infinite chessboard average!

Probability Level 4

The n 2 n^2 squares of an n × n n \times n chessboard are filled with positive integers in such a way that each integer is the average of the integers on the neighbouring squares (sharing common edge and vertex). Let k k be the number of all such distinct configurations. What is the value of lim ⁡ n → ∞ k n \displaystyle \lim_{n \to \infty} \frac{k}{n} ?


The answer is 1.

Mehul Kumar by Mehul Kumar , 23, India

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1 solution

Mehul Kumar
Jun 7, 2016

Take the MINIMUM number among all such positive integer entries. Since its the average of its neighbours which are greater than or equal to it ⟹ \implies the neighbouring numbers has to be the same as this MINIMUM number. So eventually all numbers will become same. Therefore k=n since the numbers are positive integers. ∴ lim ⁡ n → + ∞ k n \therefore \lim_{n \to +\infty} \frac{k}{n} = 1

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