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