Draw a big square. Divide it into 9 columns and 9 lines, so that you get 9X9 equal squares in it. So you have 9x9 = 81 places. Now write in 81 numbers one number into each square of course. The numbers are integers and have to start from 1 and have to end at 81. Each number must be placed only one time. Place the numbers in that way that their sum of each column and of each line and of both diagonals are the same. How do you calculate the sum? Post the formula (equation) to calculate the sum which is the same and the finished square with all numbers 1 to 81. How do you arrange the numbers that each square with an odd amount of columns / lines is working (5X5, 13X13, 21X21 for instance.) Please do not google or something, I don't know if there is a solution in the web, but I assume it is. Be a fair player and post your ideas. If you know the solution already, because somebody has shown it to you how it works, please wait a bit and let the others try. Enjoy. BTW: I remembered this because of the other maths thread. A teacher gave us that riddle when I was a child. Hint: You have to think '3 dimensional' to resolve the riddle. (This time it is pure linear and discursive thinking, different regarding the 'Self', lol.)

If you could clarify, I just want to be sure to understand... I think you are meaning that: (sum of each column) = (sum of each line/row) = (sum of any diagonal) Is that correct?

Yes, it's the sum of 9 numbers always that must be the same. It are 9 columns, 9 rows and 2 diagonals. A little more hint: The sum at the 9X9 square is 369 each.

Yeah, to calculate the sum is not difficult. One basically has to add all numbers to get the total sum: 1+2+3+4+5+....+81 = total sum. And this total sum is found when the sums of all nine rows | or nine columns are added. So finally you have to divide the total sum by the amount of rows | columns, here divided by 9. Total sum / 9 = sum of each row = sum of each column. To simplify the addition (1+2+3+4+5+6+7+.....+n) one can use Gauß formula: [n*(n+1)]/2 = totalsum. n is the last number to add = total amount of numbers. And to get the sum of each row | column: [n*(n+1)]/(2*z) = sum. z is the amount of rows | columns. Btw: One don't need to write the multiplication signs at this formula [n(n+1)]/(2z)

I love math and have always been good at it, but I've been away from things like this for almost twenty years.

It's hard to resolve I have to admit. There is no direct mathematically algorithm. How to fill up the numbers? Draw the squares and cut the big 9x9 square. Place number one in the middle of the top line. To fill up the numbers to place the next number you have to use the square which is always one line up and one column to the right (means to be placed always diagonally). This is the basic rule to fill the patterns! So number two would have to be placed in a square which is outside of the 9x9 pattern. Bend the square that it becomes a cylindrical roll, from the top to the bottom, in that way that the top and the bottom edges are together. Now you can determine the square which gets number two. (Use the square which is always one line up and one column to the right.) Now continue with number 3 until you reach again the end (here the right side). This will be here at number 5. Now bend it again, this time that way that the left and right edges come together. So you know now where to place number 6. If a square should be occupied already just use the square below the previous number in the same column and continue. This here will firstly happen when you try to place number 10. The square is already occupied by number one, so just place number 10 below the previous number in the same column, which is number 9. When finished one should end with the last number being always placed in the middle of the bottom line. Number one is in the top line and the last number in the bottom line, both located in the middle of the lines. This works with any odd numbered squares.