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There are 9 numbers from 7 to 15 . Their total is 7+8+9+10+11+12+13+14+15
= 99.  There are arrange in 3 rows and columns.  Total along  a column or a row i s the same.  So total along a row or a column = 99 / 3 = 33
SInce all the sums are to be 33 along diagonals and rows columns are 33 each. there is one center, it is inferred that center has the number which is the middle of the range 7 to 15.  that is 11. Then the filling is symmetrical.  Now choose one number to put at one place and try till its complete.

Take 7 put it in one square, then third number in that row will be 33 - 7 - 11 = 15. Here you see that you cannot put 7 in a corner. let us see why. For a number in the corner, that number is part of 3 sums to value 33 : one along diagonal, two along a row and a column. Now let us say we put 7 in the corner. 15 is put in the opposite diagonal corner. To fill the next place to 7, we have numbers 8,9,10,12,13,14.  The remaining two numbers in row with 7 must total to 26, ssince row totals to 33.  Check if there are two pairs with total of 26 each. THere are 14+12, and that is all. So 7 can be in the side middle square.
Now fill up box opposite to 7 and next to seven either side with 14,12.  The rest follows. it is simple enough.

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The answer is incomplete. Where is magic square?
The difficulty consisted in the construction, and not to write the obvious theory.
It's logical that is easy to build a magic square with central term equal to one third of sum....
you already gave the square and filled it. I explained to the student some of the principles. i donot need to draw the table again.
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