Fill a 3x3 square with the numbers from 1 to 9, so the four sides and the two diagonals share the same sum, which is to be as large as possible.
The numbers which have to be common & highest are the four corners since they are used in sides & diagonals. Hence 9, 8, 7 ,6 are at the corners. 9 has to be opposite 6 & 7 opposite 8 for balancing total.
Adding for diagonals & sides & solving we get
9 2 7
1 3 5
8 4 6
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Posted by salil
on 2006-02-14 03:42:26 |
solution belated
