Multiplication ( Traditional versus Vedic mathematics approach )

 MULTIPLICATION (  Short - cut )

The first impression we all have when we have to do multiplication of long numbers or many digits like  88 x 88 ...91 x 57 .....75 x 75....1243 x 987 etc...is where is paper and pen or calculator. 

But by using sutras from vedic mathematics we will learn to do such calculations in few minutes that to be orally or minimal calculations and the plus point is we don't have to learn tables...we just need to know basic tables till 9.

1. To find the squares ending with 5

     a.  25 x 25 = 625

     Process is first number multiply first number         plus 1 and multiplication of last two 5.

       2 × (2+1) / 5×5

       2 × 3/ 25

        625

      

    b.   35 x 35 = 1225 

           3 x ( 3 + 1) / 5×5 = 1225

    C. 45 x 45 = 2025

         4 x ( 4+1) / 5 x 5 = 2025

         And so on.............

2. To Multiple ( using  vedic sutra Nikhilam Navatascaramam means all from 9 and last from 10)

Single digit numbers    

   a.  9 × 7

  ● Suppose we have to multiply 9 by 7.

    We should take as base for our calculations,          that power of 10 which is nearest to the                    numbers 

      (10) 

      9 - 1

      7 - 3

    Now subtract the base 10 from the sum of given number I.e ( 9 + 7)

  So 16 - 10 = 6

    Or  

 Subtract the sum of two deficiencies ( 1+ 3) from base 10

     10 - 4 = 6

     Or

 Cross subtract deficiency 3 on the second row from the original  no. 9

   9 - 3 =6

   Or 

Cross subtract in the converse way 

   7 - 1 = 6

Result will be same

● Now , Vertically multiply the two deficit figures ( 1 and 3 ) . The product is 3. And this is the right hand side portion of the answer .

     Thus 9 × 7 = 63

     (10) 

      9 - 1

      7 - 3

     -----------

   9+7 - 10 / 1× 3

   16 - 10 / 3

     63 Answer.

   

b.    8 × 6

      Assuming base 10

        8 - 2

        6 - 4

     -------------

    8+ 6 -10/ 2×4

    = 48 answer

      Or

    8-4 /2×4

   = 48

This proves the correctness of the formula .The algebraic explanation for this is very simple:

 ( x - a) (x - b )= x ( x - a - b) + ab

 C.  7× 6

     (10)

     7 - 3

     6 - 4

  ------------

   3/ 12(4×3)

 As you can see product is of double digit and we have taken the base of 10 . So in this case we will only keep unit digit  I.e 2 and 1 will be carried over to the left 

 Thus 

(10)

     7 - 3

     6 - 4

  ------------

   3/ 12(4×3)

   3+1/ 2

 =    42 answer

Double  digit numbers.

 

a. 91 × 91

Here we will take 100 as base.

 91× 91

91 - 9

91 - 9

------------

91-9/9×9

82/81

= 8281 answer

b. 88 x 98

88 - 12

98 - 2

------------

88-2 / 12 ×2

86/24

= 8624 

c.  56 × 98

56 - 44

98 - 2

-------------

56 - 2 / 44x2

54/88

=5488 answer.

d. 25× 99

25 - 75

99 - 1

-----------

25 -1 / 75x1

2475 

= 2475 answer.

e.  88 x 91

88 - 12

91 - 9

--------------

88-9/ 12 x 9

79 / 108

 As we have taken 100 as base . So we will keep two place values on right side and carry the surplus digit over to the left.

 88 - 12

  91 - 9

--------------

88-9/ 12 x 9

79 / 108

79 +1 /08

80/08

 = 8008 answer.

Three digit numbers we will take base as 1000 and four digit numbers we will take base as 10000. And so on....

 Now if we have to multiply 

Special cases:::

a. 12 x 11

In this case we will take base as 10 as it is nearest to 10 but no. Is more so.

(10)

12 + 2

11 +1

-----------

12+1/2×1

132

= 132 answer

b. 17×12

17 + 7

12 + 2

------------

17+2/7×2

19/14

So again carry forward

19+1/4

=204 answer.

c. 12 x 8

12 +2

8 - 2

----------

12-2/2×(-2)

10/-4

= 96    (10/0 - 4)=96

 d. 108 x 97

108 + 8

97 -3

-------------

108-3/ 8 x ( -3)

105/ -24

105 /00

-        24

---------------

104/76

10476 answer

e.  1033 × 997

Take base as 1000

1033+33

997 - 3

-----------------

1033 -3 / 33x(-3)

1030/ -099

We have taken 0 because base is 1000 . So instead of 33 x 3 = 99 ...we have written 099

1030/000

-         099

------------------

1029/901

= 1029901 answer.