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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 :

  1. In every row, the entries add up to the same sum $S$
  2. In every column, the entries also add up to this sum of $S$
  3. 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.

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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.

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  • $\begingroup$ @RossMilikan Please could your explanation be more mathematical, because this is an Olympiad question. $\endgroup$ – Temidire Adesiji Oct 14 '17 at 14:00
  • $\begingroup$ A worked example for a $6 \times 6$ is in the Wikipedia page. If you look at it you can see that the diagonals work. Going through the example should allow you to prove that the square constructed here works. You can just construct it, exhibit the square, and show that all the required sums are correct. $\endgroup$ – Ross Millikan Oct 14 '17 at 14:03
  • $\begingroup$ I need to show valid proof and not for $5×5$...Its $10×10$. $\endgroup$ – Temidire Adesiji Oct 14 '17 at 15:02
  • $\begingroup$ Each cell of the $5 \times 5$ is replaced with a $2 \times 2$ block, which gives you $10 \times 10$. Just follow the instructions and get the square. You can then find the patterns that show it meets the requirements. $\endgroup$ – Ross Millikan Oct 16 '17 at 0:16
  • $\begingroup$ Thanks a lot...Do u know how to solve maths Olympiad questions well? $\endgroup$ – Temidire Adesiji Oct 16 '17 at 0:19

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