Chessboard problem. Olympiad question.

Place the integers (without duplication) on an chessboard, with one integer per square. Show that there exist two adjacent entries such that their difference is at least . (Adjacent means horizontally, vertically or diagonally adjacent.)

I tried using "extreme principle" but that doesn't lead me anywhere. Any ideas?

Re: Chessboard problem. Olympiad question.

Re: Chessboard problem. Olympiad question.

Quote:

Originally Posted by

**awkward** Assume the contrary, so there is a numbering of the squares in which the difference between any two adjacent entries is at most

. Suppose we move a king from one square to another. (Kings in chess move one square horizontally, vertically, or diagonally.) The number on the square he moves to increases by at most

from the previous square, so in

moves, the entries increase by at most

. Chessplayers (of the

variety, at least) know that the king can move from any square on the board to any other square in at most

moves.

So what does this say about a king moving from square

to square

?

A proof from "The Book" ... as Erdos would put it.

Thank you so much!

Re: Chessboard problem. Olympiad question.

Quote:

Originally Posted by

**abhishekkgp** Place the integers

(without duplication) on an

chessboard, with one integer per square. Show that there exist two adjacent entries such that their difference is at least

. (Adjacent means horizontally, vertically or diagonally adjacent.)

I tried using "extreme principle" but that doesn't lead me anywhere. Any ideas?

Is this really an Olympiad question, abhishekkgp???? The answer is so straight forward!!

Re: Chessboard problem. Olympiad question.

Quote:

Originally Posted by

**godelproof** Is this really an Olympiad question, abhishekkgp???? The answer is so straight forward!!

No. I don't think this is an Olympiad question. But its in an Olympiad training book. Its supposed to be among the harder ones.