Square of a number

Numbers ending in 5

The square of any number ending in 5 can be obtained by a method "Ekaadhikena Purvena", meaning one more than the previous one. 

Let us take a simple example to start with :
25²  

In traditional method, it is done as 25 x 25 ---------  125  50x --------- 625 ---------

Using Ekaadhikena Purvena method, it is 25 x 25 = (2 x 3) 52              = 625 5 is the reference here. Previous number is 2 and one more than 2 is 3.

Some more examples for better understanding :

75²  = (7x8) 5²  = 5625

45²  = (4x5) 5²  = 2025 and so on..

Now the obvious next question would be, how does it work for numbers more than 2 digits. The answer is, it works the same way.

125²  = (12x13) 5²  = 15625

765²  = (76x77) 5²  = 585225

Well, for bigger numbers doing the (N x N+1) calculation become a lengthy process again, and VM has a solution to it also using another Sutra which will be handled in a different topic. For now, be assured, there is a method.

Numbers near to decimal bases

This is obtained by a Sutra called “Nikhilam Navatascaramam Dasatah” (all from 9 and last from 10). Let us see this with an example :

12²

In traditional method, it is done as 12 x 12 ---------  24  12x --------- 144 ---------

Using Nikhilam method, 12 x 12 = (12+2) 22 = 144 12 is 2 more than 10. So 22 12 is 2 more than 10, so add 2 to 12 to get the digits left to the unit2.

Let us see some more examples for better understanding :

8²     = (8-2) 2²         = 64 ,              here 8 is 2 less than 10. So 2² and subtract 2 from 8.

13²  = (13 + 3) 3²  = 169 ,      here  13 is 3 more than 10. So 3² and add 3 to 13.

14²  = (14+ 4) 4²   = (18)(16) = (18+1)(6) = 196. Here 14 is 4 more than 10, and we do the usual of squaring 4, and adding 4 to 14. Since the base reference here is 10, we expect the result of (N²) to be single digit. So additional value is considered as carry and added to preceding digits.

Another example,

16²  = (16+6) 6²  = (22)(36) = (22+3)(6) = 256

Now, let us move to numbers near base 100. Here base is 100, so we expect the result of (N²⁾ to be 2 digit, and only if the result  is a 3 digit number we do carry forward.

91²  = (91-9) 9²        = (82)(81)   =    8281

89²  = (89-11) 11²    = (78)(121) = (78+1) (21) = 7921

106²  = (106+6) 6²  = (112)(36) = 11236

The squares of 11, 9 or 6, if needed, can be again done using nikhilam method for smaller children.

This applies to any decimal base. Let us see few examples for base 1000. Here we expect the result of (N²) to be 3 digits and only if the result is a 3 digit number we do carry forward.

1011² = (1011 + 11) 11² = (1022)(121) = 1022121

984²  = (984 – 16) 16²  = (968) (256) = 968256

 Here again, if needed, the squares of 11 and 16 can be done using nikhilam method for smaller children.  
 This can be extended to any decimal base provided the number is nearer to that so that the + or - value of that number from the base is easier to be squared. 
 

Any 2 digit number

Let us consider any 2 digit number. It is solved using “Dwadanda Yoga” (Duplex combination method). Infact this method is used to find the square any number which is a more generalised method of finding squares, which is explained in the next section of this page. For now, let us take an example of 2 digit numbers :

23² =  2²  /  2x2x3  /  3²   =        4    /  12   / 9  = (4+1)  /  2  /  9  = 529

Here, we expect only one digit from each calculation so any higher value is taken as carry and added to the preceding digit. Since we got 12 from (2x2x3), we take 1 as carry to be added to 4.

Some more examples for better understanding :

48²  =  4²  /  2x4x8  /  8²   = (16)  /  (64)  /  (64)   = (16 + 7)  /  (64+6)  / 4  = 2304

86²  = 8²  / 2x8x6  /  6²     = (64)  /  (96)  /  (36)   = (64+9)    / (96+3)   / 6   = 7396

General Method for any number

It is time to go for a generalised method which can be used to find the square of any number which may or may not fall into the  categories defined in the previous sections.

The method we use here is duplex combination. Before going into examples it would be easier if we get familiar with the formula of how to get the duplex result of a number.

a = a²      , 1 digit number

ab = 2ab , 2 digit number

abc = 2ac + b , 3 digit number

abcd = 2ad + 2bc , 4 digit number

abcde = 2ae + 2bd + c² , 5 digit number  and so on.

This is the pattern by which the duplex combination result of any digit number is obtained. Let us move on to few examples and apply this method. Remember, for a N digit number, the duplex method is to be done for N + (N-1) times. 

216²

Being a 3 digit number, we need to do the duplex method for 5 times (3+(3-1)). So write the number as  00216 and start finding the duplex combinations. For clarity, I have made a table below for the steps.

Combination	calculation	Value	Carry	Result
6	62	36	3	6
16	2x1x6	12	1	5
216	2x2x6 + 12	25	2	6
0216	2x0x6 + 2x2x1	4	0	6
00216	2x0x6 + 2x0x1 + 22	4	0	4

Answer is 46656

For clarity purpose, I have included expression involving multiplication by 0 also. Once you get more practise, that is not needed as we all know the result will be 0 only.

Let us go for a bigger number now.

7694²

Being a 4 digit number, we need to do the duplex method for 7 times. So write the number as 0007694.

Combination	calculation	Value	Carry	Result
4	42	16	1	6
94	2x9x4	72	7	3
694	2x6x4 + 92	129	13	6
7694	2x7x4 + 2x6x9	164	17	7
07694	0 + 2x7x9 + 62	162	17	9
007694	0 + 0 + 2x7x6	84	10	1
0007694	0 + 0 + 0 + 72	49	5	9

Answer : 59197636

By now, I hope you are able  to appreciate the power of VM in arithmetic calculations. I will add  more Sutras and their applications in mathematics time to time. I suggest to check regularly / follow the blog for updates.

Please drop in your comments / questions. Your feedback is always welcome.