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Round the square (Posted on 2004-07-30) Difficulty: 3 of 5
Can you place the numbers 1 to 8 on the outer squares of a 3x3 grid, so on every side the middle number is the sum of the corners next to it?

Can you manage to place the same numbers so the middle number of each side is the average of the corners next to it?

See The Solution Submitted by Federico Kereki    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 5

Yes, the numbers can be placed as follows:

6   8   2
7        5
1   4   3

This can be determined simply by trial and error, given that the numbers 7 and 8 must be placed in the middle without fail. The remaining possibilities become more and more limited, giving rise to a solution.

For the second part, recognize that 8 must be on a corner since no two numbers can sum to 16. Similarly, since no two numbers can sum to 2, 1 must be on a corner. 8 and 1 must then be on opposite corners, since an odd number will give you an x.5 solution for the average which violates the rules of the problem. To give integer values, 8 requires an even number on its adjacent corners, and 1 requires an odd number on its adjacent corners. Since this is impossible, part two has no solution. In other words, no, you can't.

 

Edited on July 30, 2004, 9:12 am
  Posted by Eric on 2004-07-30 09:11:51

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