Vedic Math - Division of large numbers (Part-II)

In previous article, we learn the procedure of doing 'division of large numbers' with divisor's first digit as '1'. In this article, we learn the procedure with divisor's first digit has all numbers except 1.

In this article too, we first go through with division of polynomials.

-4x3-7x2+9x-12 divided by 2x-4
        _________________
2x-4 | -4x3 - 7x2 + 9x - 12 | -2x2 - 15/2x - 21/2
       -(-4x3 + 8x2)
  ------------------------
        -15x2 + 9x
      -(-15x2 + 30x)
       ------------------------
                           - 21x - 12
                        -(- 21x + 42)
                     ---------------------
                                      -54
Quotient = -2x2 - 15/2x - 21/2 , Remainder = -54

See the above example in another form by making the first coefficient of the divisor as '1'. Like in following example, we divide the divisor by 2 and later divide the quotient also by 2 :

2x-4  |  -4x3 - 7x2 + 9x    - 12
  x-2
   +2             - 8    - 30     - 42
           -----------------------------
             -4   - 15   - 21    - 54

Now, divide this by 2 (2 is the first coefficient of the divisor).
Also note that we don't divide the remainder by 2, it will remain constant.

     -2   - 15/2  - 21/2  - 54
Quotient = -2x2 - 15/2x - 21/2 , Remainder = -54

One more illustration:
First, make the first coefficient of divisor as '1'

2x2 -3x +1   |  2x5 -9x4 +5x3 +16x2    -16x +36
 x2-3/2x+1/2
    +3/2 -1/2            3    -1
                                     -9    +3
                                             -15/2     +5/2
                                                     69/4 -23/4
                    ---------------------------------------------
                       2    -6     -5      23/2  15/4 +121/4
Divide by 2 )  1    -3    -5/2   23/4     15/4 +121/4

Quotient = x3-3x2-5/2x+23/4 , Remainder = 15/4x +121/4

Remember that Remainder is constant in every case

Lets take examples:

1699 divided by 223
      2 2 3     |  1  6    9   9
  2) 1 1 3/2 |
        -1-3/2 |     -1  -3/2
                              -5  -15/2
                ---------------------
                     1  5   5/2  3/2
Divide by 2)  15/2  5/2  3/2    (Don't divide the remainder as remainder is constant)
                      7 1/2  5/2  3/2    (Always write Quotient in least fraction form like here, 15/2 = 7 1/2 =
                                                    7+1/2)
Here, Quotient is in fraction (i.e. 7 + 1/2). (See how to make quotient into integer)
Here Quotient becomes 7 and will divide the actual divisor by 2 (i.e. 223/2) and add this in the remainder.
So, Remainder becomes 223/2 + 50/2 +3/2 = 276/2 = 138    (5/2 is on tens place, that is why, we
                                                                                                   write 50/2 )

Now, Quotient = 7 , Remainder = 138

7685 divided by 672
      672 / 6 = 112
      6 7 2 | 7  6    8  5
      1 1 2 |
       -1-2 |    -7  -14
                         1   2
               -----------------
                 7 -1   -5  7
           69     -43        (Quoitent= 70-1 =69, Remainder= -50+7 = -43)
Divide by 6)  69/6    -43   (Remainder remain constant)
                     11 1/2  -43   (Quotient in least fraction, 11+1/2)
                     11      293   (1/2 from quotient gets added in remainder after dividing actual divisor
                                            by 2)
Quotient = 11 and Remainder = 672/2 - 43 = 293

There is another way to avoid fraction
(i) By multipling the divisor with a suitable number to make the divisor closer to the base value, or
(ii)By multipling the divisor with a suitable number to make the divisor's first digit '1'

1334 divided by 439
      439 x 2 = 878
Taking 878 as new divisor, divide 1334 by 878, we get, quotient = 1 and remainder = 456
(Solved using technique described in article "Division by number near base")
Now, multiply quotient by 2 and remainder remain constant. Quotient = 2 , Remainder = 456 . But remainder is greater than divisor '439'
So, Quotient = 2+1 =3 , Remainder = 456-439 = 17

1177 divided by 516
      516 x 2 = 1032
Taking 1032 as new divisor, divide 1177 by 1032, we get, quotient = 1 and remainder = 145
(Solved using technique described in previous article)
Now, multiply quotient by 2 and remainder remain constant.
Quotient = 2 , Remainder = 145


Please note that quotient will by divided or multipled by same factor, which is being used to make the divisor first digit as '1' or closer to the base, i.e. if we have divide the divisor by a number, quotient will also be divided by same number later And if divisor is multiplied by a number, quotient will be multiplied by this number.  

Hope it helps. Share your experience with these techniques.