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

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

In this article, we shall learn the procedure to divide the large numbers. To make it better understandable, lets go through with the division of polynomials. It will help us to understand the procedure better.

Lets take an example, 12x²-8x-32 by x-2
         ____________
x-2 | 12x² - 8x - 32  | 12x + 16
      -(12x² - 24x)
-----------------
                16x - 32
             -(16x - 32)
               -----------------
                        0
    Here, Quotient = 12x + 16 , Remainder = 0

Take one more example, 6x⁴+13x³+39x²+37x+45 by x²-2x-9
             _______________________
x²-2x-9 | 6x⁴+ 13x³+ 39x²+ 37x+ 45  |   6x² + 25x + 143
           - (6x⁴ - 12x³- 54x²)
             --------------------------
                      25x³+ 93x²+ 37x
                   -(25x³- 50x²- 225x)
                    -------------------------
                             143x²+ 262x+ 45
                          -(143x²- 286x -1287)
                        ---------------------------
                                    548x+1332
    Here, Quotient = 6x² + 25x + 143 , Remainder = 548x + 1332

Now, see the above example in different form, which will actually be our procedure for division of large numbers

x²-2x-9 | 6x⁴+ 13x³+ 39x²  | +37x + 45 |
      2  9           12      54
                                50       225
                                       286   1287
          ----------------------------------------
               6     25     143    |  548   1332
Quotient = 6x² + 25x + 143 , Remainder = 548x + 1332

Following are the steps that we done above:

Step 1: Divisor ( x²-2x-9 )

The first step is to write the divisor. In second line, write all the coefficients except the first, changing all of those to the opposite sign of coefficients. (Here, opposite signs of '-2' and '-9' are '2' and '9' respectively and first coefficient of x² i.e. '1' is not being considered )

Step 2: Dividend ( 6x⁴+13x^3+39x²    +37x + 45 )

The second step is to rearrange the expression of the dividend, leaving space after last two expressions from right. The last two expressions are for the remainder. (Here question arises, how do we know from where we have to split. The answer is : one less than the length of the divisor expression (i.e. number of coefficients + constant). In above expression, length of divisor expression is 3, therefore we shall split after two expressions from right (3 - 1 = 2).

Step 3: Procedure

• Multiply the first coefficient of the dividend (i.e. '6' in above example) by each digit of the rewritten divisor (i.e. 2 & 9) on the left (we get, 6x2=12, 6x9=54). Write each digit one after the other leaving the first place. 
• Now see across the row, add the digits(i.e. 13+12= 25). Now repeat the same as done in Step3(i), multiply '25' by rewritten divisor (i.e. 2 & 9) we get, 25x2=50, 25x9=225. Write each digit one after the other leaving the first two places.  
• Repeating the same as done in last step, now add (39+54+50=143). Now multiply 143 by (2 & 9) and we get (286 & 1287).  
• Now add remaining columns which is the remainder, add (37+225+286=548, 45+1287=1332) and Quotient is (6 25 143)expression.

We have to repeat the process up to last digit of the dividend.

IMPORTANT: In this article, we consider only those divisors whose first digit is unity i.e. '1'. The divisors whose first digit is not '1', will be taken in the next article.