# VMO P-3 Nikhilam (Base) Multiplication: One number above and one number below the base.

We have seen how simple Nikhilam multiplication can be when applied to numbers above a base and numbers below a base. Now we can see that the same principles can be used in multiplication where one number is above and one number is below the base.

Our first example 147 x 98, with large digits such 7 8 and 9, is not easy to solve and would require using pen and paper if the conventional right to left method is employed. Nikhilam multiplication enables us to solve these kinds of problems mentally.

When one number is above and the other number is below the base, we either a) **add the excess of the number above the base to the other number**, or b) **deduct the deficiency of the number below the base from the other number**. Choose whichever is easier.

We can see that one multiplicand has an excess of **47** over the base, **100 **while the other multiplicand, **98 **is deficient by **2**. We can add **47** to **98**, but it is easier to deduct **2** from **147**, In any case we will get **145**.

The product of the excess **47** and the deficiency, **– 2** is **– 94** so we cannot simply suffix it to the first part. We must deduct **94** from **14,500 (145 with 2 zeroes) **to get** 14, 406. **

When computing mentally, we just subtract **1** from the result of the first part **145** to get 144 and suffix to it the ten’s complements of **94**.

**Proof:**

** Let x = base**

** a, b = excess/deficiency from the base**

** ( x + a) ( x – b) = x ^{2} + ( a – b) x + a(-b)**

** = (x + a – b)x – ab**

** = [(x + a) – b]x – ab **

For **108 x 96**, we can immediately see that one factor has an excess of **8** while the other has a deficiency of **4; **their numerical product is **32** and its ten’s complement is **68**. We can then mentally deduct **4 **from **108** to get **104** then deduct another **1 **to get** 103 **before attaching** 68 **to it to get the final answer of** 10,368.**

** 1 0 8 + 8 **

** X 9 6 – 4**

** 1 0 8 – 4 | 8 x -4**

** 1 0 4 | – 32 = 1 0, 3 6 8**

In cases like **113 x 89, **where the second part will yield **13 x – 11 = -143,** the first part will be **(113 – 11) – 2 = 100** while the second part is the ten’s complement of **43** or **57**. The final answer is **10, 057.**

Exercise P-3: Find the following products using base multiplication

- 12 x 9; 2) 103 x 98; 3) 102 x 97; 4) 102 x 98; 5) 103 x 97; 6) 105 x 93; 7) 75 x 103; 8) 112 x 89; 9) 1012 x 991; 10) 1125 x 995

Answers to Exercise J -2

- 9409; 2) 8649; 3) 7921; 4) 7056; 5) 6241; 6) 998,001;7) 992, 016; 8) 976, 144; 9) 99,880,036; 10) 97,970,404

Suggested Readings:

Prudente, Virgilio, *25 Math Short Cuts*, pp 67 -68.

Glover, James, *The Curious Hats of Magical Maths*, Book 2, pp. 14 – 15

Williams, Kenneth, *Vedic Mathematics Teacher’s Manual (Intermediate Level)* p. 34

IAVM, Inspirational Maths from India, pp 77-78.