Fun With 0.3

(centered so the 1's work out. Meh.)

We sort of skipped over problem 0.3 and I thought given our use of the well-ordering principle it would be fun to go over this one. 0.3 goes like so:

On an infinite chess board each square is labeled with a positive integer and each such label is the average of the four labels above, below, left, & right of it. One square is labeled 17. What is the sum of the square two above and two to the right of it?

Way 1: Well, a board of all 17s works and the problem DOES say what "IS the sum" not "what ARE the sumS." So 34.

Way 2: OK, Way 1 is evil, let's be fair. We want to know if any other combination of numbers is possible (other than a board filled with all 17s). Consider any board where the conditions are met. Now, by the well-ordering principle, pick the smallest # out, let's call it x.

To make everything simpler, let's subtract (x - 1) from all the squares. This still leads to a valid board as all the averages are shifted down (x - 1) too. (e.g. the average of (4 - 2), (5 - 2), (6 - 2), and (5 - 2) is two less than the average of 4,5,6, and 5) Now our smallest number is 1.

So 1 is somewhere on the board:

1

But what numbers can surround 1? Well, no numbers below one (the board is all positive; 1 is the smallest #). So it must be numbers >= 1. But if we have a number LARGER than 1 surrounding 1, we must have a number SMALLER than 1 by it. Oh wait, we just said we couldn't have a number smaller than 1. So 1 must be surrounded by all 1s:

1
111
1

But those 1s must be surrounded by 1s...

1
111
11111
111
1

... etc. So our board becomes all 1's, or, translating back to our original board, all x's (the smallest # on the board). But since there's a 17 all the numbers must be 17.

Hope that was educational, see you all tomorrow.