# VMO I-2 Multiplying larger numbers near a base

Nikhilam or base multiplication works well even with large numbers when they are near the base. In our example, we can readily see the multiplicands 678 and 998 are composed of large digits and therefore more difficult to compute using the conventional method. But having large digits means that their differences or deficiencies from their base of 1000 will be composed of small digits which are then easier to compute.

Using the Sutra “All from 9 and the Last from 10” which we will discuss extensively in another post, we can easily get their deficiencies as 322 and 2 easily. By cross subtraction we can get the first part of the product from 998 – 322, which is 678. But it is easier to use the simpler 678 – 2 to get the same result.

The second part of the answer is 322 x 2 = 644. The product then, is 676,644. All calculations here can be done mentally.

This type of multiplication is easier when the multiplicands are both above the base. In 1325 x 1003, we can easily see that their excesses are 325 and 3 respectively so that we can immediately say that the answer is 1325 + 3 | 325 x 3 or 1, 328, 975.

For 9,687 x 10,030 we have 9687 + 30 | – 323 x 30 = 9717 | – 9690 = 97,160,310

Exercise I-2: Find the following products:

- 996 x 989; 2) 787 x 997; 3) 9898 x 9998; 4) 1234 x 1002; 5) 1012 x 1014; 6) 12345 x 10002; 7) 998 x 1002; 8) 6767 x 1003; 9) 9998 x 12345; 10) 96,789,876 x 99,999,998.

Answers to Exercise J-3:

- 2601; 2) 3249; 3) 3221; 4) 2401; 5) 1764; 6) 1521; 7) 256,036; 8) 271,441: 9) 244,036; 10) 216,225

Suggested readings:

- Prudente, Virgilio,
*25 Math Short Cuts*, pp 61 -68. - Glover, James,
*The Curious Hats of Magical Maths*, Book 2, pp. 14 – 15 - Williams, Kenneth,
*Vedic Mathematics Teacher’s Manual (Intermediate Level)*pp. 30 – 35 - IAVM, Inspirational Maths from India, pp 74-79.