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VMO P-3 Nikhilam (Base) Multiplication: One Number Above And One Number Below The Base.

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) = x2 + ( 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

  1. 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

  1. 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.

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