Decide whether the integers $1,2,...,100$ can be arranged in cell $C(i,j)$ of $10×10$ $matrix$ (where $1 \le i,j \le 10$), such that following conditions are satisfied :
- In every row, the entries add up to the same sum $S$
- In every column, the entries also add up to this sum of $S$
- For every $k = 1,2,...,10$ the ten entries $C(i,j)$ with $i-j \equiv k(mod10)$ add up to $S$.
I have tried guessing numbers in the table. But I have given up on guessing. Am stuck so I need help from you guys, please help.
The Medgig method of constructing an even order magic square will do the trick. Construct a $5 \times 5$ square by the usual odd order shift up and right each cycle, then drop down one. Divide each cell into a $2 \times 2$ array and add $0,25,50,75$ to the original number in some repeating pattern. This gives a square that is magic on the rows, columns, and downward right diagonals.