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Infinite 3x3 Magic Square

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[This story is based on old recollections of what Arye Amitai told me over 35 years ago. While some details may be presented inaccurately, the specific mathematics details, which are of interest, are not.]

Sometime in the 1950's an Israeli periodical that was devoted to mathematics published a contest:

What numbers should be entered into a 3-by-3 magic square such that, such that the most cell combinations, when are added up, their sum equals 15? In other words, the winning solution would be that which provides the highest count of unique cell combinations that, when the cells are totaled, the sum equals 15.

[I do not recall the precise text and rules of the contest and I don't know anyone who does.]

The standard true 3´ 3 magic square is

6
1
8
7
5
3
2
9
4

It uses the numbers 1 through 9 and all the rows, columns and two long diagonals add up to 15. Therefore, this magic square has 8 that equal 15.

A trivial solution for this contest is

5
5
5
5
5
5
5
5
5

It uses only the number 5 but in it any combination of three cells sums up to 15 for the grand total of 84 such sums. Since

Amitai told me his solution. When he found out that he lost the contest and that the winning solution resulted with fewer sums than his own, he protested to the editor. When his protest was rejected on the ground his solution was not valid, he sued and during the trial a well-known mathematician testified that Amitai's solution was indeed valid.
 

 Show Amitai's solution
 
 




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Last Update: Apr. 28, 2002