MULTIPLICATION USING “NIKHILUM” SUTRA

“NIKHILUM NAVATAHA CHARMAM DASHTAHA” is used to simplify the process of multiplication. The meaning of this sutra is “all from nine and last from ten”. With the help of this sutra we get the complement of a number which is used for finding the product.

We classified the use of this sutra in the following classes-

i)when both the numbers are less than the nearest working base-

When both the numbers are less than the nearest working base, then first find the complements of multiplicand and multiplier, that should be in digits equal to multiplicand and multipliers i.e., for the number 97, compliment is 3 but it should be written in the form 0. Also put a negative sign before complement which shows that the number is how much less than the working base. Write corresponding complements against multiplicand and multipliers respectively.

Now, we would find the product of these complements, the result is right of resultant. Also, the sum of cross numbers is the left side of the resultant. In this process we should be very careful about carryovers. In the product of single digits numbers right side will with be single digit and for the product of the two digits numbers right side will with be two digits and so on, the remaining carryover will be with left side digit(by adding).

This process will be much clear in the following examples-

 Example- 1:     find the product of 91 &93.

Solution:           complements of 91 and 93 are 9 & 7 respectively (they are 9 & 7 respectively smaller than 100)

           91 -09
         * 93 -07
          _________
          84  / 63 

Now, complements are put against the numbers with negative sign. But we have to be careful here since the numbers are of two digits, so complements should also be of two digits which is made by putting zero before complement.

The product of (-09) and (-07) is 63, also cross sum is (93-09) or (91-07) equal to 84.

NOTE-The bar is put to keep both the sides separately. The importance of this bar will be understandable in the later segments.

Thus,        91*93=8463

Example- 2:     find the product of 995 and 997.

Solution:          complements of 995 & 997 are 005 & 003 respectively

                                                            995 -005

997 -003

___________

992 / 015

Complements are put against numbers with negative sign. The product of (-005) and (-003) is 15 but it should be of three digits, so we put one zero before 15. We too find the cross sum which is (997-005) or (995-003)  equal to 992.

Thus,   995*997=992015

ii)when both the numbers are greater than the nearest working base-

In those cases, in which numbers are greater than the nearest working base, everything is same but the complements of the multiplicand and multiplier are put against corresponding numbers with positive sign.

The process will be much clear with the help of the following examples-

Example-3:    find the product of 115 and 103.

Solution:         complements of 115 & 103 are 15 & 03 respectively.(they are 15 & 03 respectively greater than 100-the working base)

                                                             115  +015

103  +003

We put the complements against the numbers. The product of 015 and 003 is 045. Also, cross sum is 118.

Thus,          115*103=118045

iii)when one number is less than the working base and the other is greater than the working base-

In this case greater number(than working base) has a complement against that with positive sign and other smaller number has a complement against that with negative sign. These sign are used in cross sum and put a bar on the product of the right side. All other rules regarding digit surplus, digit deficit etc. will be exactly the same as before.

Example-4:      find 1013*986

Solution:           complements of 1013 and 986 are respectively 13 and -14 .

                                                          1013 +13

* 986 -14                      

     ____________

999  / -182

Here, left side = 986+13=999    or    1013-14=999

And                            right side =13*-14= -182

Thus,                        1013*986

= 999/-182

                                                    =999/(1000-182)

                                                    =999818

iv)when the numbers are not nearer to working base-

When the numbers are not nearer to working base 10, or the multiple of 10, then we have to take a base near the numbers but cross sum(left side of the resultant) is multiplied by the number (which is obtained by dividing assumed base by last working base). Here it is noted that this number is an integer.

      All other rules will be exactly the same as before.

     To understand the process follow the following examples-

Example-5:     find 65*57

Solution:         let the working base is 60 which is 6 times to the last working base 10. Complements of 65 & 57 are +05 & -03 respectively.

                        65   +05

                     *  57   -03

                       ___________

                        62  / -15

Left side= 62*6= 372

And right side= -15

Thus,     65*57 = 372/-15

                           =370/(20-15)

                           =370/5 = 3705

Example-6:     find 9561*8997

Solution:         let the base be 9000, which is 9 times to 1000. Complements of 9561 and 8997 are 561 and -3 respectively.

                           9561 +561

                         * 8997 -3

                         ______________

                         9558  / -1683

Left side= 9558*9= 86022

And right side= 561*(3)  = -1683

Thus, 9561*8997 = 86022/-1683

                                 =86020 / (2000- 1683)

                                = 86020 / 317   = 86020317