Atrue mathematician would know this problem. so if you know the magic square(you can look it up)i need each of a 3 by 3 square to have one of these each number in it: 5,6,7,8,9,10,11,12, and 13. you can only use one number once. but each row, column and from diagonal must all equal up to 27.

Attached is one version.

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The key is that 9 must go in the middle.

If you know how to work a 3 by 3 magic square using the digits 1 to 9, you can make any 3 by 3 magic square starting with any number.

Starting Square

8  1   6

3  5  7

4  9  2

This magic square adds up to 15. That is 3 * the center number.

That is not the magic square you want, but it can be made into yours.

Just add 4 to every number in the base magic square.

12 5  10

7   9  11

8  13  5

And the sum is 3 * 9 = 27 in all the rows columns and diagonals, just what you requested. What happens when you have 0 in the center?

That means you take away 5 from every number in the base square.

3    - 4      1

-2     0     2

-1      4     -3

It still works. Every row column and diagonal add to 0.  

answer: plz put a pic in so i can answer it

step-by-step explanation:

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d