# MSC #11 – Base Multiplication

Vedic Mathematics use base multiplication extensively. Also called “Nikhilam” multiplication because it uses “All from 9 and the Last from 10”(see MSC #2) in determining the 10’s complements of the factors. Although other bases may be used, multiplications near powers of ten are easiest to do.

While our example is extremely difficult to solve mentally using the conventional method, the answer can be given Immediately by just looking at the figures. We can immediately see that the answer will have 8 digits and 3202 and 3 are the ten’s complement of the factors.

Now for the first part of the answer, we subtract 3 from 6798 to get 6795 (note that we can also get the same answer if we subtract 3202 from 9997, but that would be harder). We can begin to announce the answer as sixty-seven million, nine hundred fifty…” then we multiply the complements to get the second part” nine thousand, six hundred six”.

For better understanding of this technique, see the following simple examples.

Example 1: 9 x 8 =

Both factors are a little less than 10 so we will use 10 as base.  Place numbers one on top of the other and indicate their deficiency from the base at the right side. In base multiplication, the product consists of two parts

9    –   1          Cross-subtract 9 – 2 or 8 – 1 = 7 to get the 1st part

8    –    2         multiply the deficiencies to get the second part, 1 x 2 = 2

7    |   2

Note that since the base, 10 has only 1 zero, the second part is allotted 1 place only.

(See pp 65-66 of 25 Math Short Cuts for more explanation)

Example 2: 102 x 107 =

Here both factors are above the base, 100 so we will place the excesses on the right

102  +   2       Cross add this time to get the 1st part, 102 + 7 or 107 + 2 = 109

107 +    7       2  places are allotted to the 2nd part

109 | 14

Example 3: 14 x 17 =

14  + 4           Adding 4 to 17 is easier to get 21

17 + 7 only 1 place allowed in 2nd part

21 |28  = 238    the 2 must be carried

Example 4: 103 x 98 =

In this case one is above and one is below the base

103  +   3       Both 103 – 2 or 98 + 3 result in 101

98  –   2       Two places are allotted to the 2nd part

101  | -06      Better to express – 06 in bar form

= 10,094        Knowledge of ten complements is advantageous