# How to Construct Multiple Squares of Odd Order

In a Magic Square there are i) equal number of rows and columns. ii) We have to fill up consecutive numbers, generally beginning with 1 and going up to nxn i.e, n square, such that the total of each row, each column, and the two main diagonals is same. So for a 5x5 squares we will have numbers from 1 to 25 and the total required would be 65 for each row, column and diagonal. For the odd order square let us first look at what is known as THE HINDU RULE.

## Steps

- 1
**The right hand side 7x7 square illustrates the rule using alphabets.**Write the first number Aa in the center of the topmost row. Next, write Ab in the lowest space of the vertical column next adjacent to the right. After, inscribe the remaining numbers in their natural order in the squares diagonally upwards towards the right that, upon reaching the right-hand margin, the inscription shall be continued from the left-hand margin in the row just above, and again, on reaching the upper margin, shall be continued from the lower margin in the column next adjacent to the right, noting that whenever we are arrested in our progress by a square already occupied we are to fill out the square next beneath the one we have filled. In this manner, for example, the 7×7 square given below has been formed:- Eb Fd Gf Aa Bc Ce Dg
- Fc Ge Ag Bb Cd Df Ea
- Gd Af Ba Cc De Eg Fb
- Ae Bg Cb Dd Ef Fa Gc
- Bf Ca Dc Ee Fg Gb Ad
- Cg Db Ed Ff Ga Ac Be
- Da Ec Fe Gg Ab Bd Cf
- Please note that here Aa means A+a, Bb means B+b, and so on.

- 2
**Give specific values to A’s and a’s.**For illustrating the method, let A=0, B=7, C=14, D=21, E=28, F=35, G=42, and a=1,b=2,c=3,d=4,e=5,f=6, and g=7, to get the square given below:- 39 48 01 10 19 28
- 47 07 09 18 27 29
- 06 08 17 26 35 37
- 14 16 25 34 36 45
- 15 24 33 42 44 04
- 23 32 41 43 03 12
- 31 40 49 02 11 20

- 3
**To put it differently, you can start with the center square in the last column and get this 7x7 square:**- 12 04 45 37 29 28
- 03 44 36 35 27 19
- 43 42 34 26 18 10
- 41 33 25 17 09 01
- 32 24 16 08 07 48
- 23 15 14 06 47 39
- 21 13 05 46 38 30
- This method is neat and quick. De La Loubere, Envoy of Louis X1V to Siam learnt of this method here.

- 4
**To get multiple squares, start with the central cell of the bottom row, central cell of the first column or the last column as preferred.**Giving A’s one of the values from 0,7,14,28,35,42, and a’s one of the values from 1,2,3,4,5,6,7 – with the provision that ‘D’ has to be given the value 21, you will be able to cover all numbers from 1 to 49 and get the magic sum of 175. It gives (7!x6!)/4 clear solutions. - 5
**Alternatively, give A’s one of the values – 1,2,3,4,5,6,7 and to a’s – 0,7,14,28,35,42 – with ‘D’ being given the value 4.**Of course, no value should be repeated; this automatically ensures that all the numbers are used. You will then be able to generate (7!x6!)/4 squares at one go by this method too. (Please note 7! Stands for 1x2x3x4x5x6x7 and 6! Is 1x2x3x4x5x6) - 6
**To apply another method which will give an even larger number of squares -- as there is no restriction on the value to be given to the central number, take the 11x11 square by this method.**The method is very general. The next line in this square starts with A’s from the 3rd column and a’s from the 4th column in the earlier line. One gets the basic square shown below:- Aa Bb Cc Dd Ee Ff Gg Hh Ii Jj Kk
- Cd De Ef Fg Gh Hi Ij Jk Ka Ab Bc
- Eg Fh Gi Hj Ik Ja Kb Ac Bd Ce Df
- Gj Hk Ia Jb Kc Ad Be Cf Dg Eh Fi
- Ib Jc Kd Ae Bf Cg Dh Ei Fj Gk Ha
- Ke Af Bg Ch Di Ej Fk Ga Hb Ic Jd
- Bh Ci Dj Ek Fa Gb Hc Id Je Kf Ag
- Dk Ea Fb Gc Hd Ie Jf Kg Ah Bi Cj
- Fc Gd He If Jg Kh Ai Bj Ck Da Eb
- Hf Ig Jh Ki Aj Bk Ca Db Ec Fd Ge
- Ji Kj Ak Ba Cb Dc Ed Fe Gf Hg Ih
- Of course Aa is to be read as A+a, and so on. A’s are to be chosen from 1 to 11 and a’s from 0, 11, 22,33,44,55,66,77,88,99,110, (i.e. multiples of 11). You thus automatically get all numbers from 1 to 121 and the number of solutions is (11!x11!)/4, with a total of 671 for each row, column and the 2 diagonals.

- 7
**Select the values that the diagonals/rows/columns total come correctly, in cases when some of the alphabets will repeat.**In case of 9x9 square it will be seen that A,D,G repeat in one diagonal and a,d,g; b,e,h and c,f,i repeat in the columns. Therefore ensure that A,D,G are so selected that they total 15, and a,d,g; b,e,h: c,f,i so that they each total 108, i.e 1/3rd of 324, which is the total of 0,9,18,27,36,45,54,63,72.. - 8
**Look at one of the 7x7 square by this method, with numbers from 1 to 49, with a total of 175.**A=1. B=2, C=3.D=4, E=5, F=6, G=7, and a=0, b=7, c=14, d=21, e=28, f=35, g=42.- 09 17 25 33 41 49
- 32 40 48 07 08 16
- 06 14 15 23 31 39
- 22 30 38 46 05 13
- 45 04 12 20 28 29
- 19 27 35 36 44 03
- 42 43 02 10 18 26

## Article Info

Categories: Mathematics