It's Magic

There are total 121 small squares and numbers are form 1 to 121 sum of these 121 numbers = 121 × 122/2 = 7381

This sum 7381 is divided in 11 columns , rows and diagonals so magic sum = 7381/11 = 671 the following arrangement is shown for 11 × 11

68 81 94 107 120 1 14 27 40 53 66

80 93 106 119 11 13 26 39 52 65 67

92 105 118 10 12 25 38 51 64 77 79

104 117 9 22 24 37 50 63 76 78 91

116 8 21 23 36 49 62 75 88 90 103

7 20 33 35 48 61 74 87 89 102 115

19 32 34 47 60 73 86 99 101 114 6

31 44 46 59 72 85 98 100 113 5 18

43 45 58 71 84 97 110 112 4 17 30

55 57 70 83 96 109 111 3 16 29 42

56 69 82 95 108 121 2 15 28 41 54

Using this technique any odd number magic square can be arranged.

let n= no. of rows let s= sum of all in 1 row normally, the sum is the middle number of no. of sq * n so, first, no. of sq = $n^{2}$ the mid number = $\frac {n^ {2}-1}{2}$ so, $s= n* \frac {n^ {2}-1}{2}$

It will surely be there in case of magic square with odd rows and columns , that there exists a centre cell , I think you could try to visualize it , which is exactly at the centre of the square. Now consider the row and column and diagnal which pass through the centre cell . We know their sum should be same . But how , let's apply some strategy or tricks by which it could be same . I have found one which we shall discuss . Let there be a number m in the centre cell , then we could write a number _ m-p _ on top of it and m+p below it and m-q on top of m-p and m+q below m+p and so on on both row and column with defferent values in place of p or q but the sum would be same as the values would get cancel. So what we will get constant value of two diagnal , a column and a row . And in this way we will by logic enter rest of the values . The value of the centre cell should be middle of the values that can be entered , so in a cell which has n rows and columns with values that can be entered be $n^{2}$ , then we have a formula for finding the middle number median that has to be entered in the centre cell , $\frac{n^{2}+1}{2}$ And so we can easily get the sum of each row , column and each of the two diagnols as , by a little , that as the sum of each column , row and diagnal could be found by just finding that of anyone , so we try for the column of centre cell . Which is got by adding , m , m-p , m+p , m-q , m+q ...... , and all p's or q's would get cancelled so this gives us the sum to be $n\times m$ and $m$ being equal to $\frac{n^{2}+1}{2}$ we have the formula for sum for each row , column or diagnal as $n\frac{n^{2}+1}{2}$ Substituting $n$ for $11$ we have answer $\boxed{671}$