Division by Nikhilam, 18 mathematics tricks and fun

Division by Nikhilam, 18 mathematics tricks and fun 

In the current system, We need to use trial digits when dividing with big divisors.in the vedic system We can 

add or subtract for division which makes it easy.

Here there are 4 methods of which argumental division

 I am omitting since it is good for algebraical divisions 

and not for arithmetical computations.

(1)Nikhilam method used for big numbers,

 (2)paravartya method for smaller digits.Best of all

 these and (3)quick one is Straight Division.

Nikhilam Sutra is used as said earlier when divisor is 

with bigger digits like 6,7,8,90.we should find the complement of this and divide.

 Here we have 3 examples. First 2 are two digit 

numbers to be divided by 9.The third one is three digit dividend. In all these divisor is 9.The complement

 of 9 is 1. We will write this as 9/1, as shown.Now We should count the digits in the divisor and that much 

digits we should count in the dividend and keep on

 the right and remaining on the left.here we have 12/1since 1 is the complement of 9. Now let us take1 keep it down. now 1×1 is 1 we put this up below 2 and add. 2+1 is 3, we will put 3 As remainder. Same way secondeg. 21/9= 21/1 bring down 2 and 2×1 is 2. 2+1

 is 3 so quotient 2 remainder 3.

In the 3rd example we have to multiply 104/9=104/1

 we will arrange as 10 on left and 4 on the right..first 

bring down the first digit 1×1=1 we write 1 below 0 and 0+1 is write this 1 down .. Then 1×1 is 1 .put this 1 

below 4 and add .this we'll give 5 .so quotient 11 and 

R 5.

 Here 3 more examples are shown. Dividend is 3 

digit,5 digit and 6 digit numbers and divisor is 9so 

we arrange according to Nikhilam method and complement 1

First eg.311/9=311/1Thenfirst bring down 3, then 3×1

 is 3 ,3+1 is 4 write 4 ,then 4×1 is 4add 4+1 is 5.so

 quotient 34 and remainder 5

Same way other two examples are done .

 Here we have 3 eg. 136/9 so we proceed doing as

 taught earlier and remainder comes as 10.since 10 

is more than 9 we can divide this again ,which gives

 1as quotient to be added to the original quotient,so 14+1is 15 as quotient and final remainder is 1.

2 nd example 23/8. Here complement is 2. As

 earlier bring down 2 as quotient then 2×2 is 4 which 

we have to add to the right side 3 which gives us remainder 7.

3rd eg.111/89=111/11, so first bring down 1 as

 quotient 1×11 is 11 which will put below 11 on the

 right side .so quotient 1 and remainder 22

 Here two eg.shown.first example is similar to the one discussed earlier.and quotient is 1 remainder 38.

2nd eg.is 110 01/ 88= 110 01/12.as usual bring down 

1 ,1×12 is 12.write 12 below 10.add 1+1 is 2.2×12 is 

24 we will write 2 below 0 on left side and 4 below 

the 0 on the right .now we 2+2 on the left gives 4. 

 4×12 is 48 which we will put below 01 on the right 

adding digits on right gives 4+48+01 gives 89. 89is

 more than 88 so dividing this gives 1 adding to 124 gives 125 as Quotient and Remainder 1.

Here also two eg.shown. 198/97= 1 98/03 and divide 

as usual. Here we get Quotient 2 reminder 4. Same 

way a bigger number is shown.as example.and we get 

Q 14 and R 34609

 This shows the weakness of division by Nikhilam

 method since divisor is big the procedure is going

 long so we can tackle using Anurupyena sutra which shows find ratio and adjust proportionately.

In this example 101 10/23 = 101 10/77 this

 complement can be reduced using Anurupyena. We multiply 23×4 gives 92 so complement is 08. Then

 We get 109 as quotient and remainder 82. We then

 multiply 109× 4 which gives 436 now remainder 82 is bigger than 23 which 3 which added to 436 gives 439

 and remainder 13.

Thank you we will continue division by paravartya method.. subscribe to my channel.