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 largest available digits are 6, 7, 8 and 9; and so these numbers are placed at the four corners as follows:
8 _ 9
_ _ _
6 _ 7
After a little trial and error, it is observed that the largest possible sum with a solution occurs at 18.
Thus, value at cell 21 = 18-8-6 =12; cell 23 = 18- 16=2;
cell 12 = 18 - 17 = 1; cell 32 = 18-6-7= 5. The structure at this
point are as follows:
8 1 9
4 _ 2
6 5 7
The only remaining number is 3, which will correspond to cell 22.
Consequently, the required square as completed is:
8 1 9
4 3 2
6 5 7
Puzzle Resolution
