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MSC #11: Base Multiplication

MSC #11: Base Multiplication

Nikhilam or base multiplication is a very fast and easy way to get the product of two numbers near powers of 10. It got its name from the sutra Nikhilam Navatascaraman Dasatah meaning All from 9 and the Last from 10 because it uses the ten’s complement of the multiplicands in computing.

This kind of multiplication can be done with a common base or with different bases. And the multiplicands can be both below or both above the base or one above and one below the base.

The most simple case is when both multiplicands are above a common base, like 107 x 109. The written solution involves placing one number on top of the other just like in the conventional solution. Then their excesses over the base are placed on the right side with a leading “plus” sign Indicating that the multiplicands are 7 and 9 units above the base.

The answer can be calculated in two parts:

  1. the left-hand part is the result of adding the excess of one factor to the other multiplicand – (107 + 9) or (109 + 7) will both give 116.
    1. The right-hand part, which is allocated with the same number of decimal places as the number of zeroes in the base, is the product of the excesses – 7 x 9 = 63.

With little practice the product 11, 663 can be obtained quickly and mentally.  

            107        + 7

            109         + 9

            109+7|7 x 9  = 116|63 = 11,663

While some may dismiss this technique as a “trick”, it can be easily proved using algebra or geometry. Please refer to page 68 of our book, 25 Math Short Cuts for the algebraic proof.

In our featured example, 6,798 x 9,997 both factors are below their base of 10,000.

From our experience, the bigger the digits are in the multiplicands, the harder it is to perform multiplication when using the traditional right to left method. But then bigger digits imply smaller complements.

By using  All from 9 and the Last From 10, it is very easy to get their deficiencies as 3202 and 0003 which we will place at the right of the multiplicands preceeded by “minus” signs.

We can get the first part by subtracting 3202 from 9997 but it is much easier to deduct 3 from 6,798. In either case, we will get 6,795.

The second part is the product of the deficiencies: 3202 x 3. Again, this can be done mentally with little effort to obtain 9606. Since the number of digits in this product is equal to the number of zeroes in the base, there is no need for “carries” or adding leading zeroes.  

            6798          –  3202

            9997          –        3             6798 – 3 |3202 x 3 = 6795|9606 = 67, 959, 606.

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