Remainder theorem and Vedic Mathematics

The one line method of division in polynomials.

Thu Jul 13, 2023

Say Yes to Vedic Mathematics and Polynomials

"Paravartya Yojayet is the Vedic sutra suitable for algebric divisions.

Paravartya Yojayet sutra means transpose and apply.

Invariable change of sign with every change of side. The "+" becomes "-" and conversely and similarly "multiplication becomes division and division in multiplication.

The Vedic one line mental answers through "Paravartya Yojayet".And the second is Dhwajank sutra or flag method. Paravartya Yojayet teaches us the one line Vedic mental mathematics.

Paravartya Yojayet tells us that :

Arrange divisor and divided both in decreasing powers of variables.

Fill the missing terms in both divisor and dividend.

No of terms reserved from the RHS for reminder will depend upon the degree of the divisor.

This division method is self-explanatory, because it is only a variation of the 'polynomial long division' method, where the Transposition is only a representation of (x - r), where x is Base and r is the Divisor.

Not only does this tactic provide a marked advantage by reducing the division to multiplications of digits - it is also a generalized tactic that will work with any n-digit Divisor.

The Sūtra: parāvartya yojayet is use for division (x ÷ y) of two numbers, especially when a remainder is required. This technique, also sometimes referred as Base-Division, is a generic method that supercedes the techniques for division used by Sūtra: nikhilaṃ navataścaramaṃ daśataḥ and Sūtra: ūrdhva tiryagbhyāṃ, which is the reason that the technique is not discussed in the afore-mentioned Sūtras.


We can use Nikhilam ,Urdhva tiryak and Paravartya. But the Paravartya Yojayet is the general method, there is no special case. So we emphasize on this.

As an illustration, let us use this Sūtra for:
97 ÷ 12

Steps97 ÷ 12
1.Find a Base, that is near the Divisor - that is a power of 10, that is 10n.In this case, the close power of 10 that is nearest to 12 is 10, which is the Base.
2.Find the Transposition of the Divisor - as (Base - Divisor), and represent as Vinculum, if required. Make sure that the Transposition has the same number of digits, as the Zeroes of the Base, 10nIn this case, the transposition is:
10 - 12 = -2 = 2
3.Arrange the numbers in a manner, that the Divisor is slightly raised, from its Transposition, and the Dividend on the right-hand side.In this case,
12 2 ) 97
4.Leave some space between the digits of the Dividend, and partition it into 2 parts:
1st Part is for Quotient, and the 2nd Part is for Remainder, such that the digits for the Remainder should be same - as the Zeroes of the Base, 10n.		In this case,
12 2 ) 9 | 7
|
---------
5.Drop the first digit (or the cummulative sum of any other number in the same column) of the Dividend, and multiply it (the 'Dropped' digit)with each digits of the Transposition - placing them in rows, next to the place of the 'Dropped' digit.In this case, the first digit is dropped:
12 2 ) 9 | 7
|
---------
9 |
And, Multiplying the dropped digit, with each digits of the Transposition:
12 2 ) 9 | 7
| 18
---------
9 |

Note: 9 × 2 = 9 × -2 = -18 = 18
5.Drop the next digit (or the cummulative sum of any other number in the same column) of the Dividend, and repeat above step (Step 4) till all the places for every digit of the Dividend is exhausted.In this case, the next digit is dropped:
12 2 ) 9 | 7
| 18
---------
9 | 11
All places for the digits in the Dividend is exhausted, so we stop.

Note: 7 + (-18) = -11 = 11
6.Finalize the Quotient and Remainder parts as:
1. Remove any viniculum number from Quotient;
2. Remove any viniculum number from Remainder;
3. If the Remainder is negative, take appropriate numbers of Quotient to make it positive;
4. If the Remainder is not less than the Dividor, give appropriate numbers to Quotient to make it lesser.

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.

The following examples will illustrate this more clearly.

239479 divided by 11203:

1 1 2 0 3 | 2 3 9 4 7 9
-1-2 0-3 | -2 -4 0 -6
-1 -2 0 -3
----------------------
2 1 4 2 1 6
Quotient= 21, Remainder = 4216
  • Start with writing the new divisor in next line, which is -1 -2 0 -3.
  • Rearrange the dividend leaving space after 4 digits from right.(One less than the length of the divisor, 5-1=4 )
  • Multiply (-1 -2 -0 -3) by 2 (the first digit of the dividend) and write the answer (-2 -4 0 -6) down in the next row (Row 2), from column 2 onwards.
  • Add the numbers in column 2 to get the next multiplication factor which is (3 + (-2) = 1) in this example.
  • Multiply the new divisor by this multiplication factor and write the answer (-1 -2 0 -3) to the right, i.e Row 3, Columns 3 onwards.
  • Add all the columns to get the answer. The quotient is on the left and remainder on the right. (Note: Here the remainder is in last 4 columns)
  • Remainder = 4216 (i.e. 4000+200+10+6, '4' on thousandth place, '2' on hundredth place, '1' on tenth place & '6' on ones place) Similarly, Quotient = 21 (i.e. 20+1, '2' on tenth place & '1' on ones place)

1248 divided by 160:
1 6 0 | 1 2 4 8
-6 0 -6 0
24 0
------------------
1-4 28 8
6 28 8

7 12 8 (Remainder is greater than divisor, hence Remainder=288-160=128, Quotient=6+1=7 carry over)

Quotient= 7, Remainder= 128

  • Start with writing down the new divisor which is -6 -0.
  • Rearrange the dividend leaving space after 2 digits from right.(One less than the length of the divisor, 3-1=2 )
  • Multiply this (-6 0) by 1 (the first digit of the dividend) and write the answer (-6 0) in the next row (Row 2), in column 2 and 3.
  • Add the numbers in column 2 to get the next multiplication factor which is (2 + (-6)=-4 in this example.
  • Multiply the new divisor (-6 0) by this multiplication factor (-4) and write the answer (24 0) to the right, i.e. Row 3, Columns 3 and 4.
  • Add all the columns to get the answer. The quotient is on the left and remainder on the right. (Note: Here the remainder is in last 2 columns). So the left side is 10-4=6. ('1' on tenth place & '-4' on ones place) The right side is (280+8)=288. ('28' on tenth place & '8' on ones place) Remainder is bigger than 160 so we subtract the divisor from it and carry over 1 to quotient.


Another example:
12349 divided by 1133
1 1 3 3 | 1 2 3 4 9
-1-3-3 | -1 -3-3
-1-3-3
------------------
1 1 -1-2+6
1 1 -114
-----------------
1 0 1019

Quotient= 10+1 = 11, Remainder= -100-20+6 = -114 { -1 = -100(hundredth place), -2=-20(tenth place), +6=+6(ones place)}
Since remainder is negative, so we subtract 1 from quotient and subtract the current remainder from divisor
Quotient= 11-1=10, Remainder= 1133-114 = 1019
Quotient= 10, Remainder= 1019

I hope this interesting technique will help in solving large divisions. In next article, we shall discuss about those cases where divisor's first digit is not '1'.

Keep practicing and keep sharing your experience with awesome Vedic Math!!
In this case,
The Quotient part contains: 9
And, the Remainder part contains: 11 = -11

The Remainder is negative. So, we take one from the Quotient: 9 - 1 = 8
And give it to Remainder: 1 × 12 - 11 = 1

So, we get: Quotient = 8, and Remainder = 1, which is the answer!





Shalinee
A dedicated person for Vedic Mathematics.