' values can not exceed M's value, all the neighbors must be equal to M in order for M to be the average of its neighbors. Extend this argument for the entire chessboard and we'll see that all cells have to be equal.

Great.

I'm hoping my next puzzle will not be cracked so fast.