mix 3

DIVISION

In the conventional procedure for division, the process is of the following form.

                          Quotient
                          _______
              Divisor ) Dividend        or    Divisor ) Dividend ( Quotient
                          ----------                          ----------
                          ----------                          ----------
                          _________                        _________
                          Remainder                          Remainder
        But in the Vedic process, the format is

                            Divisor ) Dividend
                                         --------
                                         --------
                                __________________
                                Quotient / Remainder
The conventional method is always the same irrespective of the divisor. But Vedic methods are different depending on the nature of the divisor.
Example 1: Consider the division 1235 ÷ 89.
i) Conventional method:

                    89 ) 1235 ( 13
                             89
                          _____
                           345
                           267     Thus Q = 13 and R = 78.
                          _____
                            78
ii) Nikhilam method:
This method is useful when the divisor is nearer and less than the base. Since for 89, the base is 100 we can apply the method. Let us recall the nikhilam division already dealt.
Step (i):
    Write the dividend and divisor as in the conventional method. Obtain the modified divisor (M.D.) applying the Nikhilam formula. Write M.D. just below the actual divisor.
Thus for the divisor 89, the M.D. obtained by using Nikhilam is 11 in the last from 10 and the rest from 9. Now Step 1 gives

                89 ) 1235
                __
                11
Step (ii):
    Bifurcate the dividend by by a slash so that R.H.S of dividend contains the number of digits equal to that of M.D. Here M.D. contains 2 digits hence

                89 ) 12 / 35
                __
                11
Step (iii): Multiply the M.D. with first column digit of the dividend. Here it is 1. i.e. 11 x 1 = 11. Write this product place wise under the 2nd and 3rd columns of the dividend.

                89 ) 12 / 35
                __
                11     1   1
Step (iv):
    Add the digits in the 2nd column and multiply the M.D. with that result i.e. 2+1=3 and 11x3=33. Write the digits of this result column wise as shown below, under 3^rd and 4^th columns. i.e.

                89 ) 12 / 35
                __
                11     1   1
                             33
                      _______
                      13 / 78
Now the division process is complete, giving Q = 13 and R = 78.
Example 2: Find Q and R for 121134 ÷ 8988.
Steps (1+2):

                    8988 ) 12 / 1134
                    ____
                    1012
Step (3):

                    8988 ) 12 / 1134
                    ____
                    1012     1    012
Step(4):

                    8988 ) 12 / 1134
                    ____
                    1012      1  012     [ 2 + 1 = 3 and 3x1012 = 3036 ]

                                    3036
Now final Step

                    8988 ) 12 / 1134
                    ____
                    1012    1     012

                                    3036(Column wise addition)
                             _________
                              13 / 4290
Thus 121134¸ 8988 gives Q = 13 and R = 4290.
iii) Paravartya method: Recall that this method is suitable when the divisor is nearer but more than the base.
Example 3: 32894 ÷ 1028.

The divisor has 4 digits. So the last 3 digits of the dividend are set apart for the remainder and the procedure follows.

Now the remainder contains -19, -12 i.e. negative quantities. Observe that 32 is quotient. Take 1 over from the quotient column i.e. 1x1028 = 1028 over to the right side and proceed thus: 32 - 1 = 31 becomes the Q and R = 1028+200 - 190 - 12 =1028-2 =1026.
Thus 3289 ÷ 1028 gives Q = 31 and R = 1026.

The same problem can be presented or thought of in any one of the following forms.

*Converting the divisor 1028 into vinculum number we get 1028 = 1032 Now

__
*Converting dividend into vinculum number 32894 = 33114 and proceeding we get

Now we take another process of division based on the combination of Vedic sutras urdhva-tiryak and Dhvjanka. The word Dhvjanka means " on the top of the flag"
Example 4: 43852 ÷ 54.
Step1: Put down the first digit (5) of the divisor (54) in the divisor column as operator and the other digit (4) as flag digit. Separate the dividend into two parts where the right part has one digit. This is because the falg digit is single digit. The representation is as follows.

                    4 : 4 3 8 5 : 2
                  5
Step2: i) Divide 43 by the operator 5. Now Q= 8 and R = 3. Write this Q=8 as the 1st Quotient - digit and prefix R=3, before the next digit i.e. 8 of the dividend, as shown below. Now 38 becomes the gross-dividend ( G.D. ) for the next step.

                    4 : 4 3 8 5 : 2
                  5   :     3
                ________________
                       :   8
ii) Subtract the product of falg digit (4) and first quotient digit (8) from the G.D. (38) i.e. 38-(4X8)=38-32=6. This is the net - dividend (N.D) for the next step.
Step3: Now N.D Operator gives Q and R as follows. 6 ÷ 5, Q = 1, R = 1. So Q = 1, the second quotient-digit and R - 1, the prefix for the next digit (5) of the dividend.

                   4 : 4 3 8 5 : 2
                 5   :     3 1
                ________________
                      :   8 1
Step4: Now G.D = 15; product of flag-digit (4) and 2nd quotient - digit (1) is 4X1=4 Hence N.D=15-4=11 divide N.D by 5 to get 11 ÷ 5, Q = 2, R= 1. The representation is

                4 : 4 3 8 5 : 2
              5   :     3   1 :1
            ________________
                   :   8 1 2 :
Step5: Now the R.H.S part has to be considered. The final remainder is obtained by subtracting the product of falg-digit (4)and third quotient digit (2) form 1² i.e., 12:
Final remainder = 12 - (4 X 2) = 12 - 8 = 4. Thus the division ends into

                4 : 4 3 8 5 : 2
              5   :     3 1   :1
             ________________
                   :   8 1 2 : 4
Thus 43852 ÷ 54 gives Q = 812 and R = 4.
Consider the algebraic proof for the above problem. The divisor 54 can be represented by 5x+4, where x=10
The dividend 43852 can be written algebraically as 43x³ + 8x² + 5x + 2
since x³ = 10³ = 1000, x² = 10² = 100.
Now the division is as follows.

            5x + 4 ) 43x³+ 8x²+ 5x + 2 ( 8x²+ x + 2
                        43x³+ 32x²
                       _________________
                         3x³– 24x²
                      = 6x²+ 5x        (

3x³ = 3 . x . x²
                         5x²+ 4x                    = 3 . 10x²= 30 x²)
                       _________________
                           x²+ x
                      = 11x + 2        (

x²= x . x = 10x )
                         10x + 8
                       __________________
                            x – 6

                             = 10 – 6
                             = 4.
Observe the following steps:
1. 43x³ ÷ 5x gives first quotient term 8x² , remainder = 3x³ - 24x² which really mean 30x² + 8x² - 32x² = 6x².
    Thus in step 2 of the problem 43852 ÷ 54, we get Q= 8 and N.D = 6.
2. 6x² ÷ 5x gives second quotient term x, remainder = x² + x which really mean 10x + x = 11x.
    Thus in step 3 & Step 4, we get Q=1and N.D =11.
3. 11x ÷ 5x gives third quotient term 2, remainder = x - 6 , which really mean the final remainder 10-6=4.
Example 5: Divide 237963 ÷ 524
Step1: We take the divisor 524 as 5, the operator and 24, the flag-digit and proceed as in the above example. We now seperate the dividend into two parts where the RHS part contains two digits for Remainder.
Thus
            24 : 2 3 7 9 : 63
           5
Step2:
    i) 23÷5 gives Q = 4 and R = 3, G.D = 37.

ii) N.D is obtained as

= 37 – ( 8 + 0)
                              = 29.
Representation
                            24 : 2 3 7 9 : 63
                           5          3
                          _________________
                                : 4
Step3:
    i) N.D ÷ Operator = 29 ÷ 5 gives Q = 5, R = 4 and G.D = 49.

ii) N.D is obtained as

                               = 49 – (10 + 16)
                                = 49 – 26
                                  = 23.

            i.e.,
                   24 : 2 3 7 9 : 63
                  5    :     3 4   :
                 _________________
                        : 4   5     :
Step 4:
    i) N.D ÷ Operator = 23 ÷ 5 gives Q = 4, R = 3 and G.D = 363.
    Note that we have reached the remainder part thus 363 is total sub–remainder.

                    24 : 2 3 7 9 : 63
                   5    :     3 4   :3
                  _________________
                         :   4 5 4 :
Step 5: We find the final remainder as follows. Subtract the cross-product of the two, falg-digits and two last quotient-digits and then vertical product of last flag-digit with last quotient-digit from the total sub-remainder.

i.e.,,

Note that 2, 4 are two falg digits: 5, 4 are two last quotient digits:

represents the last flag - digit and last quotient digit.

Thus the division 237963 ÷ 524 gives Q = 454 and R = 67.

Thus the Vedic process of division which is also called as Straight division is a simple application of urdhva-tiryak together with dhvajanka. This process has many uses along with the one-line presentation of the answer.

MISCELLANEOUS ITEMS

1. Straight Squaring:
We have already noticed methods useful to find out squares of numbers. But the methods are useful under some situations and conditions only. Now we go to a more general formula.
The sutra Dwandwa-yoga (Duplex combination process) is used in two different meanings. They are i) by squaring ii) by cross-multiplying.
We use both the meanings of Dwandwa-yoga in the context of finding squares of numbers as follows:
We denote the Duplex of a number by the symbol D. We define for a single digit ‘a’, D =a². and for a two digit number of the form ‘ab’, D =2( a x b ). If it is a 3 digit number like ‘abc’, D =2( a x c ) + b².
For a 4 digit number ‘abcd’, D = 2( a x d ) + 2( b x c ) and so on. i.e. if the digit is single central digit, D represents ‘square’: and for the case of an even number of digits equidistant from the two ends D represent the double of the cross- product.
Consider the examples:

Number		DuplexD

3		3²= 9
6		6²= 36
23		2 (2 x 3) = 12
64		2 (6 x 4) = 48
128		2 (1 x 8) + 2² = 16 + 4 = 20
305		2 (3 x 5) + 0² = 30 + 0 = 30
4231		2 (4 x 1) + 2 (2 x 3) = 8 + 12 = 20
7346		2 (7 x 6) + 2 (3 x 4) = 84 + 24 = 108

Further observe that for a n- digit number, the square of the number contains 2n or 2n-1 digits. Thus in this process, we take extra dots to the left one less than the number of digits in the given numbers.
Examples:1 62² Since number of digits = 2, we take one extra dot to the left. Thus

.62         for 2, D = 2² = 4
____
644         for 62, D = 2 x 6 x 2 = 24

32 for 62, D = 2(0 x 2) + 6²

^_____= 36
3844

62² = 3844.

Examples:2  234² Number of digits = 3. extradots =2 Thus

                ..234    for 4, D = 4² = 16
               _____
               42546    for 34, D = 2 x 3 x 4 = 24
               1221      for 234, D = 2 x 2 x 4 + 3² = 25
               _____
               54756     for .234, D = 2.0.4 + 2.2.3 = 12
                            for ..234, D = 2.0.4 + 2.0.3 + 2² = 4
Examples:3 1426². Number of digits = 4, extra dots = 3

                i.e
                       ...1426            6, D = 36
                     ________
                     1808246           26, D = 2.2.6 = 24
                       22523           426, D = 2.4.6 + 2² = 52
                    _________
                     2033476        1426, D = 2.1.6 + 2.4.2 = 28
                                         .1426, D = 2.0.6 + 2.1.2 + 4² = 20
                                        ..1426, D = 2.0.6 + 2.0.2 + 2.1.4 = 8
                                       ...1426, D = 1² = 1

                    Thus 1426² = 2033476.
With a little bit of practice the results can be obtained mentally as a single line answer.
Algebraic Proof:
Consider the first example 62²

        Now 62² = (6 x 10 + 2)² = (10a + b)² where a = 6, b = 2
                                             = 100a² + 2.10a.b + b²
                                             = a² (100) + 2ab (10) + b²
i.e. b² in the unit place, 2ab in the 10^th place and a² in the 100^th place i.e. 2² = 4 in units place, 2.6.2 = 24 in the 10^th place (4 in the 10^th place and with carried over to 100^th place). 6²=36 in the 100^th place and with carried over 2 the 100^th place becomes 36+2=38.
    Thus the answer 3844.

2.CUBING
Take a two digit number say 14.

i) Find the ratio of the two digits i.e. 1:4

ii) Now write the cube of the first digit of the number i.e. 1³

iii) Now write numbers in a row of 4 terms in such a way that the first one is the cube of the first digit and remaining three are obtained in a geometric progression with common ratio as the ratio of the original two digits (i.e. 1:4) i.e. the row is

                          1    4    16    64.

iv) Write twice the values of 2^nd and 3^rd terms under the terms respectively in second row.

i.e.,
        1    4    16    64
              8    32           (

2 x 4 = 8, 2 x 16 = 32)

v) Add the numbers column wise and follow carry over process.

1 4 16 64

Since 16 + 32 + 6 (carryover) = 54
              8    32           4 written and 5 (carryover) + 4 + 8 = 17
        ______________
        2    7     4     4

7 written and 1 (carryover) + 1 = 2.

This 2744 is nothing but the cube of the number 14

Example 1: Find 18³

Example 2: Find 33³

Algebraic Proof:
Let a and b be two digits.
Consider the row a³     a²b     ab²     b³ the first is a³ and the numbers are in the ratio a:b
    since a³:a²b=a²b:b³=a:b
Now twice of a²b, ab² are 2a²b, 2ab²

            a³+   a²b + ab²+ b³
                   2a²b + 2ab^{2
              ________________________________}
            a³+ 3a²b + 3ab²+ b³ = (a + b)³.
Thus cubes of two digit numbers can be obtained very easily by using the vedic sutra ‘anurupyena’. Now cubing can be done by using the vedic sutra ‘Yavadunam’.
Example 3:   Consider 106³.
i) The base is 100 and excess is 6. In this context we double the excess and then add.
    i.e. 106 + 12 = 118. (

2 X 6 =12 )
    This becomes the left - hand - most portion of the cube.
    i.e. 106³ = 118 / - - - -
ii) Multiply the new excess by the initial excess

         i.e. 18 x 6 = 108 (excess of 118 is 18)
    Now this forms the middle portion of the product of course 1 is carried over, 08 in the middle.

         i.e. 106³ = 118 / 08 / - - - - -
                             1
iii) The last portion of the product is cube of the initial excess.
    i.e. 6³ = 216.
    16 in the last portion and 2 carried over.

         i.e. 106³ = 118 / 081 / 16 = 1191016
                             1      2
Example 4:  Find 1002³.
i) Base = 1000. Excess = 2. Left-hand-most portion of the cube becomes 1002+(2x2)=1006.
ii) New excess x initial excess = 6 x 2 = 12.
    Thus 012 forms the middle portion of the cube.
iii) Cube of initial excess = 2³ = 8.
    So the last portion is 008.
    Thus 1002³= 1006 / 012 / 008 = 1006012008.
Example 5:  Find 94³.
i) Base = 100, deficit = -6. Left-hand-most portion of the cube becomes 94+(2x-6)=94-12=82.
ii) New deficit x initial deficit = -(100-82)x(-6)=-18x-6=108
    Thus middle potion of the cube = 08 and 1 is carried over.
iii) Cube of initial deficit = (-6)³ = -216
                                          __                  __
            Now 94³= 82 / 08 / 16 = 83 / 06 / 16
                                     _
                              1     2
                                            = 83 / 05 / (100 – 16)
                                            = 830584.

3. Equation of Straight line passing through two given points:
To find the equation of straight line passing through the points (x₁, y₁) and (x₂, y₂) , we generally consider one of the following methods.
1. General equation y = mx + c.
    It is passing through (x₁, y₁) then y₁ = mx₁ + c.
    It is passing through (x₂, y₂) also, then y₂ = mx₂ + c.
    Solving these two simultaneous equations, we get ‘m’ and ‘c’ and so the equation.
2. The formula
                                             (y₂ - y₁)
                                y – y₁ = ________ (x – x₁)   and substitution.
                                              (x₂ - x₁)
Some sequence of steps gives the equation. But the paravartya sutra enables us to arrive at the conclusion in a more easy way and convenient to work mentally.
Example1:  Find the equation of the line passing through the points (9,7) and (5,2).
Step1: Put the difference of the y - coordinates as the x - coefficient and vice - versa.
          i.e. x coefficient = 7 - 2 = 5
          y coefficient = 9 - 5 = 4.
          Thus L.H.S of equation is 5x - 4y.
Step 2: The constant term (R.H.S) is obtained by substituting the co-ordinates of either of the given points in
        L.H.S (obtained through step-1)
        i.e. R.H.S of the equation is

             5(9) - 4(7) = 45 - 28 = 17

            or 5(5) - 4(2) = 25 - 8 = 17.

            Thus the equation is 5x - 4y = 17.
Example 2: Find the equation of the line passing through (2, -3) and (4,-7).
Step 1 : x[-3-(-7)] –y[2-4] = 4x + 2y.
Step 2 : 4(2) + 2(-3) = 8 –6 = 2.
Step 3 : Equation is 4x + 2y =2 or 2x +y = 1.
Example 3 : Equation of the line passing through the points (7,9) and (3,-7).
Step 1 : x[9 - (-7)] – y(7 - 3) = 16x - 4y.
Step 2 : 16(7) - 4(9) = 112 – 36 = 76
Step 3 : 16x- 4y = 76 or 4x – y = 19

Find the equation of the line passing through the points using Vedic methods.

1. (1, 2), (4,-3)

2. (5,-2), (5,-4)

3. (-5, -7), (13,2)

4. (a, o) , (o,b)

GUNÌTA SAMUCCAYAH - SAMUCCAYA GUŅÌTAH

In connection with factorization of quadratic expressions a sub-Sutra, viz. 'Gunita samuccayah-Samuccaya Gunitah' is useful. It is intended for the purpose of verifying the correctness of obtained answers in multiplications, divisions and factorizations. It means in this context:
'The product of the sum of the coefficients sc in the factors is equal to the sum of the coefficients sc in the product'
Symbolically we represent as sc of the product = product of the sc (in the factors)
Example 1:     (x + 3) (x + 2) = x2 + 5x + 6

        Now ( x + 3 ) ( x + 2 ) = 4 x 3 = 12 : Thus verified.
Example 2:     (x – 4) (2x + 5) = 2x² – 3x – 20

        Sc of the product 2 – 3 – 20 = - 21

        Product of the Sc = (1 – 4) (2 + 5) = (-3) (7) = - 21. Hence verified.

        In case of cubics, biquadratics also the same rule applies.

        We have (x + 2) (x + 3) (x + 4) = x³ + 9x² + 26x + 24

        Sc of the product = 1 + 9 + 26 + 24 = 60

        Product of the Sc = (1 + 2) (1 + 3) (1 + 4)

                                = 3 x 4 x 5 = 60. Verified.
Example 3:   (x + 5) (x + 7) (x – 2) = x³ + 10x² + 11x – 70

                        (1 + 5) (1 + 7) (1 – 2) = 1 + 10 + 11 – 70

        i.e.,     6 x 8 x –1 = 22 – 70
        i.e.,     -48 = -48 Verified.
We apply and interpret So and Sc as sum of the coefficients of the odd powers and sum of the coefficients of the even powers and derive that So = Sc gives (x + 1) is a factor for thee concerned expression in the variable x. Sc = 0 gives (x - 1) is a factor.

Verify whether the following factorization of the expressions are correct or not by the Vedic check:
     i.e. Gunita. Samuccayah-Samuccaya Gunitah:

    1.     (2x + 3) (x – 2) = 2x² – x - 6

    2.     12x² – 23xy + 10y² = ( 3x – 2y ) ( 4x – 5y )

    3.     12x² + 13x – 4 = ( 3x – 4 ) ( 4x + 1 )

    4.     ( x + 1 ) ( x + 2 ) ( x + 3 ) = x³ + 6x² + 11x + 6

    5.     ( x + 2 ) ( x + 3 ) ( x + 8 ) = x³ + 13x² + 44x + 48

LOPANASTHĀPANĀBHYĀM

Lopana sthapanabhyam means 'by alternate elimination and retention'.
Consider the case of factorization of quadratic equation of type ax² + by² + cz² + dxy + eyz + fzx This is a homogeneous equation of second degree in three variables x, y, z. The sub-sutra removes the difficulty and makes the factorization simple. The steps are as follows:

i) Eliminate z by putting z = 0 and retain x and y and factorize thus obtained a quadratic in x and y by means of ‘adyamadyena’ sutra.;

ii) Similarly eliminate y and retain x and z and factorize the quadratic in x and z.

iii) With these two sets of factors, fill in the gaps caused by the elimination process of z and y respectively. This gives actual factors of the expression.

Example 1: 3x² + 7xy + 2y² + 11xz + 7yz + 6z².

Step (i):   Eliminate z and retain x, y; factorize
                3x² + 7xy + 2y² = (3x + y) (x + 2y)

Step (ii):   Eliminate y and retain x, z; factorize
                3x² + 11xz + 6z² = (3x + 2z) (x + 3z)

Step (iii):   Fill the gaps, the given expression
                 = (3x + y + 2z) (x + 2y + 3z)

Example 2: 12x² + 11xy + 2y² - 13xz - 7yz + 3z².

Step (i):   Eliminate z i.e., z = 0; factorize
                12x² + 11xy + 2y² = (3x + 2y) (4x + y)

Step (ii): Eliminate y i.e., y = 0; factorize
              12x² - 13xz + 3z² = (4x -3z) (3x – z)

Step (iii): Fill in the gaps; the given expression
                = (4x + y – 3z) (3x + 2y – z)

Example 3: 3x²+6y²+2z²+11xy+7yz+6xz+19x+22y+13z+20

Step (i): Eliminate y and z, retain x and independent term
              i.e., y = 0, z = 0 in the expression (E).
        Then E = 3x² + 19x + 20 = (x + 5) (3x + 4)

Step (ii): Eliminate z and x, retain y and independent term
               i.e., z = 0, x = 0 in the expression.
        Then E = 6y² + 22y + 20 = (2y + 4) (3y + 5)

Step (iii): Eliminate x and y, retain z and independent term
                i.e., x = 0, y = 0 in the expression.
        Then E = 2z² + 13z + 20 = (z + 4) (2z + 5)

Step (iv):   The expression has the factors (think of independent terms)
                 = (3x + 2y + z + 4) (x + 3y + 2z + 5).

In this way either homogeneous equations of second degree or general equations of second degree in three variables can be very easily solved by applying ‘adyamadyena’ and ‘lopanasthapanabhyam’ sutras.

Solve the following expressions into factors by using appropriate sutras:

1.    x² + 2y² + 3xy + 2xz + 3yz + z².
2.    3x² + y² - 4xy - yz - 2z² - zx.
3.    2p² + 2q² + 5pq + 2p – 5q - 12

4.    u² + v² – 4u + 6v – 12.
5.    x² - 2y² + 3xy + 4x - y + 2.
6.    3x² + 4y² + 7xy - 2xz - 3yz - z² + 17x + 21y – z + 20.

Highest common factor:
To find the Highest Common Factor i.e. H.C.F. of algebraic expressions, the factorization method and process of continuous division are in practice in the conventional system. We now apply' Lopana - Sthapana' Sutra, the 'Sankalana vyavakalanakam' process and the 'Adyamadya' rule to find out the H.C.F in a more easy and elegant way.
Example 1:   Find the H.C.F. of x² + 5x + 4 and x² + 7x + 6.

      1. Factorization method:

                x² + 5x + 4 = (x + 4) (x + 1)
                x² + 7x + 6 = (x + 6) (x + 1)
                H.C.F. is ( x + 1 ).

     2. Continuous division process.

         x²+ 5x + 4 ) x²+ 7x + 6 ( 1
                             x²+ 5x + 4
                          ___________
                                  2x + 2 ) x²+ 5x + 4 ( ½x
                                              x²+ x
                                             __________
                                                    4x + 4 ) 2x + 2 ( ½
                                                                2x + 2
                                                                ______
                                                                    0

            Thus 4x + 4 i.e., ( x + 1 ) is H.C.F.
3. Lopana - Sthapana process i.e. elimination and retention or alternate destruction of the highest and the lowest powers is as below:

i.e.,, (x + 1) is H.C.F

Example 2: Find H.C.F. of 2x² – x – 3 and 2x² + x – 6

Example 3: x³ – 7x – 6 and x³ + 8x² + 17x + 10.

Now by Lopana - Sthapana and Sankalana – Vyavakalanabhyam

Example 4: x³ + 6x² + 5x – 12 and x³ + 8x² + 19x + 12.

(or)

Example 5:    2x³ + x² – 9 and x⁴ + 2x² + 9.

        By Vedic sutras:

                Add: (2x³ + x² – 9) + (x⁴ + 2x² + 9)
                         = x⁴ + 2x³ + 3x².

                  ÷ x² gives      x² + 2x + 3 ------ (i)

       Subtract after multiplying the first by x and the second by 2.

            Thus   (2x⁴ + x³ – 9x) - (2x⁴ + 4x² + 18)
                      = x³ - 4x² – 9x – 18 ------ ( ii )

       Multiply (i) by x and subtract from (ii)

                x³ – 4x² – 9x – 18 – (x³ + 2x² + 3x)
                 = - 6x² – 12x – 18

                 ÷ - 6 gives     x² + 2x + 3.
Thus ( x² + 2x + 3 ) is the H.C.F. of the given expressions.
Algebraic Proof:
Let P and Q be two expressions and H is their H.C.F. Let A and B the Quotients after their division by H.C.F.

               P                   Q
        i.e., __ = A and   __ = B     which gives P = A.H and Q = B.H
               H                   H

        P + Q = AH + BH and P – Q = AH – BH
                 = (A+B).H                 = (A–B).H
Thus we can write P ± Q = (A ± B).H
Similarly MP = M.AH and NQ = N.BH gives MP ± NQ = H (MA ± NB)
This states that the H.C.F. of P and Q is also the H.C.F. of P±Q or MA±NB.
i.e. we have to select M and N in such a way that highest powers and lowest powers (or independent terms) are removed and H.C.F appears as we have seen in the examples.

Find the H.C.F. in each of the following cases using Vedic sutras:

        1.   x² + 2x – 8, x² – 6x + 8

        2.   x³ – 3x² – 4x + 12, x³ – 7x² + 16x - 12

        3.   x³ + 6x² + 11x + 6, x³ – x² - 10x - 8

        4.   6x⁴ – 11x³ + 16x² – 22x + 8,
              6x⁴ – 11x³ – 8x² + 22x – 8.

ANTYAYOREVA

'Atyayoreva' means 'only the last terms'. This is useful in solving simple equations of the following type.
The type of equations are those whose numerator and denominator on the L.H.S. bearing the independent terms stand in the same ratio to each other as the entire numerator and the entire denominator of the R.H.S. stand to each other.
Let us have a look at the following example.
Example 1:

                        x²+ 2x + 7          x + 2
                      __________    =   _____
                        x²+ 3x + 5          x + 3
In the conventional method we proceed as

                        x²+ 2x + 7          x + 2
                      __________    =   _____
                        x²+ 3x + 5          x + 3

                    (x + 3) (x²+ 2x + 7) = (x + 2) (x²+ 3x + 5)
        x³+ 2x²+ 7x + 3x²+ 6x + 21 = x³+ 3x²+ 5x + 2x²+ 6x + 10
                     x³+ 5x²+ 13x + 21 = x³+ 5x²+ 11x + 10

        Canceling like terms on both sides

                 13x + 21 = 11x + 10
                 13x – 11x = 10 – 21
                           2x = -11
                            x = -11 / 2
Now we solve the problem using anatyayoreva.

                         x²+ 2x + 7          x + 2
                      __________    =   _____
                        x²+ 3x + 5          x + 3

     Consider
                          x²+ 2x + 7         x + 2
                      __________    =   _____
                        x²+ 3x + 5           x + 3

     Observe that

                 x²+ 2x      x (x + 2)       x + 2
                 ______ = ________ = _____
                 x²+ 3x      x (x + 3)       x + 3

     This is according to the condition in the sutra. Hence from the sutra

                    x + 2         7
                  _____   = __
                    x + 3         5

                5x + 10 = 7x + 21
                7x – 5x = -21 + 10
                       2x = -11
                        x = -11 / 2
Algebraic Proof:

        Consider the equation

                AC + D         A
                ______   = ___   ------------- (i)
                BC + E         B
    This satisfies the condition in the sutra since

                    AC         A
                    ___   = ___
                    BC          B
    Now cross–multiply the equation (i)

                B (AC + D) = A (BC + E)
                  BAC + BD = ABC + AE
                           BD = AE which gives

                        A        D
                       __   = __    --------(ii)
                        B        E
i.e., the result obtained in solving equation (i) is same as the result obtained in solving equation (ii).
Example 2: solve
                                2x²+ 3x + 10        2x + 3
                                ___________   =    _____
                                3x²+ 4x + 14        3x + 4
Since
                2x²+ 3x        x (2x + 3)        2x+3
                _______    = ________    =   ____
                3x²+ 4x        x (3x + 4)        3x+4
We can apply the sutra.

                2x + 3         10
                _____    =    __
                3x+4           14
Cross–multiplying

            28x + 42 = 30x + 40
            28x – 30x = 40 – 42
                    -2x = -2

x = -2 / -2 = 1.
Let us see the application of the sutra in another type of problem.
Example 3: (x + 1) (x + 2) (x + 9) = (x + 3) (x + 4) (x + 5)

        Re–arranging the equation, we have

                (x + 1) (x + 2)          x + 3
                ____________    =    _____
                (x + 4) (x + 5)          x + 9
    i.e.,
                x²+ 3x + 2x + 3
            = ______________
                x²+ 9x + 20x + 9
Now
         x²+ 3x      x (x + 3)       x + 3
         ______  = _______   = _____    gives the solution by antyayoreva
         x²+ 9x      x (x + 9)       x + 9
Solution is obtained from

                    x + 3          2
                    ____    =    __
                    x + 9         20

                20x + 60 = 2x + 18
                20x – 2x = 18 – 60
                       18x = -42

x = -42 / 18 = -7 / 3.

    Once again look into the problem

        (x + 1) (x + 2) (x + 9) = (x + 3) (x + 4) (x + 5)

    Sum of the binomials on each side

            x + 1 + x + 2 + x + 9 = 3x + 12
            x + 3 + x + 4 + x + 5 = 3x + 12
It is same. In such a case the equation can be adjusted into the form suitable for application of antyayoreva.
Example 4: (x + 2) (x + 3) (x + 11) = (x + 4) (x + 5) (x + 7)

        Sum of the binomials on L.H.S. = 3x + 16
        Sum of the binomials on R.H.S. = 3x + 16

        They are same. Hence antyayoreva can be applied. Adjusting we get

        (x + 2) (x + 3)        x + 5         2 x 3           6
        ____________    = _____   =  _____   =   ___
        (x + 4) (x + 7)        x + 11       4 x 7          28

            28x + 140 = 6x + 66
            28x – 6x = 66 – 140
                   22x = -74

                    -74        -37
               x = ___    = ___
                     22         11

Solve the following problems using ‘antyayoreva’

        1.
                3x²+ 5x + 8             3x + 5
                __________     =     ______
                5x²+ 6x +12            5x + 6

        2.
               4x²+ 5x + 3              4x + 5
               __________       =    ______
               3x²+ 2x + 4              3x + 2

        3.     (x + 3) (x + 4) (x + 6) = (x + 5) (x + 1) (x + 7)

        4.     (x + 1) (x + 6) (x + 9) = (x + 4) (x + 5) (x + 7)

        5.
               2x²+ 3x + 9            2x + 3
               __________     =    ______
               4x²+5x+17             4x + 5

YĀVADŨNAM TĀVADŨNĪKŖTYA
VARGAÑCA YOJAYET

The meaning of the Sutra is 'what ever the deficiency subtract that deficit from the number and write along side the square of that deficit'.
This Sutra can be applicable to obtain squares of numbers close to bases of powers of 10.
Method-1 : Numbers near and less than the bases of powers of 10.

     Eg 1: 9² Here base is 10.

        The answer is separated in to two parts by a’/’

        Note that deficit is 10 - 9 = 1

        Multiply the deficit by itself or square it

        1² = 1. As the deficiency is 1, subtract it from the number i.e., 9–1 = 8.

        Now put 8 on the left and 1 on the right side of the vertical line or slash i.e., 8/1.

        Hence 81 is answer.
Eg. 2: 96² Here base is 100.

        Since deficit is 100-96=4 and square of it is 16 and the deficiency subtracted from the number 96 gives 96-4 = 92, we get the answer 92 / 16 Thus 96² = 9216.
Eg. 3: 994² Base is 1000

        Deficit is 1000 - 994 = 6. Square of it is 36.

        Deficiency subtracted from 994 gives 994 - 6 = 988

        Answer is 988 / 036     [since base is 1000]
Eg. 4:   9988²   Base is 10,000.

        Deficit = 10000 - 9988 = 12.

        Square of deficit = 12² = 144.

        Deficiency subtracted from number = 9988 - 12 = 9976.

        Answer is 9976 / 0144     [since base is 10,000].
Eg. 5:   88²   Base is 100.

        Deficit = 100 - 88 = 12.

        Square of deficit = 12² = 144.

        Deficiency subtracted from number = 88 - 12 = 76.

        Now answer is 76 / ₁44 = 7744    [since base is 100]
Algebraic proof:
The numbers near and less than the bases of power of 10 can be treated as (x-y), where x is the base and y, the deficit.

        Thus
            (1) 9 = (10 -1) (2) 96 = ( 100-4) (3) 994 = (1000-6)

            (4) 9988 = (10000-12 ) (v) 88 = (100-12)

            ( x – y )² = x² – 2xy + y²
                         = x ( x – 2y ) + y²
                         = x ( x – y – y ) + y²
                         = Base ( number – deficiency ) + ( deficit )²
        Thus

            985² = ( 1000 – 15 )²
                   = 1000 ( 985 – 15 ) + (15)²
                   = 1000 ( 970 ) + 225
                   = 970000 + 225
                   = 970225.
or we can take the identity a² - b² = (a + b) ( a - b) and proceed as

                  a² - b² = (a + b) ( a - b).

        gives          a²= (a + b) ( a - b) + b²

    Thus for a = 985 and b = 15;

        a²= (a + b) ( a - b) + b²

    985² = ( 985 + 15 ) ( 985 - 15 ) + (15)²

           = 1000 ( 970 ) + 225

           = 970225.
Method. 2 : Numbers near and greater than the bases of powers of 10.

      Eg.(1): 13² .

        Instead of subtracting the deficiency from the number we add and proceed as in Method-1.

        for 13² , base is 10, surplus is 3.

        Surplus added to the number = 13 + 3 = 16.

        Square of surplus = 3² = 9

        Answer is 16 / 9 = 169.

      Eg.(2):   112²Base = 100, Surplus = 12,

        Square of surplus = 12² = 144

        add surplus to number = 112 + 12 = 124.

        Answer is 124 / ₁44 = 12544

    Or think of identity a² = (a + b) (a – b) + b²   for a = 112, b = 12:

                112² = (112 + 12) (112 – 12) + 12²
                        = 124 (100) + 144
                        = 12400 + 144
                        = 12544.

            (x + y)² = x² + 2xy + y²
                         = x ( x + 2y ) + y²
                         = x ( x + y + y ) + y²

        = Base ( Number + surplus ) + ( surplus )²

       gives
            112² = 100 ( 112 + 12 ) + 12²
                    = 100 ( 124 ) + 144
                    = 12400 + 144
                    = 12544.

     Eg. 3:   10025²

                = ( 10025 + 25 ) / 25²

                = 10050 / 0625     [ since base is 10,000 ]

                = 100500625.
Method - 3: This is applicable to numbers which are near to multiples of 10, 100, 1000 .... etc. For this we combine the upa-Sutra 'anurupyena' and 'yavadunam tavadunikritya varganca yojayet' together.
Example 1:   388² Nearest base = 400.

        We treat 400 as 4 x 100. As the number is less than the base we proceed as follows

        Number 388, deficit = 400 - 388 = 12

        Since it is less than base, deduct the deficit

            i.e. 388 - 12 = 376.

        multiply this result by 4 since base is 4 X 100 = 400.

                    376 x 4 = 1504

        Square of deficit = 12² = 144.

        Hence answer is 1504 / ₁44 = 150544    [since we have taken multiples of 100].

Example 2: 485² Nearest base = 500.

Treat 500 as 5 x 100 and proceed

Example 3:   67² Nearest base = 70

Example 4: 416² Nearest ( lower ) base = 400

Here surplus = 16 and 400 = 4 x 100

Example 5: 5012² Nearest lower base is 5000 = 5 x 1000

Surplus = 12

Apply yavadunam to find the following squares.

            1. 7²              2. 98²            3. 987²           4. 14²

            5. 116²         6. 1012²        7. 19²             8. 475²

            9. 796²       10. 108²        11. 9988²      12. 6014².

So far we have observed the application of yavadunam in finding the squares of number. Now with a slight modification yavadunam can also be applied for finding the cubes of numbers.

     Cubing of Numbers:

      Example : Find the cube of the number 106.

      We proceed as follows:

i)   For 106, Base is 100. The surplus is 6.

    Here we add double of the surplus i.e. 106+12 = 118.

    (Recall in squaring, we directly add the surplus)

    This makes the left-hand -most part of the answer.

            i.e. answer proceeds like 118 / - - - - -

ii) Put down the new surplus i.e. 118-100=18 multiplied by the initial surplus

            i.e. 6=108.

Since base is 100, we write 108 in carried over form 108 i.e. .

As this is middle portion of the answer, the answer proceeds like 118 / ₁08 /....

iii) Write down the cube of initial surplus i.e. 6³ = 216 as the last portion

          i.e. right hand side last portion of the answer.

     Since base is 100, write 216 as ₂16 as 2 is to be carried over.

     Answer is 118 / ₁08 / ₂16

    Now proceeding from right to left and adjusting the carried over, we get the answer

                    119 / 10 / 16 = 1191016.
Eg.(1):   102³ = (102 + 4) / 6 X 2 / 2³

                          =   106      =    12 = 08

                          = 1061208.

        Observe initial surplus = 2, next surplus =6 and base = 100.
Eg.(2):   94³

        Observe that the nearest base = 100. Here it is deficit contrary to the above examples.

i) Deficit = -6. Twice of it -6 X 2 = -12
add it to the number = 94 -12 =82.

ii) New deficit is -18.
Product of new deficit x initial deficit = -18 x -6 = 108

iii) deficit³ = (-6)³ = -216.

__
Hence the answer is 82 / ₁08 / -₂16

Since 100 is base 1 and -2 are the carried over. Adjusting the carried over in order, we get the answer

            ( 82 + 1 ) / ( 08 – 03 ) / ( 100 – 16 )

                = 83   /    = 05      /     = 84          = 830584
      __
     16 becomes 84 after taking1 from middle most portion i.e. 100. (100-16=84).
                                                                                     _
        Now 08 - 01 = 07 remains in the middle portion, and 2 or 2 carried to it makes the middle as 07 - 02 = 05. Thus we get the above result.
Eg.(3):
              998³ Base = 1000; initial deficit = - 2.

              998³= (998 – 2 x 2) / (- 6 x – 2) / (- 2)³

                      =    994         /     = 012      / = -008

                      =    994 / 011 / 1000 - 008

                     =     994 / 011 / 992

                     =     994011992.

Find the cubes of the following numbers using yavadunam sutra.

1. 105 2. 114 3. 1003 4. 10007 5. 92

6. 96 7. 993 8. 9991 9. 1000008 10. 999992.

EKANYŨŅENA PŨRVENA

The Sutra Ekanyunena purvena comes as a Sub-sutra to Nikhilam which gives the meaning 'One less than the previous' or 'One less than the one before'.

     1) The use of this sutra in case of multiplication by 9,99,999.. is as follows .

    Method :
a) The left hand side digit (digits) is ( are) obtained by applying the ekanyunena purvena i.e. by deduction 1 from the left side digit (digits) .
         e.g. ( i ) 7 x 9; 7 – 1 = 6 ( L.H.S. digit )
b) The right hand side digit is the complement or difference between the multiplier and the left hand side digit (digits) . i.e. 7 X 9 R.H.S is 9 - 6 = 3.
c) The two numbers give the answer; i.e. 7 X 9 = 63.
Example 1:   8 x 9  Step ( a ) gives 8 – 1 = 7 ( L.H.S. Digit )
                                  Step ( b ) gives 9 – 7 = 2 ( R.H.S. Digit )
                                  Step ( c ) gives the answer 72
Example 2:  15 x 99   Step ( a ) : 15 – 1 = 14
                                      Step ( b ) : 99 – 14 = 85 ( or 100 – 15 )
                                      Step ( c ) : 15 x 99 = 1485
Example 3:   24 x 99
        Answer :

Example 4: 356 x 999
Answer :

Example 5:     878 x 9999
        Answer :

Note the process : The multiplicand has to be reduced by 1 to obtain the LHS and the rightside is mechanically obtained by the subtraction of the L.H.S from the multiplier which is practically a direct application of Nikhilam Sutra.

    Now by Nikhilam

                           24 – 1 = 23    L.H.S.
                        x 99 – 23 = 76    R.H.S. (100–24)
                        _____________________________
                                   23 / 76                             = 2376
Reconsider the Example 4:

                          356 – 1    = 355    L.H.S.
                       x 999 – 355 = 644    R.H.S.
                      ________________________
                                 355 / 644                    =   355644
and in Example 5: 878 x 9999 we write

                    0878 – 1   = 877     L.H.S.
                 x 9999 – 877 = 9122    R.H.S.
                __________________________
                              877 / 9122                = 8779122
Algebraic proof :
As any two digit number is of the form ( 10x + y ), we proceed
                   ( 10x + y ) x 99
                 = ( 10x + y ) x ( 100 – 1 )
                 = 10x . 10² – 10x + 10² .y – y
                 = x . 10³ + y . 10² – ( 10x + y )
                 = x . 10³ + ( y – 1 ) . 10² + [ 10² – ( 10x + y )]
    Thus the answer is a four digit number whose 1000^th place is x, 100^th place is ( y - 1 ) and the two digit number which makes up the 10^th and unit place is the number obtained by subtracting the multiplicand from 100.(or apply nikhilam).

        Thus in 37 X 99. The 1000^th place is x i.e. 3

        100^th place is ( y - 1 ) i.e. (7 - 1 ) = 6

        Number in the last two places 100-37=63.

        Hence answer is 3663.

    Apply Ekanyunena purvena to find out the products

        1. 64 x 99           2. 723 x 999           3. 3251 x 9999

        4. 43 x 999         5. 256 x 9999         6. 1857 x 99999

We have dealt the cases
   i) When the multiplicand and multiplier both have the same number of digits
        ii) When the multiplier has more number of digits than the multiplicand.
In both the cases the same rule applies. But what happens when the multiplier has lesser digits?
i.e. for problems like 42 X 9, 124 X 9, 26325 X 99 etc.,
For this let us have a re-look in to the process for proper understanding.
Multiplication table of 9.
                                        a    b
                             2 x 9 =    1    8
                             3 x 9 =    2    7
                             4 x 9 =    3    6
                             - - - - - - - - - -
                             8 x 9 =    7    2
                             9 x 9 =    8    1
                           10 x 9 =    9    0
Observe the left hand side of the answer is always one less than the multiplicand (here multiplier is 9) as read from Column (a) and the right hand side of the answer is the complement of the left hand side digit from 9 as read from Column (b)
Multiplication table when both multiplicand and multiplier are of 2 digits.
                          a     b
               11 x 99 = 10    89 = (11–1) / 99 – (11–1) = 1089
               12 x 99 = 11    88 = (12–1) / 99 – (12–1) = 1188
               13 x 99 = 12    87 = (13–1) / 99 – (13–1) = 1287
              -------------------------------------------------
               18 x 99 = 17    82 ----------------------------
               19 x 99 = 18    81
               20 x 99 = 19    80 = (20–1) / 99 – (20–1) = 1980
The rule mentioned in the case of above table also holds good here
Further we can state that the rule applies to all cases, where the multiplicand and the multiplier have the same number of digits.
Consider the following Tables.
(i)
                                       a    b
                        11 x 9 =   9    9
                        12 x 9 = 10    8
                        13 x 9 = 11    7
                   ----------------------
                        18 x 9 = 16    2
                        19 x 9 = 17    1
                        20 x 9 = 18    0
(ii)
                        21 x 9 = 18    9
                        22 x 9 = 19    8
                        23 x 9 = 20    7
                   -----------------------
                        28 x 9 = 25    2
                        29 x 9 = 26    1
                        30 x 9 = 27    0
(iii)
                        35 x 9 = 31    5
                        46 x 9 = 41    4
                        53 x 9 = 47    7
                        67 x 9 = 60    3
                  -------------------------so on.
From the above tables the following points can be observed:
1) Table (i) has the multiplicands with 1 as first digit except the last one. Here L.H.S of products are uniformly 2 less than the multiplicands. So also with 20 x 9
2) Table (ii) has the same pattern. Here L.H.S of products are uniformly 3 less than the multiplicands.
3) Table (iii) is of mixed example and yet the same result i.e. if 3 is first digit of the multiplicand then L.H.S of product is 4 less than the multiplicand; if 4 is first digit of the multiplicand then, L.H.S of the product is 5 less than the multiplicand and so on.
4) The right hand side of the product in all the tables and cases is obtained by subtracting the R.H.S. part of the multiplicand by Nikhilam.
Keeping these points in view we solve the problems:
Example1 : 42 X 9
i) Divide the multiplicand (42) of by a Vertical line or by the Sign : into a right hand portion consisting of as many digits as the multiplier.

                        i.e. 42 has to be written as 4/2 or 4:2
ii) Subtract from the multiplicand one more than the whole excess portion on the left. i.e. left portion of multiplicand is 4.

                        one more than it 4 + 1 = 5.

                We have to subtract this from multiplicand
                        i.e. write it as
                                        4 : 2
                                            :-5
                                  ---------------
                                        3 : 7

            This gives the L.H.S part of the product.
This step can be interpreted as "take the ekanyunena and sub tract from the previous" i.e. the excess portion on the left.
iii) Subtract the R.H.S. part of the multiplicand by nikhilam process.
i.e. R.H.S of multiplicand is 2

                    its nikhilam is 8

                It gives the R.H.S of the product

                i.e. answer is 3 : 7 : 8 = 378.

                Thus 42 X 9 can be represented as

                            4 : 2
                               :-5 : 8
                     ------------------
                            3 : 7 : 8 = 378.
Example 2 :     124 X 9

               Here Multiplier has one digit only .

                        We write 12 : 4

              Now step (ii), 12 + 1 = 13

                          i.e.    12 : 4
                                  -1 : 3
                              ------------
              Step ( iii ) R.H.S. of multiplicand is 4. Its Nikhilam is 6

                 124 x 9 is    12 : 4
                                    -1 : 3 : 6
                            -----------------
                                   11 : 1 : 6    =    1116

        The process can also be represented as
        124 x 9 = [ 124 – ( 12 + 1 ) ] : ( 10 – 4 ) = ( 124 – 13 ) : 6 = 1116
Example 3:     15639 x 99

     Since the multiplier has 2 digits, the answer is
     [15639 – (156 + 1)] : (100 – 39) = (15639 – 157) : 61 = 1548261

Find the products in the following cases.

1. 58 x 9 2. 62 x 9 3. 427 x 99

4. 832 x 9 5. 24821 x 999 6. 111011 x 99

Ekanyunena Sutra is also useful in Recurring Decimals. We can take up this under a separate treatment.
Thus we have a glimpse of majority of the Sutras. At some places some Sutras are mentioned as Sub-Sutras. Any how we now proceed into the use of Sub-Sutras. As already mentioned the book on Vedic Mathematics enlisted 13 Upa-Sutras.
But some approaches in the Vedic Mathematics book prompted some serious research workers in this field to mention some other Upa-Sutras. We can observe those approaches and developments also.

PŨRANĀPŨRAŅĀBHYĀM

The Sutra can be taken as Purana - Apuranabhyam which means by the completion or non - completion. Purana is well known in the present system. We can see its application in solving the roots for general form of quadratic equation.
We have : ax² + bx + c = 0

                   x²+ (b/a)x + c/a = 0    ( dividing by a )

                    x²+ (b/a)x = - c/a

    completing the square ( i.e.,, purana ) on the L.H.S.

                x²+ (b/a)x + (b²/4a²) = -c/a + (b²/4a²)

                  [x + (b/2a)]² = (b² - 4ac) / 4a²
                                                                           ________
                                                                 - b ± √ b² – 4ac
        Proceeding in this way we finally get x =   _______________
                                                                          2a
    Now we apply purana to solve problems.
Example 1.   x³ + 6x² + 11 x + 6 = 0.

                    Since (x + 2 )³ = x³ + 6x² + 12x + 8
                        Add ( x + 2 ) to both sides
                    We get x³ + 6x² + 11x + 6 + x + 2 = x + 2
                        i.e.,, x³ + 6x² + 12x + 8 = x + 2
                        i.e.,, ( x + 2 )³ = ( x + 2 )
                    this is of the form y³ = y for y = x + 2
                    solution y = 0, y = 1, y = - 1
                                i.e.,, x + 2 = 0,1,-1
                    which gives x = -2,-1,-3
Example 2:     x³ + 8x² + 17x + 10 = 0

            We know ( x + 3 )³ = x³ + 9x² + 27x + 27
       So adding on the both sides, the term ( x² + 10x + 17 ), we get
                  x³ + 8x² + 17x + x² + 10x + 17 = x² + 10x + 17
                  i.e.,, x³ + 9x² + 27x + 27 = x² + 6x + 9 + 4x + 8
                  i.e.,, ( x + 3 )³ = ( x + 3 )² + 4 ( x + 3 ) – 4
                      y³ = y² + 4y – 4 for y = x + 3
                        y = 1, 2, -2.
    Hence x = -2, -1, -5
         Thus purana is helpful in factorization.
         Further purana can be applied in solving Biquadratic equations also.

Solve the following using purana – apuranabhyam.

            1.     x³ – 6x² + 11x – 6 = 0
            2.     x³ + 9x² + 23x + 15 = 0
            3.     x² + 2x – 3 = 0
            4.     x⁴ + 4x³ + 6x² + 4x – 15 = 0

SAŃKALANA – VYAVAKALANĀBHYAM

This Sutra means 'by addition and by subtraction'. It can be applied in solving a special type of simultaneous equations where the x - coefficients and the y - coefficients are found interchanged.
Example 1:
                            45x – 23y = 113
                            23x – 45y = 91
In the conventional method we have to make equal either the coefficient of x or coefficient of y in both the equations. For that we have to multiply equation ( 1 ) by 45 and equation ( 2 ) by 23 and subtract to get the value of x and then substitute the value of x in one of the equations to get the value of y or we have to multiply equation ( 1 ) by 23 and equation ( 2 ) by 45 and then subtract to get value of y and then substitute the value of y in one of the equations, to get the value of x. It is difficult process to think of.
From Sankalana – vyavakalanabhyam

        add them,
                i.e., ( 45x – 23y ) + ( 23x – 45y ) = 113 + 91
                i.e., 68x – 68y = 204

x – y = 3

      subtract one from other,
                i.e., ( 45x – 23y ) – ( 23x – 45y ) = 113 – 91
                i.e., 22x + 22y = 22

x + y = 1

      and repeat the same sutra, we get x = 2 and y = - 1
      Very simple addition and subtraction are enough, however big the coefficients may be.
Example 2:
                        1955x –   476y = 2482
                          476x – 1955y = -4913
Oh ! what a problem ! And still

                just add, 2431( x – y ) = - 2431

x – y = -1

subtract, 1479 ( x + y ) = 7395

x + y = 5

once again add, 2x = 4

x = 2

subtract - 2y = - 6

y = 3

Solve the following problems using Sankalana – Vyavakalanabhyam.

            1.     3x + 2y = 18
                    2x + 3y = 17

            2.     5x – 21y = 26
                    21x – 5y = 26

            3.     659x + 956y = 4186
                    956x + 659y = 3889

ĀNURŨPYE ŚŨNYAMANYAT

The Sutra Anurupye Sunyamanyat says : 'If one is in ratio, the other one is zero'.
We use this Sutra in solving a special type of simultaneous simple equations in which the coefficients of 'one' variable are in the same ratio to each other as the independent terms are to each other. In such a context the Sutra says the 'other' variable is zero from which we get two simple equations in the first variable (already considered) and of course give the same value for the variable.
Example 1:
                            3x + 7y = 2
                            4x + 21y = 6
Observe that the y-coefficients are in the ratio 7 : 21 i.e., 1 : 3, which is same as the ratio of independent terms i.e., 2 : 6 i.e., 1 : 3. Hence the other variable x = 0 and 7y = 2 or 21y = 6 gives y = 2 / 7
Example 2:
                            323x + 147y = 1615
                            969x + 321y = 4845
The very appearance of the problem is frightening. But just an observation and anurupye sunyamanyat give the solution x = 5, because coefficient of x ratio is
        323 : 969 = 1 : 3 and constant terms ratio is 1615 : 4845 = 1 : 3.
             y = 0 and 323 x = 1615 or 969 x = 4845 gives x = 5.

Solve the following by anurupye sunyamanyat.

        1.   12x + 78y = 12                     2.    3x + 7y = 24
              16x + 96y = 16                           12x + 5y = 96

        3.    4x – 6y = 24                         4.    ax + by = bm
               7x – 9y = 36                                 cx + dy = dm

In solving simultaneous quadratic equations, also we can take the help of the ‘sutra’ in the following way:
Example 3 :
                   Solve for x and y
                                                                 x + 4y = 10
                                     x² + 5xy + 4y² + 4x - 2y = 20

              x² + 5xy + 4y² + 4x - 2y = 20 can be written as
                ( x + y ) ( x + 4y ) + 4x – 2y = 20
             10 ( x + y ) + 4x – 2y = 20 ( Since x + 4y = 10 )
               10x + 10y + 4x – 2y = 20
                              14x + 8y = 20

                  Now x + 4y = 10
             14x + 8y = 20 and 4 : 8 :: 10 : 20

      from the Sutra, x = 0 and 4y = 10, i.e.,, 8y = 20 y = 10/4 = 2½
         Thus x = 0 and y = 2½ is the solution.

vedic solutions

SŨNYAM SĀMYASAMUCCAYE

The Sutra 'Sunyam Samyasamuccaye' says the 'Samuccaya is the same, that Samuccaya is Zero.' i.e., it should be equated to zero. The term 'Samuccaya' has several meanings under different contexts.

i) We interpret, 'Samuccaya' as a term which occurs as a common factor in all the terms concerned and proceed as follows.

Example 1: The equation 7x + 3x = 4x + 5x has the same factor ‘ x ‘ in all its terms. Hence by the sutra it is zero, i.e., x = 0.

Otherwise we have to work like this:

7x + 3x = 4x + 5x

10x = 9x

10x – 9x = 0

x = 0

This is applicable not only for ‘x’ but also any such unknown quantity as follows.

Example 2: 5(x+1) = 3(x+1)

No need to proceed in the usual procedure like

5x + 5 = 3x + 3

5x – 3x = 3 – 5

2x = -2 or x = -2 ÷ 2 = -1

Simply think of the contextual meaning of ‘ Samuccaya ‘

Now Samuccaya is ( x + 1 )

x + 1 = 0 gives x = -1

ii) Now we interpret ‘Samuccaya ‘as product of independent terms in expressions like (x+a) (x+b)

Example 3: ( x + 3 ) ( x + 4) = ( x – 2) ( x – 6 )

Here Samuccaya is 3 x 4 = 12 = -2 x -6

Since it is same , we derive x = 0

This example, we have already dealt in type ( ii ) of Paravartya in solving simple equations.

iii) We interpret ‘ Samuccaya ‘ as the sum of the denominators of two fractions having the same numerical numerator.

Consider the example.

1 1

____ + ____ = 0

3x-2 2x-1

for this we proceed by taking L.C.M.

(2x-1)+(3x–2)

____________ = 0

(3x–2)(2x–1)

5x–3

__________ = 0

(3x–2)(2x–1)

5x – 3 = 0 5x = 3

x = __

Instead of this, we can directly put the Samuccaya i.e., sum of the denominators

i.e., 3x – 2 + 2x - 1 = 5x - 3 = 0

giving 5x = 3 x = 3 / 5

It is true and applicable for all problems of the type

m m

____ + _____ = 0

ax+b cx+d

Samuccaya is ax+b+cx+d and solution is ( m ≠ 0 )

- ( b + d )

x = _________

( a + c )

iii) We now interpret ‘Samuccaya’ as combination or total.

If the sum of the numerators and the sum of the denominators be the same, then that sum = 0.

Consider examples of type

ax + b ax + c

_____ = ______

ax + c ax + b

In this case, (ax+b) (ax+b) = (ax+c) (ax+c)

a2x2 + 2abx + b2 = a2x2 + 2acx + c2

2abx – 2acx = c2 – b2

x ( 2ab – 2ac ) = c2 – b2

c2–b2 (c+b)(c-b) -(c+b)

x = ______ = _________ = _____

2a(b-c) 2a(b-c) 2a

As per Samuccaya (ax+b) + (ax+c) = 0

2ax+b+c = 0

2ax = -b-c

-(c+b)

x = ______

2a Hence the statement.

Example 4:

3x + 4 3x + 5

______ = ______

3x + 5 3x + 4

Since N1 + N2 = 3x + 4 + 3x + 5 = 6x + 9 ,

And D1 + D2 = 3x + 4 + 3x + 5 = 6x + 9

We have N1 + N2 = D1 + D2 = 6x + 9

Hence from Sunya Samuccaya we get 6x + 9 = 0

6x = -9

-9 -3

x = __ = __

6 2

Example 5:

5x + 7 5x + 12

_____ = _______

5x +12 5x + 7

Hence N1 + N2 = 5x + 7 + 5x + 12 = 10x + 19

And D1 + D2 = 5x + 12 + 5x + 7 = 10x + 19

N1 + N2 = D1 + D2 gives 10x + 19 = 0

10x = -19

-19

x = ____

Consider the examples of the type, where N1 + N2 = K ( D1 + D2 ), where K is a numerical constant, then also by removing the numerical constant K, we can proceed as above.

Example 6:

2x + 3 x + 1

_____ = ______

4x + 5 2x + 3

Here N1 + N2 = 2x + 3 + x + 1 = 3x + 4

D1 + D2 = 4x + 5 + 2x + 3 = 6x + 8

= 2 ( 3x + 4 )

Removing the numerical factor 2, we get 3x + 4 on both sides.

3x + 4 = 0 3x = -4 x = - 4 / 3.

v) ‘Samuccaya‘ with the same meaning as above, i.e., case (iv), we solve the problems leading to quadratic equations. In this context, we take the problems as follows;

If N1 + N2 = D1 + D2 and also the differences

N1 ~ D1 = N2 ~ D2 then both the things are equated to zero, the solution gives the two values for x.

Example 7:

3x + 2 2x + 5

_____ = ______

2x + 5 3x + 2

In the conventional text book method, we work as follows :

3x + 2 2x + 5

_____ = ______

2x + 5 3x + 2

( 3x + 2 ) ( 3x + 2 ) = ( 2x + 5 ) ( 2x + 5 )

9x2 + 12x + 4 = 4x2 + 20x + 25

9x2 + 12x + 4 - 4x2 - 20x – 25 = 0

5x2 – 8x – 21 = 0

5x2 – 15x + 7x – 21 = 0

5x ( x – 3 ) + 7 ( x – 3 ) = 0

(x – 3 ) ( 5x + 7 ) = 0

x – 3 = 0 or 5x + 7 = 0

x = 3 or - 7 / 5

Now ‘Samuccaya’ sutra comes to help us in a beautiful way as follows :

Observe N1 + N2 = 3x + 2 + 2x + 5 = 5x + 7

D1 + D2 = 2x + 5 + 3x + 2 = 5x + 7

Further N1 ~ D1 = ( 3x + 2 ) – ( 2x + 5 ) = x – 3

N2 ~ D2 = ( 2x + 5) – ( 3x + 2 ) = - x + 3 = - ( x – 3 )

Hence 5x + 7 = 0 , x – 3 = 0

5x = -7 , x = 3

i.e., x = -7 / 5 , x = 3

Note that all these can be easily calculated by mere observation.

Example 8:

3x + 4 5x + 6

______ = _____

6x + 7 2x + 3

Observe that

N1 + N2 = 3x + 4 + 5x + 6 = 8x + 10

and D1 + D2 = 6x + 7 + 2x + 3 = 8x + 10

Further N1 ~ D1 = (3x + 4) – (6x + 7)

= 3x + 4 – 6x – 7

= -3x – 3 = -3 ( x + 1 )

N2 ~ D2 = (5x + 6) – (2x + 3) = 3x + 3 = 3( x + 1)

By ‘Sunyam Samuccaye’ we have

8x + 10 = 0 3( x + 1 ) = 0

8x = -10 x + 1 = 0

x = - 10 / 8 x = -1

= - 5 / 4

vi)‘Samuccaya’ with the same sense but with a different context and application .

Example 9:

1 1 1 1

____ + _____ = ____ + ____

x - 4 x – 6 x - 2 x - 8

Usually we proceed as follows.

x–6+x-4 x–8+x-2

___________ = ___________

(x–4) (x–6) (x–2) (x-8)

2x-10 2x-10

_________ = _________

x2–10x+24 x2–10x+16

( 2x – 10 ) ( x2 – 10x + 16 ) = ( 2x – 10 ) ( x2 – 10x + 24)

2x3–20x2+32x–10x2+100x–160 = 2x3–20x2+48x–10x2+100x-240

2x3 – 30x2 + 132x – 160 = 2x3 – 30x2 + 148x – 240

132x – 160 = 148x – 240

132x – 148x = 160 – 240

– 16x = - 80

x = - 80 / - 16 = 5

Now ‘Samuccaya’ sutra, tell us that, if other elements being equal, the sum-total of the denominators on the L.H.S. and their total on the R.H.S. be the same, that total is zero.

Now D1 + D2 = x – 4 + x – 6 = 2x – 10, and

D3 + D4 = x – 2 + x – 8 = 2x – 10

By Samuccaya, 2x – 10 gives 2x = 10

x = __ = 5

Example 10:

1 1 1 1

____ + ____ = ____ + _____

x - 8 x – 9 x - 5 x – 12

D1 + D2 = x – 8 + x – 9 = 2x – 17, and

D3 + D4 = x – 5 + x –12 = 2x – 17

Now 2x – 17 = 0 gives 2x = 17

x = __ = 8½

Example 11:

1 1 1 1

____ - _____ = ____ - _____

x + 7 x + 10 x + 6 x + 9

This is not in the expected form. But a little work regarding transposition makes the above as follows.

1 1 1 1

____ + ____ = ____ + _____

x + 7 x + 9 x + 6 x + 10

Now ‘Samuccaya’ sutra applies

D1 + D2 = x + 7 + x + 9 = 2x + 16, and

D3 + D4 = x + 6 + x + 10 = 2x + 16

Solution is given by 2x + 16 = 0 i.e., 2 x = - 16.

x = - 16 / 2 = - 8.

Solve the following problems using Sunyam Samya-Samuccaye process.

1. 7 ( x + 2 ) + 3 ( x + 2 ) = 6 ( x + 2 ) + 5 ( x + 2 )

2. ( x + 6 ) ( x + 3 ) = ( x – 9 ) ( x – 2 )

3. ( x - 1 ) ( x + 14 ) = ( x + 2 ) ( x – 7 )

1 1

4. ______ + ____ = 0

4 x - 3 x – 2

4 4

5. _____ + _____ = 0

3x + 1 5x + 7

2x + 11 2x+5

6. ______ = _____

2x+ 5 2x+11

3x + 4 x + 1

7. ______ = _____

6x + 7 2x + 3

4x - 3 x + 4

8. ______ = _____

2x+ 3 3x - 2

1 1 1 1

9. ____ + ____ = ____ + _____

x - 2 x - 5 x - 3 x - 4

1 1 1 1

10. ____ - ____ = _____ - _____

x - 7 x - 6 x - 10 x - 9

Sunyam Samya Samuccaye in Certain Cubes:

Consider the problem ( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5 )3. For the solution by the traditional method we follow the steps as given below:

( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5 )3

x3 – 12x2 + 48x – 64 + x3 – 18x2 + 108x – 216

= 2 ( x3 – 15x2 + 75x – 125 )

2x3 – 30x2 + 156x – 280 = 2x3 – 30x2 + 150x – 250

156x – 280 = 150x – 250

156x – 150x = 280 – 250

6x = 30

x = 30 / 6 = 5

But once again observe the problem in the vedic sense

We have ( x – 4 ) + ( x – 6 ) = 2x – 10. Taking out the numerical factor 2, we have ( x – 5 ) = 0, which is the factor under the cube on R.H.S. In such a case “Sunyam samya Samuccaye” formula gives that x – 5 = 0. Hence x = 5

Think of solving the problem (x–249)3 + (x+247)3 = 2(x–1)3

The traditional method will be horrible even to think of.

But ( x – 249 ) + ( x + 247 ) = 2x – 2 = 2 ( x – 1 ). And x – 1. on R.H.S. cube, it is enough to state that x – 1 = 0 by the ‘sutra’.

x = 1 is the solution. No cubing or any other mathematical operations.

Algebraic Proof :

Consider ( x – 2a )3 + ( x – 2b )3 = 2 ( x – a – b )3 it is clear that

x – 2a + x – 2b = 2x – 2a – 2b

= 2 ( x – a – b )

Now the expression,

x3 - 6x2a + 12xa2 – 8a3 + x3 – 6x2b + 12xb2 – 8b3 =

2(x3–3x2a–3x2b+3xa2+3xb2+6axb–a3–3a2b–3ab2–b3)

= 2x3–6x2a–6x2b+6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3

cancel the common terms on both sides

12xa2+12xb2–8a3–8b3 = 6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3

6xa2 + 6xb2 – 12xab = 6a3 + 6b3 – 6a2b – 6ab2

6x ( a2 + b2 – 2ab ) = 6 [ a3 + b3 – ab ( a + b )]

x ( a – b )2 = [ ( a + b ) ( a2 + b2 –ab ) – ( a + b )ab]

= ( a + b ) ( a2 + b2 – 2ab )

= ( a + b ) ( a – b )2

since x = a + b

Solve the following using “Sunyam Samuccaye” process :

1. ( x – 3 )3 + ( x – 9 )3 = 2 ( x – 6 )3

2. ( x + 4 )3 + ( x – 10 )3 = 2 ( x – 3 )3

3. ( x + a + b – c )3 + ( x + b + c – a )3 = 2 ( x + b )3

Example :

(x + 2)3 x + 1

______ = _____

(x + 3)3 x + 4

with the text book procedures we proceed as follows

x3 + 6x2 + 12x +8 x + 1

_______________ = _____

x3 + 9x2 + 27x +27 x + 4

Now by cross multiplication,

( x + 4 ) ( x3 + 6x2 + 12x + 8 ) = ( x + 1 ) ( x3 + 9x2 + 27x + 27 )

x4 + 6x3 + 12x2 + 8x + 4x3 + 24x2 + 48x + 32 =

x4 + 9x3 + 27x2 + 27x + x3 + 9x2 + 27x + 27

x4 + 10x3 + 36x2 + 56x + 32 = x4 + 10x3 + 36x2 + 54x + 27

56x + 32 = 54x + 27

56x – 54x = 27 – 32

2x = - 5

x = - 5 / 2

Observe that ( N1 + D1 ) with in the cubes on

L.H.S. is x + 2 + x + 3 = 2x + 5 and

N2 + D2 on the right hand side

is x + 1 + x + 4 = 2x + 5.

By vedic formula we have 2x + 5 = 0 x = - 5 / 2.

Solve the following by using vedic method :

(x + 3)3 x+1

______ = ____

(x + 5)3 x+7

(x - 5)3 x - 3

______ = ____

(x - 7)3 x - 9

vedic solutions

PARĀVARTYA – YOJAYET

‘Paravartya – Yojayet’ means 'transpose and apply'

(i) Consider the division by divisors of more than one digit, and when the divisors are slightly greater than powers of 10.

Example 1 : Divide 1225 by 12.

Step 1 : (From left to right ) write the Divisor leaving the first digit, write the other digit or digits using negative (-) sign and place them below the divisor as shown.

-2

¯¯¯¯

Step 2 : Write down the dividend to the right. Set apart the last digit for the remainder.

i.e.,, 12 122 5

- 2

Step 3 : Write the 1st digit below the horizontal line drawn under the dividend. Multiply the digit by –2, write the product below the 2nd digit and add.

i.e.,, 12 122 5

-2 -2

¯¯¯¯¯ ¯¯¯¯

Since 1 x –2 = -2 and 2 + (-2) = 0

Step 4 : We get second digits’ sum as ‘0’. Multiply the second digits’ sum thus obtained by –2 and writes the product under 3rd digit and add.

12 122 5

- 2 -20

¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯

102 5

Step 5 : Continue the process to the last digit.

i.e., 12 122 5

- 2 -20 -4

¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯

102 1

Step 6: The sum of the last digit is the Remainder and the result to its left is Quotient.

Thus Q = 102 and R = 1

Example 2 : Divide 1697 by 14.

14 1 6 9 7

- 4 -4–8–4

¯¯¯¯ ¯¯¯¯¯¯¯

1 2 1 3

Q = 121, R = 3.

Example 3 : Divide 2598 by 123.

Note that the divisor has 3 digits. So we have to set up the last two digits of the dividend for the remainder.

1 2 3 25 98 Step ( 1 ) & Step ( 2 )

-2-3

¯¯¯¯¯ ¯¯¯¯¯¯¯¯

Now proceed the sequence of steps write –2 and –3 as follows :

1 2 3 2 5 9 8

-2-3 -4 -6

¯¯¯¯¯ -2–3

¯¯¯¯¯¯¯¯¯¯

2 1 1 5

Since 2 X (-2, -3)= -4 , -6; 5 – 4 = 1

and (1 X (-2,-3); 9 – 6 – 2 = 1; 8 – 3 = 5.

Hence Q = 21 and R = 15.

Example 4 : Divide 239479 by 11213. The divisor has 5 digits. So the last 4 digits of the dividend are to be set up for Remainder.

1 1 2 1 3 2 3 9 4 7 9

-1-2-1-3 -2 -4-2-6 with 2

¯¯¯¯¯¯¯¯ -1-2-1-3 with 1

¯¯¯¯¯¯¯¯¯¯¯¯¯

2 1 4 0 0 6

Hence Q = 21, R = 4006.

Example 5 : Divide 13456 by 1123

1 1 2 3 1 3 4 5 6

-1–2–3 -1-2-3

¯¯¯¯¯¯¯ -2-4 –6

¯¯¯¯¯¯¯¯¯¯¯¯¯

1 2 0–2 0

Note that the remainder portion contains –20, i.e.,, a negative quantity. To over come this situation, take 1 over from the quotient column, i.e.,, 1123 over to the right side, subtract the remainder portion 20 to get the actual remainder.

Thus Q = 12 – 1 = 11, and R = 1123 - 20 = 1103.

Find the Quotient and Remainder for the problems using paravartya – yojayet method.

1) 1234 ÷ 112 2) 11329 ÷ 1132

3) 12349 ÷ 133 4) 239479 ÷ 1203

Now let us consider the application of paravartya – yojayet in algebra.

Example 1 : Divide 6x2 + 5x + 4 by x – 1

X - 1 6x2 + 5x + 4

¯¯¯¯¯¯

1 6 + 11

¯¯¯¯¯¯¯¯¯¯¯¯

6x + 11 + 15 Thus Q = 6x+11, R=15.

Example 2 : Divide x3 – 3x2 + 10x – 4 by x - 5

X - 5 x3 – 3x2 + 10x – 4

¯¯¯¯¯

5 5 + 10 100

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

x2 + 2x + 20, + 96

Thus Q= x2 + 2x + 20, R = 96.

The procedure as a mental exercise comes as follows :

i) x3 / x gives x2 i.e.,, 1 the first coefficient in the Quotient.

ii) Multiply 1 by + 5,(obtained after reversing the sign of second term in the Quotient) and add to the next coefficient in the dividend. It gives 1 X( +5) = +5, adding to the next coefficient, i.e.,, –3 + 5 = 2. This is next coefficient in Quotient.

iii) Continue the process : multiply 2 by +5, i.e.,, 2 X +5 =10, add to the next coefficient 10 + 10 = 20. This is next coefficient in Quotient. Thus Quotient is x2 + 2x + 20

iv) Now multiply 20 by + 5 i.e.,, 20 x 5 = 100. Add to the next (last) term,

100 + (-4) = 96, which becomes R, i.e.,, R =9.

Example 3:

x4 – 3x3 + 7x2 + 5x + 7

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

x + 4

Now thinking the method as in example ( 1 ), we proceed as follows.

x + 4 x4 - 3x3 + 7x2 + 5x + 7

¯¯¯¯¯

-4 - 4 + 28 - 140 + 540

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

x3 - 7x2 + 35x - 135 547

Thus Q = x3 – 7x2 + 35x – 135 and R = 547.

or we proceed orally as follows:

x4 / x gives 1 as first coefficient.

i) -4 X 1 = - 4 : add to next coefficient – 4 + (-3) = - 7 which gives next coefficient in Q.

ii) – 7 X - 4 = 28 : then 28 + 7 = 35, the next coefficient in Q.

iii) 35 X - 4 = - 140 : then – 140 + 5 = - 135, the next coefficient in Q.

iv) - 135 X - 4 = 540 : then 540 + 7 = 547 becomes R.

Thus Q = x3 – 7x2 + 35x – 135 , R = 547.

Note :

1. We can follow the same procedure even the number of terms is more.

2. If any term is missing, we have to take the coefficient of the term as zero and proceed.

Now consider the divisors of second degree or more as in the following example.

Example :4 2x4 – 3x3 – 3x + 2 by x2 + 1.

Here x2 term is missing in the dividend. Hence treat it as 0 . x2 or 0 . And the x term in divisor is also absent we treat it as 0 . x. Now

x2 + 1 2x4 - 3x3 + 0 . x2 - 3x + 2

x2 + 0 . x + 1 0 - 2

¯¯¯¯¯¯¯¯¯¯¯¯

0 - 1 0 + 3

0 + 2

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

2 - 3 - 2 0 4

Thus Q = 2x2 - 3x - 2 and R = 0 . x + 4 = 4.

Example 5 : 2x5 – 5x4 + 3x2 – 4x + 7 by x3 – 2x2 + 3.

We treat the dividend as 2x5 – 5x4 + 0. x3 + 3x2 – 4x + 7 and divisor as x3 – 2x2 + 0 . x + 3 and proceed as follows :

x3 – 2x2 + 0 . x + 3 2x5 – 5x4 + 0.x3 + 3x2 – 4x + 7

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

2 0 - 3 4 0 - 6

-2 0 + 3

- 4 0 + 6

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

2 - 1 - 2 - 7 - 1 +13

Thus Q = 2x2 – x – 2, R = - 7 x2 – x + 13.

You may observe a very close relation of the method paravartya in this aspect with regard to REMAINDER THEOREM and HORNER PROCESS of Synthetic division. And yet paravartya goes much farther and is capable of numerous applications in other directions also.

Apply paravartya – yojayet to find out the Quotient and Remainder in each of the following problems.

1) (4x2 + 3x + 5) ÷ (x+1)

2) (x3 – 4x2 + 7x + 6) ÷ (x – 2)

3) (x4 – x3 + x2 + 2x + 4) ÷ (x2 - x – 1)

4) (2x5 + x3 – 3x + 7) ÷ (x3 + 2x – 3)

5) (7x6 + 6x5 – 5x4 + 4x3 – 3x2 + 2x – 1) ÷ (x-1)

Paravartya in solving simple equations :

Recall that 'paravartya yojayet' means 'transpose and apply'. The rule relating to transposition enjoins invariable change of sign with every change of side. i.e., + becomes - and conversely ; and X becomes ÷ and conversely. Further it can be extended to the transposition of terms from left to right and conversely and from numerator to denominator and conversely in the concerned problems.

Type ( i ) :

Consider the problem 7x – 5 = 5x + 1

7x – 5x = 1 + 5

i.e.,, 2x = 6 x = 6 ÷ 2 = 3.

Observe that the problem is of the type ax + b = cx + d from which we get by ‘transpose’ (d – b), (a – c) and

d - b.

x = ¯¯¯¯¯¯¯¯

a - c

In this example a = 7, b = - 5, c = 5, d = 1

Hence 1 – (- 5) 1+5 6

x = _______ = ____ = __ = 3

7 – 5 7-5 2

Example 2: Solve for x, 3x + 4 = 2x + 6

d - b 6 - 4 2

x = _____ = _____ = __ = 2

a - c 3 - 2 1

Type ( ii ) : Consider problems of the type (x + a) (x+b) = (x+c) (x+d). By paravartya, we get

cd - ab

x = ______________

(a + b) – (c + d)

It is trivial form the following steps

(x + a) (x + b) = (x + c) (x + d)

x2 + bx + ax + ab = x2 + dx + cx + cd

bx + ax – dx – cx = cd – ab

x( a + b – c – d) = cd – ab

cd – ab cd - ab

x = ____________ x = _________________

a + b – c – d ( a + b ) – (c + d.)

Example 1 : (x – 3) (x – 2 ) = (x + 1 ) (x + 2 ).

By paravartya

cd – ab 1 (2) – (-3) (-2)

x = __________ = ______________

a + b – c –d - 3 – 2 – 1 – 2

2 - 6 - 4 1

= _______ = ___ = __

- 8 - 8 2

Example 2 : (x + 7) (x – 6) = (x +3) (x – 4).

Now cd - ab (3) (-4) – (7) (-6)

x = ___________ = ________________

a + b – c – d 7 + (-6) – 3 - (-4)

- 12 + 42 30

= ____________ = ___ = 15

7 – 6 – 3 + 4 2

Note that if cd - ab = 0 i.e.,, cd = ab, i.e.,, if the product of the absolute terms be the same on both sides, the numerator becomes zero giving x = 0.

For the problem (x + 4) (x + 3) = (x – 2 ) ( x – 6 )

Solution is x = 0 since 4 X 3 = - 2 X - 6. = 12

Type ( iii) :

Consider the problems of the type ax + b m

______ = __

cx + d n

By cross – multiplication,

n ( ax + b) = m (cx + d)

nax + nb = mcx + md

nax - mcx = md – nb

x( na – mc ) = md – nb

md - nb

x = ________

na - mc.

Now look at the problem once again

ax + b m

_____ = __

cx + d n

paravartya gives md - nb, na - mc and

md - nb

x = _______

na - mc

Example 1: 3x + 1 13

_______ = ___

4x + 3 19

md - nb 13 (3) - 19(1) 39 - 19 20

x = ______ = ____________ = _______ = __

na - mc 19 (3) - 13(4) 57 - 52 5

= 4

Example 2: 4x + 5 7

________ = __

3x + 13/2 8

(7) (13/2) - (8)(5)

x = _______________

(8) (4) - (7)(3)

(91/2) - 40 (91 - 80)/2 11 1

= __________ = _________ = ______ = __

32 – 21 32 – 21 2 X 11 2

Type (iv) : Consider the problems of the type m n

_____ + ____ = 0

x + a x + b

Take L.C.M and proceed.

m(x+b) + n (x+a)

______________ = 0

(x + a) (x +b)

mx + mb + nx + na

________________ = 0

(x + a)(x + b)

(m + n)x + mb + na = 0 (m + n)x = - mb - na

-mb - na

x = ________

(m + n)

Thus the problem m n

____ + ____ = 0, by paravartya process

x + a x + b

gives directly

-mb - na

x = ________

(m + n)

Example 1 : 3 4

____ + ____ = 0

x + 4 x – 6

gives -mb - na

x = ________ Note that m = 3, n = 4, a = 4, b = - 6

(m + n)

-(3)(-6) – (4) (4) 18 - 16 2

= _______________ = ______ = __

( 3 + 4) 7 7

Example 2 :

5 6

____ + _____ = 0

x + 1 x – 21

gives -(5) (-21) - (6) (1) 105 - 6 99

x = ________________ = ______ = __ = 9

5 + 6 11 11

I . Solve the following problems using the sutra Paravartya – yojayet.

1) 3x + 5 = 5x – 3 6) (x + 1) ( x + 2) = ( x – 3) (x – 4)

2) (2x/3) + 1 = x - 1 7) (x – 7) (x – 9) = (x – 3) (x – 22)

3) 7x + 2 5 8) (x + 7) (x + 9) = (x + 3 ) (x + 21)

______ = __

3x - 5 8

4) x + 1 / 3

_______ = 1

3x - 1

5) 5 2

____ + ____ = 0

x + 3 x – 4

II)

1. Show that for the type of equations

m n p

____ + ____ + ____ = 0, the solution is

x + a x + b x + c

- mbc – nca – pab

x = ________________________ , if m + n + p =0.

m(b + c) + n(c+a) + p(a + b)

2. Apply the above formula to set the solution for the problem

Problem 3 2 5

____ + ____ - ____ = 0

x + 4 x + 6 x + 5

some more simple solutions :

m n m + n

____ + ____ = _____

x + a x + b x + c

Now this can be written as,

m n m n

____ + ____ = _____ + _____

x + a x + b x + c x + c

m m n n

____ - ____ = _____ - _____

x + a x + c x + c x + b

m(x +c) – m(x + a) n(x + b) – n(x + c)

________________ = ________________

(x + a) (x + c) (x + c) (x + b)

mx + mc – mx – ma nx + nb – nx – nc

________________ = _______________

(x + a) (x + c) (x +c ) (x + b)

m (c – a) n (b –c)

____________ = ___________

x +a x + b

m (c - a).x + m (c - a).b = n (b - c). x + n(b - c).a

x [ m(c - a) - n(b - c) ] = na(b - c) – mb (c - a)

or x [ m(c - a) + n(c - b) ] = na(b - c) + mb (a - c)

Thus mb(a - c) + na (b - c)

x = ___________________

m(c-a) + n(c-b).

By paravartya rule we can easily remember the formula.

Example 1 : solve 3 4 7

____ + _____ = ____

x + 1 x + 2 x + 3

In the usual procedure, we proceed as follows.

3 4 7

____ + ____ = ____

x + 1 x + 2 x + 3

3(x + 2) + 4(x + 1) 7

________________ = _____

(x + 1) (x + 2) x + 3

3x + 6 + 4x + 4 7

_____________ = ____

x2 + 2x + x + 2 x + 3

7x + 10 7

_________ = ____

x2 + 3x + 2 x + 3

(7x + 10) (x + 3) = 7(x2 + 3x + 2)

7x2 + 21x + 10x + 30 = 7x2 + 21x + 14.

31x + 30 = 21x + 14

31 x – 21 x = 14 – 30

i.e.,, 10x = - 16

x = - 16 / 10 = - 8 / 5

Now by Paravartya process

3 4 7

____ + ____ = ____ ( ... N1 + N2 = 3+4 = 7 = N3)

x + 1 x + 2 x + 3

mb( a – c ) + na ( b – c )

x = _____________________ here N1 = m = 3 , N2 = n = 4 ;

m ( c – a ) + n ( c – b ) a = 1, b = 2, c = 3

3 . 2 ( 1 – 3 ) + 4 . 1 . ( 2 – 3)

= __________________________

3 ( 3 – 1 ) + 4 ( 3 – 2 )

6 ( -2)+ 4 (-1) - 12 – 4 - 16 - 8

= _____________ = _______ = ____ = ___

3 (2) + 4(1) 6 + 4 10 5

Example 2 :

3 5 8

____ + ____ = _____ Here N1 + N2 = 3 + 5 = 8.

x - 2 x – 6 x + 3

mb ( a – c ) + na ( b – c)

x = _____________________

m ( c – a ) + n ( c – b )

3 . ( -6 ) ( - 2 - 3 ) + 5 .( -2 ) ( -6 – 3 )

= __________________________________

3 ( 3 – ( -2 ) ) + 5 ( 3 – ( - 6 ) )

3 ( - 6 ) ( - 5 ) + 5 ( - 2 ) ( - 9 )

= ____________________________

3( 3 + 2 ) + 5 ( 3 + 6 )

90 + 90

= _______ = 180 / 60 = 3.

15 + 45

Solve the problems using the methods explained above.

1) 2 3 5

____ + ____ = ____

x + 2 x + 3 x + 5

2) 4 6 10

____ + ____ = ____

x + 1 x + 3 x + 4

3) 5 2 3

____ + ___ = ____

x - 2 3 - x x – 4

4) 4 9 15

_____ + _____ = _____

2x + 1 3x + 2 3x + 1

Note : The problem ( 4 ) appears to be not in the model said above.

But 3 (4) 2 (9) 2(15)

________ + ________ = _______ gives

3(2x + 1) 2( 3x + 2) 2(3x + 1)

12 18 30

_____ + _____ = _____ Now proceed.

6x + 3 6x + 4 6x + 2

Simultaneous simple equations:

By applying Paravartya sutra we can derive the values of x and y which are given by two simultaneous equations. The values of x and y are given by ration form. The method to find out the numerator and denominator of the ratio is given below.

Example 1: 2x + 3y = 13, 4x + 5y = 23.

i) To get x, start with y coefficients and the independent terms and cross-multiply forward, i.e.,, right ward. Start from the upper row and multiply across by the lower one, and conversely, the connecting link between the two cross-products being a minus. This becomes numerator.

i.e.,, 2x + 3y = 13

4x + 5y = 23

Numerator of the x – value is 3 x 23 – 5 x 13 = 69 – 65 = 4

ii) Go from the upper row across to the lower one, i.e.,, the x- coefficient but backward, i.e.,, leftward.

Denominator of the x – value is 3 x 4 – 2 x 5 = 12 – 10 = 2

Hence value of x = 4 ÷ 2 = 2.

iii) To get y, follow the cyclic system, i.e.,, start with the independent term on the upper row towards the x–coefficient on the lower row. So numerator of the y–value is

13 x 4 – 23 x 2 = 52 – 46 = 6.

iv) The denominator is the same as obtained in Step(ii) i.e.,, 2. Hence value of y is 6÷2=3.

Thus the solution to the given equation is x = 2 and y = 3.

Example 2: 5x – 3y = 11

6x – 5y = 09

Now Nr. of x is (-3) (9) – (5) (11) = - 27 + 55 = 28

Dr. of x is (-3) (6) – (5) (-5) = - 18 + 25 = 07

x = Nr ÷ Dr = 28 ÷ 7 = 4

and for y, Nr is (11) (6) – (9)(5) = 66 – 45 = 21

Dr is 7

Hence y = 21 ÷ 7 = 3.

Example 3: solve 3x + y = 5

4x – y = 9

Now we can straight away write the values as follows:

(1)(9) – (-1)(5) 9 + 5 14

x = _____________ = _____ = ___ = 2

(1)(4) – (3)(-1) 4 + 3 7

(5)(4) – (9)(3) 20 – 27 -7

y = ____________ = _______ = ___ = -1

(1)(4) – (3)(-1) 4 + 3 7

Hence x = 2 and y = -1 is the solution.

Algebraic Proof:

ax + by = m ……… ( i )

cx + dy = n ………. ( ii )

Multiply ( i ) by d and ( ii ) by b, then subtract

adx + bdy = m.d

cbx + dby = n.b

____________________

( ad – cb ) .x = md – nb

md - nb bn - md

x = ______ = ______

ad - cb bc - ad

Multiply ( i ) by c and ( ii ) by a, then subtract

acx + bcy = m.c

cax + day = n.a

_____________________

( bc – ad ) . y = mc - na

mc - na

y = ______

bc - ad

You feel comfort in the Paravartya process because it avoids the confusion in multiplication, change of sign and such other processes.

Find the values of x and y in each of the following problems using Paravartya process.

1. 2x + y = 5 2. 3x – 4y = 7

3x – 4y = 2 5x + y = 4

3. 4x + 3y = 8 4. x + 3y = 7

6x - y = 1 2x + 5y = 11

Algebraic proof :
    a) Let the two 2 digit numbers be (ax+b) and (cx+d). Note that x = 10. Now consider the product
       (ax + b) (cx + d) = ac.x² + adx + bcx + b.d
                                = ac.x² + (ad + bc)x + b.d
Observe that

i) The first term i.e., the coefficient of x² (i.e., 100, hence the digit in the 100^th place) is obtained by vertical multiplication of a and c i.e., the digits in 10^th place (coefficient of x) of both the numbers;

ii) The middle term, i.e., the coefficient of x (i.e., digit in the 10^th place) is obtained by cross wise multiplication of a and d; and of b and c; and the addition of the two products;

iii) The last (independent of x) term is obtained by vertical multiplication of the independent terms b and d.

b) Consider the multiplication of two 3 digit numbers.
Let the two numbers be (ax² + bx + c) and (dx² + ex + f). Note that x=10
Now the product is
                            ax² + bx + c
                          x dx² + ex + f
          ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
       ad.x⁴+bd.x³+cd.x²+ae.x³+be.x²+ce.x+af.x²+bf.x+cf
= ad.x⁴ + (bd + ae). x³ + (cd + be + af).x² + (ce + bf)x + cf
Note the following points :

i) The coefficient of x⁴ , i.e., ad is obtained by the vertical multiplication of the first coefficient from the left side :

ii)The coefficient of x³ , i.e., (ae + bd) is obtained by the cross –wise multiplication of the first two coefficients and by the addition of the two products;

iii) The coefficient of x² is obtained by the multiplication of the first coefficient of the multiplicand (ax²+bx +c) i.e., a; by the last coefficient of the multiplier (dx² +ex +f) i.e.,f ; of the middle one i.e., b of the multiplicand by the middle one i.e., e of the multiplier and of the last one i.e., c of the multiplicand by the first one i.e., d of the multiplier and by the addition of all the three products i.e., af + be +cd :

iv) The coefficient of x is obtained by the cross wise multiplication of the second coefficient i.e., b of the multiplicand by the third one i.e., f of the multiplier, and conversely the third coefficient i.e., c of the multiplicand by the second coefficient i.e., e of the multiplier and by addition of the two products, i.e., bf + ce ;

v) And finally the last (independent of x) term is obtained by the vertical multiplication of the last coefficients c and f i.e., cf

Thus the process can be put symbolically as (from left to right)

Consider the following example
124 X 132.
Proceeding from right to left

i) 4 X 2 = 8. First digit = 8

ii) (2 X 2) + (3 X 4) = 4 + 12 = 16. The digit 6 is retained and 1 is carried over to left side. Second digit = 6.

iii) (1 X 2) + (2 X 3) + (1 X 4) = 2 + 6 + 4 =12. The carried over 1 of above step is added i.e., 12 + 1 = 13. Now 3 is retained and 1 is carried over to left side. Thus third digit = 3.

iv) ( 1X 3 ) + ( 2 X 1 ) = 3 + 2 = 5. the carried over 1 of above step is added i.e., 5 + 1 = 6 . It is retained. Thus fourth digit = 6

v) ( 1 X 1 ) = 1. As there is no carried over number from the previous step it is retained. Thus fifth digit = 1

124 X 132 = 16368.

Let us work another problem by placing the carried over digits under the first row and proceed.

                             234
                               x 316
                               ¯¯¯¯¯¯¯
                                61724
                                1222
                               ¯¯¯¯¯¯¯
                                73944

i) 4 X 6 = 24 : 2, the carried over digit is placed below the second digit.

ii) (3 X 6) + (4 x 1) = 18 + 4 = 22 ; 2, the carried over digit is placed below third digit.

iii) (2 X 6) + (3 X 1) + (4 X 3) = 12 + 3 + 12 = 27 ; 2, the carried over digit is placed below fourth digit.

iv) (2 X 1) + ( 3 X 3) = 2 + 9 = 11; 1, the carried over digit is placed below fifth digit.

v) ( 2 X 3 ) = 6.

vi) Respective digits are added.

Note :

1. We can carry out the multiplication in urdhva - tiryak process from left to right or right to left.

2. The same process can be applied even for numbers having more digits.

3. urdhva –tiryak process of multiplication can be effectively used in multiplication regarding algebraic expressions.

Example 1 : Find the product of (a+2b) and (3a+b).

Example 2 :
                             3a² + 2a + 4
                          x 2a² + 5a + 3
                         ¯¯¯¯¯¯¯¯¯¯¯¯¯¯

i)     4 X 3   = 12

ii)   (2 X 3) + ( 4 X 5 ) = 6 + 20 = 26   i.e., 26a

iii) (3 X 3) + ( 2 X 5 ) + ( 4 X 2 ) = 9 + 10 + 8 = 27 i.e., 27a²

iv) (3 X 5) + ( 2 X 2 ) = 15 + 4 = 19  i.e., 19 a³

v)    3 X 2   =   6     i.e., 6a⁴

Hence the product is 6a⁴ + 19a³ + 27a² + 26a + 12
Example 3 :     Find    (3x² + 4x + 7) (5x +6)
            Now         3.x² + 4x + 7
                                0.x² + 5x + 6
                               ¯¯¯¯¯¯¯¯¯¯¯¯

i) 7 X 6 = 42

ii) (4 X 6) + (7 X 5) = 24 + 35 = 59 i.e., 59x

iii) (3 X 6) + (4 X 5) + (7 X 0) = 18 + 20 + 0 = 38 i.e., 38x²

iv) (3 X 5) + (0 X 4) = 15 + 0 = 15 i.e., 15x³

v) 3 X 0 = 0

Hence the product is 15x³ + 38x² + 59x + 42

        Find the products using urdhva tiryagbhyam process.

        1) 25 X 16                  2) 32 X 48             3) 56 X 56

        4) 137 X 214              5) 321 X 213        6) 452 X 348

        7) (2x + 3y) (4x + 5y)                           8) (5a² + 1) (3a² + 4)

        9) (6x² + 5x + 2 ) (3x² + 4x +7)         10) (4x² + 3) (5x + 6)

Urdhva – tiryak in converse for division process:
As per the statement it an used as a simple argumentation for division process particularly in algebra.
Consider the division of (x³ + 5x² + 3x + 7) by (x – 2) process by converse of urdhva – tiryak :

    i)   x³ divided by x gives                       x² . x³+ 5x² + 3x + 7
             It is the first term of the Quotient.    ___________________
                                                                           x – 2
                               Q = x² + - - - - - - - - - - -

ii) x² X – 2 = - 2x² . But 5x² in the dividend hints 7x² more since 7x² – 2x² = 5x² . This ‘more’ can be obtained from the multiplication of x by 7x. Hence second term of Q is 7x.

        x³ + 5x² + 3x + 7
           ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯   gives Q = x² + 7x + - - - - - - - -
                      x – 2

iii)We now have – 2 X 7x = -14x. But the 3rd term in the dividend is 3x for which ‘17x more’ is required since 17x – 14x =3x.

Now multiplication of x by 17 gives 17x. Hence third term of quotient is 17

        Thus
                    x³ + 5x² + 3x + 7
                  _________________    gives Q= x² + 7x +17
                            x – 2

iv) Now last term of Q, i.e., 17 multiplied by –2 gives 17X–2 = -34 but the relevant term in dividend is 7. So 7 + 34 = 41 ‘more’ is required. As there no more terms left in dividend, 41 remains as the remainder.

             x³ + 5x² + 3x + 7
            ________________   gives Q= x² + 7x +17 and R = 41.
                      x – 2

Find the Q and R in the following divisions by using the converse process of urdhva – tiryagbhyam method :

1)  3x² – x – 6                        2) 16x² + 24x +9
      ¯¯¯¯¯¯¯¯¯                            ¯¯¯¯¯¯¯¯¯¯¯¯
          3x – 7                                          4x+3

3) x³+ 2x² +3x + 5             4) 12x⁴ – 3x² – 3x + 12
      ¯¯¯¯¯¯¯¯¯¯¯¯¯¯                ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
              x - 3                                           x² + 1

vedic solutions

EKĀDHIKENA PŪRVEŅA

The Sutra (formula) Ekādhikena Pūrvena means: “By one more than the previous one”.

i) Squares of numbers ending in 5 :

Now we relate the sutra to the ‘squaring of numbers ending in 5’. Consider the example 252.

Here the number is 25. We have to find out the square of the number. For the number 25, the last digit is 5 and the 'previous' digit is 2. Hence, 'one more than the previous one', that is, 2+1=3. The Sutra, in this context, gives the procedure 'to multiply the previous digit 2 by one more than itself, that is, by 3'. It becomes the L.H.S (left hand side) of the result, that is, 2 X 3 = 6. The R.H.S (right hand side) of the result is 52, that is, 25.

Thus 252 = 2 X 3 / 25 = 625.

In the same way,

352= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;

652= 6 X 7 / 25 = 4225;

1052= 10 X 11/25 = 11025;

1352= 13 X 14/25 = 18225;

Apply the formula to find the squares of the numbers 15, 45, 85, 125, 175 and verify the answers.

Algebraic proof:

a) Consider (ax + b)2 Ξ a2. x2 + 2abx + b2.

This identity for x = 10 and b = 5 becomes

(10a + 5) 2 = a2 . 102 + 2. 10a . 5 + 52

= a2 . 102 + a. 102 + 52

= (a 2+ a ) . 102 + 52

= a (a + 1) . 10 2 + 25.

Clearly 10a + 5 represents two-digit numbers 15, 25, 35, -------,95 for the values a = 1, 2, 3, -------,9 respectively. In such a case the number (10a + 5)2 is of the form whose L.H.S is a (a + 1) and R.H.S is 25, that is, a (a + 1) / 25.

Thus any such two digit number gives the result in the same fashion.

Example: 45 = (40 + 5)2, It is of the form (ax+b)2 for a = 4, x=10

and b = 5. giving the answer a (a+1) / 25

that is, 4 (4+1) / 25 + 4 X 5 / 25 = 2025.

b) Any three digit number is of the form ax2+bx+c for x = 10, a ≠ 0, a, b, c Є W.

Now (ax2+bx+ c) 2 = a2 x4 + b2x2 + c2 + 2abx3 + 2bcx + 2cax2

= a2 x4+2ab. x3+ (b2 + 2ca)x2+2bc . x+ c2.

This identity for x = 10, c = 5 becomes (a . 102 + b .10 + 5) 2

= a2.104 + 2.a.b.103 + (b2 + 2.5.a)102 + 2.b.5.10 + 52

= a2.104 + 2.a.b.103 + (b2 + 10 a)102 + b.102+ 52

= a2.104 + 2ab.103 + b2.102 + a . 103 + b 102 + 52

= a2.104 + (2ab + a).103 + (b2+ b)102 +52

= [ a2.102 + 2ab.10 + a.10 + b2 + b] 102+ 52

= (10a + b) ( 10a+b+1).102 + 25

= P (P+1) 102 + 25, where P = 10a+b.

Hence any three digit number whose last digit is 5 gives the same result as in (a) for P=10a + b, the ‘previous’ of 5.

Example : 1652 = (1 . 102 + 6 . 10 + 5) 2.

It is of the form (ax2 +bx+c)2 for a = 1, b = 6, c = 5 and x = 10. It gives the answer P(P+1) / 25, where P = 10a + b = 10 X 1 + 6 = 16, the ‘previous’. The answer is 16 (16+1) / 25 = 16 X 17 / 25 = 27225.

Apply Ekadhikena purvena to find the squares of the numbers 95, 225, 375, 635, 745, 915, 1105, 2545.

ii) Vulgar fractions whose denominators are numbers ending in NINE :

We now take examples of 1 / a9, where a = 1, 2, -----, 9. In the conversion of such vulgar fractions into recurring decimals, Ekadhika process can be effectively used both in division and multiplication.

a) Division Method : Value of 1 / 19.

The numbers of decimal places before repetition is the difference of numerator and denominator, i.e.,, 19 -1=18 places.

For the denominator 19, the purva (previous) is 1.

Hence Ekadhikena purva (one more than the previous) is 1 + 1 = 2.

The sutra is applied in a different context. Now the method of division is as follows:

Step. 1 : Divide numerator 1 by 20.

i.e.,, 1 / 20 = 0.1 / 2 = .10 ( 0 times, 1 remainder)

Step. 2 : Divide 10 by 2

i.e.,, 0.005( 5 times, 0 remainder )

Step. 3 : Divide 5 by 2

i.e.,, 0.0512 ( 2 times, 1 remainder )

Step. 4 : Divide 12 i.e.,, 12 by 2

i.e.,, 0.0526 ( 6 times, No remainder )

Step. 5 : Divide 6 by 2

i.e.,, 0.05263 ( 3 times, No remainder )

Step. 6 : Divide 3 by 2

i.e.,, 0.0526311(1 time, 1 remainder )

Step. 7 : Divide 11 i.e.,, 11 by 2

i.e.,, 0.05263115 (5 times, 1 remainder )

Step. 8 : Divide 15 i.e.,, 15 by 2

i.e.,, 0.052631517 ( 7 times, 1 remainder )

Step. 9 : Divide 17 i.e.,, 17 by 2

i.e.,, 0.05263157 18 (8 times, 1 remainder )

Step. 10 : Divide 18 i.e.,, 18 by 2

i.e.,, 0.0526315789 (9 times, No remainder )

Step. 11 : Divide 9 by 2

i.e.,, 0.0526315789 14 (4 times, 1 remainder )

Step. 12 : Divide 14 i.e.,, 14 by 2

i.e.,, 0.052631578947 ( 7 times, No remainder )

Step. 13 : Divide 7 by 2

i.e.,, 0.05263157894713 ( 3 times, 1 remainder )

Step. 14 : Divide 13 i.e.,, 13 by 2

i.e.,, 0.052631578947316 ( 6 times, 1 remainder )

Step. 15 : Divide 16 i.e.,, 16 by 2

i.e.,, 0.052631578947368 (8 times, No remainder )

Step. 16 : Divide 8 by 2

i.e.,, 0.0526315789473684 ( 4 times, No remainder )

Step. 17 : Divide 4 by 2

i.e.,, 0.05263157894736842 ( 2 times, No remainder )

Step. 18 : Divide 2 by 2

i.e.,, 0.052631578947368421 ( 1 time, No remainder )

Now from step 19, i.e.,, dividing 1 by 2, Step 2 to Step. 18 repeats thus giving

0 __________________ . .

1 / 19 = 0.052631578947368421 or 0.052631578947368421

Note that we have completed the process of division only by using ‘2’. Nowhere the division by 19 occurs.

b) Multiplication Method: Value of 1 / 19

First we recognize the last digit of the denominator of the type 1 / a9. Here the last digit is 9.

For a fraction of the form in whose denominator 9 is the last digit, we take the case of 1 / 19 as follows:

For 1 / 19, 'previous' of 19 is 1. And one more than of it is 1 + 1 = 2.

Therefore 2 is the multiplier for the conversion. We write the last digit in the numerator as 1 and follow the steps leftwards.

Step. 1 : 1

Step. 2 : 21(multiply 1 by 2, put to left)

Step. 3 : 421(multiply 2 by 2, put to left)

Step. 4 : 8421(multiply 4 by 2, put to left)

Step. 5 : 168421 (multiply 8 by 2 =16, 1 carried over, 6 put to left)

Step. 6 : 1368421 ( 6 X 2 =12,+1 [carry over]

= 13, 1 carried over, 3 put to left )

Step. 7 : 7368421 ( 3 X 2, = 6 +1 [Carryover]

= 7, put to left)

Step. 8 : 147368421 (as in the same process)

Step. 9 : 947368421 ( Do – continue to step 18)

Step. 10 : 18947368421

Step. 11 : 178947368421

Step. 12 : 1578947368421

Step. 13 : 11578947368421

Step. 14 : 31578947368421

Step. 15 : 631578947368421

Step. 16 : 12631578947368421

Step. 17 : 52631578947368421

Step. 18 : 1052631578947368421

Now from step 18 onwards the same numbers and order towards left continue.

Thus 1 / 19 = 0.052631578947368421

It is interesting to note that we have

i) not at all used division process

ii) instead of dividing 1 by 19 continuously, just multiplied 1 by 2 and continued to multiply the resultant successively by 2.

Observations :

a) For any fraction of the form 1 / a9 i.e.,, in whose denominator 9 is the digit in the units place and a is the set of remaining digits, the value of the fraction is in recurring decimal form and the repeating block’s right most digit is 1.

b) Whatever may be a9, and the numerator, it is enough to follow the said process with (a+1) either in division or in multiplication.

c) Starting from right most digit and counting from the right, we see ( in the given example 1 / 19)

Sum of 1st digit + 10th digit = 1 + 8 = 9

Sum of 2nd digit + 11th digit = 2 + 7 = 9

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

Sum of 9th digit + 18th digit = 9+ 0 = 9

From the above observations, we conclude that if we find first 9 digits, further digits can be derived as complements of 9.

i) Thus at the step 8 in division process we have 0.052631517 and next step. 9 gives 0.052631578

Now the complements of the numbers

0, 5, 2, 6, 3, 1, 5, 7, 8 from 9

9, 4, 7, 3, 6, 8, 4, 2, 1 follow the right order

i.e.,, 0.052631578947368421

Now taking the multiplication process we have

Step. 8 : 147368421

Step. 9 : 947368421

Now the complements of 1, 2, 4, 8, 6, 3, 7, 4, 9 from 9

i.e.,, 8, 7, 5, 1, 3, 6, 2, 5, 0 precede in successive steps, giving the answer.

0.052631578947368421.

d) When we get (Denominator – Numerator) as the product in the multiplicative process, half the work is done. We stop the multiplication there and mechanically write the remaining half of the answer by merely taking down complements from 9.

e) Either division or multiplication process of giving the answer can be put in a single line form.

Algebraic proof :

Any vulgar fraction of the form 1 / a9 can be written as

1 / a9 = 1 / ( (a + 1 ) x - 1 ) where x = 10

= ________________________

( a + 1 ) x [1 - 1/(a+1)x ]

= ___________ [1 - 1/(a+1)x]-1

( a + 1 ) x

= __________ [1 + 1/(a+1)x + 1/(a+1)x2+ ----------]

( a + 1 ) x

= 1/(a+1)x + 1/(a+1)2x2 +1/(a+1)3x3+ ---- ad infinitum

= 10-1(1/(a+1))+10-2(1/(a+1)2)+10-3(1/(a+1)3) + ---ad infinitum

This series explains the process of ekadhik.

Now consider the problem of 1 / 19. From above we get

1 / 19 = 10-1 (1/(1+1)) + 10-2 (1/(1+1)2) + 10-3 (1/(1+1)3) + ----

( since a=1)

= 10-1 (1/2) + 10-2 (1/2)2 + 10-3 (1/3)3 + ----------

= 10-1 (0.5) + 10-2 (0.25) + 10-3 (0.125)+ ----------

= 0.05 + 0.0025 + 0.000125 + 0.00000625+ - - - - - - -

= 0.052631 - - - - - - -

Example1 :

1. Find 1 / 49 by ekadhikena process.

Now ‘previous’ is 4. ‘One more than the previous’ is 4 + 1 = 5.

Now by division right ward from the left by ‘5’.

1 / 49 = .10 - - - - - - - - - - - -(divide 1 by 50)

= .02 - - - - - - - - - (divide 2 by 5, 0 times, 2 remainder )

= .0220 - - - - - - --(divide 20 by 5, 4 times)

= .0204 - - - - - - -( divide 4 by 5, 0 times, 4 remainder )

= .020440 -- - -- - ( divide 40 by 5, 8 times )

= .020408 - - - - - (divide 8 by 5, 1 time, 3 remainder )

= .02040831 - - - -(divide 31 by 5, 6 times, 1 remainder )

= .02040811 6 - - - - - - - continue

= .0204081613322615306111222244448 - -- - - - -

On completing 21 digits, we get 48

i.e.,,Denominator - Numerator = 49 – 1 = 48 stands.

i.e, half of the process stops here. The remaining half can be obtained as complements from 9.

Thus 1 / 49 = 0.020408163265306122448

979591836734693877551

Now finding 1 / 49 by process of multiplication left ward from right by 5, we get

1 / 49 = ----------------------------------------------1

= ---------------------------------------------51

= -------------------------------------------2551

= ------------------------------------------27551

= ---- 483947294594118333617233446943383727551

i.e.,,Denominator – Numerator = 49 – 1 = 48 is obtained as 5X9+3

( Carry over ) = 45 + 3 = 48. Hence half of the process is over. The remaining half is automatically obtained as complements of 9.

Thus 1 / 49 = ---------------979591836734693877551

= 0.020408163265306122448

979591836734693877551

Example 2: Find 1 / 39 by Ekadhika process.

Now by multiplication method, Ekadhikena purva is 3 + 1 = 4

1 / 39 = -------------------------------------1

= -------------------------------------41

= ----------------------------------1641

= ---------------------------------25641

= --------------------------------225641

= -------------------------------1025641

Here the repeating block happens to be block of 6 digits. Now the rule predicting the completion of half of the computation does not hold. The complete block has to be computed by ekadhika process.

Now continue and obtain the result. Find reasons for the non–applicability of the said ‘rule’.

Find the recurring decimal form of the fractions 1 / 29, 1 / 59,

1 / 69, 1 / 79, 1 / 89 using Ekadhika process if possible. Judge whether the rule of completion of half the computation holds good in such cases.

vedic solutions

NIKHILAM NAVATAS’CARAMAM DASATAH

The formula simply means : “all from 9 and the last from 10”

The formula can be very effectively applied in multiplication of numbers, which are nearer to bases like 10, 100, 1000 i.e., to the powers of 10 . The procedure of multiplication using the Nikhilam involves minimum number of steps, space, time saving and only mental calculation. The numbers taken can be either less or more than the base considered.

The difference between the number and the base is termed as deviation. Deviation may be positive or negative. Positive deviation is written without the positive sign and the negative deviation, is written using Rekhank (a bar on the number). Now observe the following table.

Number

Base

Number – Base

Deviation

14 - 10

8 - 10

-2 or 2

100

97 - 100

-03 or 03

112

100

112 - 100

___

993

1000

993 - 1000

-007 or 007

1011

1000

1011 - 1000

011

Some rules of the method (near to the base) in Multiplication

a) Since deviation is obtained by Nikhilam sutra we call the method as Nikhilam multiplication.

Eg : 94. Now deviation can be obtained by ‘all from 9 and the last from 10’ sutra i.e., the last digit 4 is from 10 and remaining digit 9 from 9 gives 06.

b) The two numbers under consideration are written one below the other. The deviations are written on the right hand side.

Eg : Multiply 7 by 8.

Now the base is 10. Since it is near to both the numbers,

we write the numbers one below the other. 8

-----

Take the deviations of both the numbers from

the base and represent _

7 3

Rekhank or the minus sign before the deviations 8 2

------

or 7 -3

8 -2

-------

or remainders 3 and 2 implies that the numbers to be multiplied are both less than 10

c) The product or answer will have two parts, one on the left side and the other on the right. A vertical or a slant line i.e., a slash may be drawn for the demarcation of the two parts i.e.,

(or)

d) The R.H.S. of the answer is the product of the deviations of the numbers. It shall contain the number of digits equal to number of zeroes in the base.

i.e., 7 3

8 2

_____________

/ (3x2) = 6

Since base is 10, 6 can be taken as it is.

e) L.H.S of the answer is the sum of one number with the deviation of the other. It can be arrived at in any one of the four ways.

i) Cross-subtract deviation 2 on the second row from the original number 7 in the first row i.e., 7-2 = 5.

ii) Cross–subtract deviation 3 on the first row from the original number8 in the second row (converse way of (i))

i.e., 8 - 3 = 5

iii) Subtract the base 10 from the sum of the given numbers.

i.e., (7 + 8) – 10 = 5

iv) Subtract the sum of the two deviations from the base.

i.e., 10 – ( 3 + 2) = 5

Hence 5 is left hand side of the answer.

Thus 7 3

8 2

¯¯¯¯¯¯¯¯¯¯¯¯

5 /

Now (d) and (e) together give the solution

7 3 7

8 2 i.e., X 8

¯¯¯¯¯¯¯ ¯¯¯¯¯¯

5 / 6 56

f) If R.H.S. contains less number of digits than the number of zeros in the base, the remaining digits are filled up by giving zero or zeroes on the left side of the R.H.S. If the number of digits are more than the number of zeroes in the base, the excess digit or digits are to be added to L.H.S of the answer.

The general form of the multiplication under Nikhilam can be shown as follows :

Let N1 and N2 be two numbers near to a given base in powers of 10, and D1 and D2 are their respective deviations from the base. Then N1 X N2 can be represented as

Case (i) : Both the numbers are lower than the base. We have already considered the example 7 x 8 , with base 10.

Now let us solve some more examples by taking bases 100 and 1000 respectively.

Ex. 1: Find 97 X 94. Here base is 100. Now following the rules, the working is as follows:

Ex. 2: 98 X 97 Base is 100.

Ex. 3: 75X95. Base is 100.

Ex. 4: 986 X 989. Base is 1000.

Ex. 5: 994X988. Base is 1000.

Ex. 6: 750X995.

Case ( ii) : Both the numbers are higher than the base.

The method and rules follow as they are. The only difference is the positive deviation. Instead of cross – subtract, we follow cross – add.

Ex. 7: 13X12. Base is 10

Ex. 8: 18X14. Base is 10.

Ex. 9: 104X102. Base is 100.

104 04

102 02

¯¯¯¯¯¯¯¯¯¯¯¯

106 / 4x2 = 10608 ( rule - f )

¯¯¯¯¯¯¯¯¯¯¯¯

Ex. 10: 1275X1004. Base is 1000.

1275 275

1004 004

¯¯¯¯¯¯¯¯¯¯¯¯

1279 / 275x4 = 1279 / 1100 ( rule - f )

____________ = 1280100

Case ( iii ): One number is more and the other is less than the base.

In this situation one deviation is positive and the other is negative. So the product of deviations becomes negative. So the right hand side of the answer obtained will therefore have to be subtracted. To have a clear representation and understanding a vinculum is used. It proceeds into normalization.

Ex.11: 13X7. Base is 10

Note : Conversion of common number into vinculum number and vice versa.

Eg :

9 = 10 – 1 = 11

98 = 100 – 2 = 102

196 = 200 – 4 = 204

32 = 30 – 2 = 28

145 = 140 – 5 = 135

322 = 300 – 22 = 278. etc

The procedure can be explained in detail using Nikhilam Navatascaram Dasatah, Ekadhikena purvena, Ekanyunena purvena in the foregoing pages of this book.]

Ex. 12: 108 X 94. Base is 100.

Ex. 13: 998 X 1025. Base is 1000.

Algebraic Proof:

Case ( i ):

Let the two numbers N1 and N2 be less than the selected base say x.

N1 = (x-a), N2 = (x-b). Here a and b are the corresponding deviations of the numbers N1 and N2 from the base x. Observe that x is a multiple of 10.

Now N1 X N2 = (x-a) (x-b) = x.x – x.b – a.x + ab

= x (x – a – b ) + ab. [rule – e(iv), d ]

= x [(x – a) – b] + ab = x (N1–b) + ab[rule–e(i),d]

or = x [(x – b) – a] = x (N2 – a) + ab. [rule –e (ii),d]

x (x – a – b) + ab can also be written as

x[(x – a) + (x – b) – x] + ab = x[N1+N2 – x] + ab [rule – e(iii),d].

A difficult can be faced, if the vertical multiplication of the deficit digits or deviations i.e., a.b yields a product consisting of more than the required digits. Then rule-f will enable us to surmount the difficulty.

Case ( ii ) :

When both the numbers exceed the selected base, we have N1 = x + a, N2 = x + b, x being the base. Now the identity (x+a) (x+b) = x(x+a+b) + a.b holds good, of course with relevant details mentioned in case (i).

Case ( iii ) :

When one number is less and another is more than the base, we can use (x-a)(x+b) = x(x–a+ b)–ab. and the procedure is evident from the examples given.

Find the following products by Nikhilam formula.

1) 7 X 4 2) 93 X 85 3) 875 X 994

4) 1234 X 1002 5) 1003 X 997 6) 11112 X 9998

7) 1234 X 1002 8) 118 X 105

Nikhilam in Division

Consider some two digit numbers (dividends) and same divisor 9. Observe the following example.

i) 13 ÷ 9 The quotient (Q) is 1, Remainder (R) is 4.

since 9 ) 13 ( 1

____

ii) 34 ÷ 9, Q is 3, R is 7.

iii) 60 ÷ 9, Q is 6, R is 6.

iv) 80 ÷ 9, Q is 8, R is 8.

Now we have another type of representation for the above examples as given hereunder:

i) Split each dividend into a left hand part for the Quotient and right - hand part for the remainder by a slant line or slash.

Eg. 13 as 1 / 3, 34 as 3 / 4 , 80 as 8 / 0.

ii) Leave some space below such representation, draw a horizontal line.

Eg. 1 / 3 3 / 4 8 / 0

______ , ______ , ______

iii) Put the first digit of the dividend as it is under the horizontal line. Put the same digit under the right hand part for the remainder, add the two and place the sum i.e., sum of the digits of the numbers as the remainder.

Eg.

1 / 3 3 / 4 8 / 0

1 3 8

______ , ______ , ______

1 / 4 3 / 7 8 / 8

Now the problem is over. i.e.,

13 ÷ 9 gives Q = 1, R = 4

34 ÷ 9 gives Q = 3, R = 7

80 ÷ 9 gives Q = 8, R = 8

Proceeding for some more of the two digit number division by 9, we get

a) 21 ÷ 9 as

9) 2 / 1 i.e Q=2, R=3

¯¯¯¯¯¯

2 / 3

b) 43 ÷ 9 as

9) 4 / 3 i.e Q = 4, R = 7.

¯¯¯¯¯¯

4 / 7

The examples given so far convey that in the division of two digit numbers by 9, we can mechanically take the first digit down for the quotient – column and that, by adding the quotient to the second digit, we can get the remainder.

Now in the case of 3 digit numbers, let us proceed as follows.

9 ) 104 ( 11 9 ) 10 / 4

99 1 / 1

¯¯¯¯¯¯ as ¯¯¯¯¯¯¯

5 11 / 5

ii)

9 ) 212 ( 23 9 ) 21 / 2

207 2 / 3

¯¯¯¯¯ as ¯¯¯¯¯¯¯

5 23 / 5

iii)

9 ) 401 ( 44 9 ) 40 / 1

396 4 / 4

¯¯¯¯¯ as ¯¯¯¯¯¯¯¯

5 44 / 5

Note that the remainder is the sum of the digits of the dividend. The first digit of the dividend from left is added mechanically to the second digit of the dividend to obtain the second digit of the quotient. This digit added to the third digit sets the remainder. The first digit of the dividend remains as the first digit of the quotient.

Consider 511 ÷ 9

Add the first digit 5 to second digit 1 getting 5 + 1 = 6. Hence Quotient is 56. Now second digit of 56 i.e., 6 is added to third digit 1 of dividend to get the remainder i.e., 1 + 6 = 7

Thus

9 ) 51 / 1

5 / 6

¯¯¯¯¯¯¯

56 / 7

Q is 56, R is 7.

Extending the same principle even to bigger numbers of still more digits, we can get the results.

Eg : 1204 ÷ 9

i) Add first digit 1 to the second digit 2. 1 + 2 = 3

ii) Add the second digit of quotient 13. i.e., 3 to third digit ‘0’ and obtain the Quotient. 3 + 0 = 3, 133

iii) Add the third digit of Quotient 133 i.e.,3 to last digit ‘4’ of the dividend and write the final Quotient and Remainder. R = 3 + 4 = 7, Q = 133

In symbolic form 9 ) 120 / 4

13 / 3

¯¯¯¯¯¯¯¯

133 / 7

Another example.

9 ) 13210 / 1 132101 ÷ 9

gives

1467 / 7 Q = 14677, R = 8

¯¯¯¯¯¯¯¯¯¯

14677 / 8

In all the cases mentioned above, the remainder is less than the divisor. What about the case when the remainder is equal or greater than the divisor?

Eg.

9 ) 3 / 6 9) 24 / 6

3 2 / 6

¯¯¯¯¯¯ or ¯¯¯¯¯¯¯¯

3 / 9 (equal) 26 / 12 (greater).

We proceed by re-dividing the remainder by 9, carrying over this Quotient to the quotient side and retaining the final remainder in the remainder side.

9 ) 3 / 6 9 ) 24 / 6

/ 3 2 / 6

¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯

3 / 9 26 / 12

¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯

4 / 0 27 / 3

Q = 4, R = 0 Q = 27, R = 3.

When the remainder is greater than divisor, it can also be represented as

9 ) 24 / 6

2 / 6

¯¯¯¯¯¯¯¯

26 /1 / 2

/ 1

¯¯¯¯¯¯¯¯

1 / 3

¯¯¯¯¯¯¯¯

27 / 3

Now consider the divisors of two or more digits whose last digit is 9,when divisor is 89.

We Know 113 = 1 X 89 + 24, Q =1, R = 24

10015 = 112 X 89 + 47, Q = 112, R = 47.

Representing in the previous form of procedure, we have

89 ) 1 / 13 89 ) 100 / 15

/ 11 12 / 32

¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯

1 / 24 112 / 47

But how to get these? What is the procedure?

Now Nikhilam rule comes to rescue us. The nikhilam states “all from 9 and the last from 10”. Now if you want to find 113 ÷ 89, 10015 ÷ 89, you have to apply nikhilam formula on 89 and get the complement 11.Further while carrying the added numbers to the place below the next digit, we have to multiply by this 11.

89 ) 1 / 13 89 ) 100 / 15

¯¯

/ 11 11 11 / first digit 1 x 11

¯¯¯¯¯¯¯¯

1 / 24 1 / 1 total second is 1x11

22 total of 3rd digit is 2 x 11

¯¯¯¯¯¯¯¯¯¯

112 / 47

What is 10015 ÷ 98 ? Apply Nikhilam and get 100 – 98 = 02. Set off the 2 digits from the right as the remainder consists of 2 digits. While carrying the added numbers to the place below the next digit, multiply by 02.

Thus

98 ) 100 / 15

¯¯

02 02 / i.e., 10015 ÷ 98 gives

0 / 0 Q = 102, R = 19

/ 04

¯¯¯¯¯¯¯¯¯¯

102 / 19

In the same way

897 ) 11 / 422

¯¯¯

103 1 / 03

/ 206

¯¯¯¯¯¯¯¯¯

12 / 658

gives 11,422 ÷ 897, Q = 12, R=658.

In this way we have to multiply the quotient by 2 in the case of 8, by 3 in the case of 7, by 4 in the case of 6 and so on. i.e., multiply the Quotient digit by the divisors complement from 10. In case of more digited numbers we apply Nikhilam and proceed. Any how, this method is highly useful and effective for division when the numbers are near to bases of 10.

* Guess the logic in the process of division by 9.

* Obtain the Quotient and Remainder for the following problems.

1) 311 ÷ 9 2) 120012 ÷ 9 3) 1135 ÷ 97

4) 2342 ÷ 98 5) 113401 ÷ 997

6) 11199171 ÷ 99979

Observe that by nikhilam process of division, even lengthier divisions involve no division or no subtraction but only a few multiplications of single digits with small numbers and a simple addition. But we know fairly well that only a special type of cases are being dealt and hence many questions about various other types of problems arise. The answer lies in Vedic Methods.

oooooo

SŨNYAM SĀMYASAMUCCAYE

The Sutra 'Sunyam Samyasamuccaye' says the 'Samuccaya is the same, that Samuccaya is Zero.' i.e., it should be equated to zero. The term 'Samuccaya' has several meanings under different contexts.
i) We interpret, 'Samuccaya' as a term which occurs as a common factor in all the terms concerned and proceed as follows.
Example 1: The equation 7x + 3x = 4x + 5x has the same factor ‘ x ‘ in all its terms. Hence by the sutra it is zero, i.e., x = 0.
Otherwise we have to work like this:

                    7x + 3x = 4x + 5x
                         10x = 9x
                  10x – 9x = 0
                            x = 0
    This is applicable not only for ‘x’ but also any such unknown quantity as follows.
Example 2:      5(x+1) = 3(x+1)
No need to proceed in the usual procedure like

                        5x + 5 = 3x + 3
                        5x – 3x = 3 – 5
                    2x = -2     or     x = -2 ÷ 2 = -1
    Simply think of the contextual meaning of ‘ Samuccaya ‘
    Now Samuccaya is     ( x + 1 )
                                    x + 1 = 0     gives     x = -1

ii) Now we interpret ‘Samuccaya ‘as product of independent terms in expressions like (x+a) (x+b)

Example 3:     ( x + 3 ) ( x + 4) = ( x – 2) ( x – 6 )

                Here Samuccaya is     3 x 4 = 12 = -2 x -6
                Since it is same ,   we derive x = 0
   This example, we have already dealt in type ( ii ) of Paravartya in solving simple equations.

iii) We interpret ‘ Samuccaya ‘ as the sum of the denominators of two fractions having the same numerical numerator.

Consider the example.

                               1            1
                            ____   +  ____   =   0
                            3x-2        2x-1
            for this we proceed by taking L.C.M.

                    (2x-1)+(3x–2)
                   ____________   =   0
                    (3x–2)(2x–1)

                          5x–3
                    __________   =   0
                    (3x–2)(2x–1)

                        5x – 3 = 0        5x = 3

                                3
                        x = __
                                5
    Instead of this, we can directly put the Samuccaya i.e., sum of the denominators
                i.e., 3x – 2 + 2x - 1 = 5x - 3 = 0
                giving 5x = 3         x = 3 / 5
    It is true and applicable for all problems of the type

                    m           m
                 ____ +  _____   =   0
                 ax+b       cx+d
    Samuccaya is ax+b+cx+d and solution is ( m ≠ 0 )

                        - ( b + d )
                 x =   _________
                           ( a + c )
    iii) We now interpret ‘Samuccaya’ as combination or total.

         If the sum of the numerators and the sum of the denominators be the same, then that sum = 0.

    Consider examples of type

                       ax + b         ax + c
                       _____   =   ______
                       ax + c         ax + b
        In this case, (ax+b) (ax+b) = (ax+c) (ax+c)
                        a²x² + 2abx + b² = a²x² + 2acx + c²
                                  2abx – 2acx = c² – b²
                              x ( 2ab – 2ac ) = c² – b²

                        c²–b²        (c+b)(c-b)       -(c+b)
             x =   ______   = _________ =   _____
                       2a(b-c)         2a(b-c)            2a
            As per Samuccaya (ax+b) + (ax+c) = 0
                                                     2ax+b+c = 0
                                                         2ax = -b-c

                                     -(c+b)
                              x = ______
                                         2a            Hence the statement.
Example 4:
                             3x + 4        3x + 5
                             ______ =   ______
                             3x + 5        3x + 4
        Since N₁ + N₂ = 3x + 4 + 3x + 5 = 6x + 9 ,
             And   D₁ + D₂ = 3x + 4 + 3x + 5 = 6x + 9
             We have N₁ + N₂ = D₁ + D₂ = 6x + 9
                    Hence from Sunya Samuccaya we get 6x + 9 = 0
                     6x = -9

                                 -9        -3
                          x = __   =   __
                                6          2
Example 5:
                        5x + 7       5x + 12
                        _____   = _______
                        5x +12       5x + 7
       Hence N₁ + N₂ = 5x + 7 + 5x + 12 = 10x + 19
            And     D₁ + D₂ = 5x + 12 + 5x + 7 = 10x + 19
                      N₁ + N₂ = D₁ + D₂ gives 10x + 19 = 0
                                    10x = -19

                                         -19
                                  x = ____
                                           10
Consider the examples of the type, where N₁ + N₂ = K ( D₁ + D₂ ), where K is a numerical constant, then also by removing the numerical constant K, we can proceed as above.
Example 6:

                        2x + 3        x + 1
                        _____   =   ______
                        4x + 5        2x + 3
        Here N₁ + N₂ = 2x + 3 + x + 1 = 3x + 4
                    D₁ + D₂ = 4x + 5 + 2x + 3 = 6x + 8
                                                             = 2 ( 3x + 4 )
        Removing the numerical factor 2, we get 3x + 4 on both sides.

            3x + 4 = 0     3x = -4     x = - 4 / 3.
v) ‘Samuccaya‘ with the same meaning as above, i.e., case (iv), we solve the problems leading to quadratic equations. In this context, we take the problems as follows;
       If N₁ + N₂ = D₁ + D₂ and also the differences
               N₁ ~ D₁ = N₂ ~ D₂ then both the things are equated to zero, the solution gives the two values for x.
Example 7:

                     3x + 2     2x + 5
                     _____ = ______
                     2x + 5     3x + 2
    In the conventional text book method, we work as follows :

                        3x + 2            2x + 5
                        _____      =    ______
                        2x + 5            3x + 2

                    ( 3x + 2 ) ( 3x + 2 ) = ( 2x + 5 ) ( 2x + 5 )
                            9x² + 12x + 4 = 4x² + 20x + 25
                     9x² + 12x + 4 - 4x² - 20x – 25 = 0
                                            5x² – 8x – 21 = 0
                                    5x² – 15x + 7x – 21 = 0
                              5x ( x – 3 ) + 7 ( x – 3 ) = 0
                                      (x – 3 ) ( 5x + 7 ) = 0
                                     x – 3 = 0 or 5x + 7 = 0
                                          x = 3 or - 7 / 5
    Now ‘Samuccaya’ sutra comes to help us in a beautiful way as follows :

                    Observe N₁ + N₂ = 3x + 2 + 2x + 5 = 5x + 7
                                D₁ + D₂ = 2x + 5 + 3x + 2 = 5x + 7

        Further N₁ ~ D₁ = ( 3x + 2 ) – ( 2x + 5 ) = x – 3
                    N₂ ~ D₂ = ( 2x + 5) – ( 3x + 2 ) = - x + 3 = - ( x – 3 )

        Hence 5x + 7 = 0 , x – 3 = 0
                      5x = -7 , x = 3
                            i.e., x = -7 / 5 , x = 3

        Note that all these can be easily calculated by mere observation.
Example 8:

                            3x + 4       5x + 6
                            ______ = _____
                            6x + 7       2x + 3

        Observe that
                    N₁ + N₂ = 3x + 4 + 5x + 6 = 8x + 10
              and D₁ + D₂ = 6x + 7 + 2x + 3 = 8x + 10

        Further     N₁ ~ D₁ = (3x + 4) – (6x + 7)
                                   = 3x + 4 – 6x – 7
                                   = -3x – 3 = -3 ( x + 1 )
                      N₂ ~ D₂ = (5x + 6) – (2x + 3) = 3x + 3 = 3( x + 1)
    By ‘Sunyam Samuccaye’ we have

                    8x + 10 = 0             3( x + 1 ) = 0
                           8x = -10               x + 1 = 0
                             x = - 10 / 8              x = -1
                               = - 5 / 4
vi)‘Samuccaya’ with the same sense but with a different context and application .

         Example 9:

                         1           1            1           1
                      ____ + _____ = ____ + ____
                      x - 4      x – 6       x - 2      x - 8
        Usually we proceed as follows.

                 x–6+x-4                 x–8+x-2
               ___________   =    ___________
                (x–4) (x–6)             (x–2) (x-8)

                    2x-10                 2x-10
                _________     =     _________
                x²–10x+24            x²–10x+16

              ( 2x – 10 ) ( x² – 10x + 16 ) = ( 2x – 10 ) ( x² – 10x + 24)
           2x³–20x²+32x–10x²+100x–160 = 2x³–20x²+48x–10x²+100x-240
               2x³ – 30x² + 132x – 160 = 2x³ – 30x² + 148x – 240
                                 132x – 160 = 148x – 240
                               132x – 148x = 160 – 240
                                        – 16x = - 80
                                   x = - 80 / - 16 = 5
Now ‘Samuccaya’ sutra, tell us that, if other elements being equal, the sum-total of the denominators on the L.H.S. and their total on the R.H.S. be the same, that total is zero.
    Now D₁ + D₂ = x – 4 + x – 6 = 2x – 10, and
                D₃ + D₄ = x – 2 + x – 8 = 2x – 10
By Samuccaya, 2x – 10 gives 2x = 10

                           10
                    x = __ = 5
                            2
Example 10:

                1         1            1           1
             ____ + ____ =   ____ + _____
             x - 8     x – 9      x - 5      x – 12

            D₁ + D₂ = x – 8 + x – 9 = 2x – 17, and
            D₃ + D₄ = x – 5 + x –12 = 2x – 17
                    Now 2x – 17 = 0 gives 2x = 17

                                    17
                            x = __ = 8½
                                     2
Example 11:

                1           1           1          1
             ____ - _____ =   ____ - _____
             x + 7     x + 10      x + 6      x + 9

     This is not in the expected form. But a little work regarding transposition makes the above as follows.

                1          1            1           1
             ____ + ____ =  ____ + _____
             x + 7     x + 9      x + 6      x + 10

     Now ‘Samuccaya’ sutra applies

            D₁ + D₂ = x + 7 + x + 9 = 2x + 16, and
            D₃ + D₄ = x + 6 + x + 10 = 2x + 16
   Solution is given by 2x + 16 = 0 i.e., 2 x = - 16.
                                                             x = - 16 / 2 = - 8.

Solve the following problems using Sunyam Samya-Samuccaye process.

        1.     7 ( x + 2 ) + 3 ( x + 2 ) = 6 ( x + 2 ) + 5 ( x + 2 )

        2.     ( x + 6 ) ( x + 3 ) = ( x – 9 ) ( x – 2 )

        3.     ( x - 1 ) ( x + 14 ) = ( x + 2 ) ( x – 7 )

                      1                 1
        4.       ______   + ____    =   0
                  4 x - 3          x – 2

                       4                 4
        5.       _____    +   _____    =   0
                  3x + 1         5x + 7

                    2x + 11         2x+5
        6.         ______   =   _____
                     2x+ 5          2x+11

                    3x + 4            x + 1
        7.         ______   =   _____
                    6x + 7           2x + 3

                    4x - 3             x + 4
        8.         ______   =   _____
                     2x+ 3           3x - 2


                   1            1             1              1
        9.     ____ + ____   =   ____ + _____
                 x - 2        x - 5       x - 3         x - 4

                   1            1             1               1
        10.     ____ -   ____   =   _____ - _____
                   x - 7        x - 6       x - 10        x - 9

Sunyam Samya Samuccaye in Certain Cubes:
Consider the problem ( x – 4 )³ + ( x – 6 )³ = 2 ( x – 5 )³. For the solution by the traditional method we follow the steps as given below:

                 ( x – 4 )³ + ( x – 6 )³ = 2 ( x – 5 )³
          x³ – 12x² + 48x – 64 + x³ – 18x² + 108x – 216
                                                = 2 ( x³ – 15x² + 75x – 125 )
             2x³ – 30x² + 156x – 280 = 2x³ – 30x² + 150x – 250
                                156x – 280 = 150x – 250
                               156x – 150x = 280 – 250
                                            6x = 30
                                              x = 30 / 6 = 5
But once again observe the problem in the vedic sense
We have ( x – 4 ) + ( x – 6 ) = 2x – 10. Taking out the numerical factor 2, we have ( x – 5 ) = 0, which is the factor under the cube on R.H.S. In such a case “Sunyam samya Samuccaye” formula gives that x – 5 = 0. Hence x = 5
Think of solving the problem (x–249)³ + (x+247)³ = 2(x–1)³
The traditional method will be horrible even to think of.
But ( x – 249 ) + ( x + 247 ) = 2x – 2 = 2 ( x – 1 ). And x – 1. on R.H.S. cube, it is enough to state that x – 1 = 0 by the ‘sutra’.
x = 1 is the solution. No cubing or any other mathematical operations.
Algebraic Proof :

        Consider ( x – 2a )³ + ( x – 2b )³ = 2 ( x – a – b )³ it is clear that
                                x – 2a + x – 2b = 2x – 2a – 2b
                                                     = 2 ( x – a – b )
        Now the expression,

         x³ - 6x²a + 12xa² – 8a³ + x³ – 6x²b + 12xb² – 8b³ =
                 2(x³–3x²a–3x²b+3xa²+3xb²+6axb–a³–3a²b–3ab²–b³)
             = 2x³–6x²a–6x²b+6xa²+6xb²+12xab–2a³–6a²b–6ab²–2b³
   cancel the common terms on both sides

       12xa²+12xb²–8a³–8b³ = 6xa²+6xb²+12xab–2a³–6a²b–6ab²–2b³
            6xa² + 6xb² – 12xab = 6a³ + 6b³ – 6a²b – 6ab²
              6x ( a² + b² – 2ab ) = 6 [ a³ + b³ – ab ( a + b )]
                  x ( a – b )² = [ ( a + b ) ( a² + b² –ab ) – ( a + b )ab]
                                  = ( a + b ) ( a² + b² – 2ab )
                                  = ( a + b ) ( a – b )²

                             since x = a + b

Solve the following using “Sunyam Samuccaye” process :

            1.     ( x – 3 )³ + ( x – 9 )³ = 2 ( x – 6 )³

            2.     ( x + 4 )³ + ( x – 10 )³ = 2 ( x – 3 )³

            3.     ( x + a + b – c )³ + ( x + b + c – a )³ = 2 ( x + b )³

Example :

                        (x + 2)³x + 1
                         ______ =   _____
                        (x + 3)³x + 4
    with the text book procedures we proceed as follows

                x³+ 6x²+ 12x +8          x + 1
                _______________    =   _____
                x³+ 9x²+ 27x +27        x + 4
    Now by cross multiplication,

    ( x + 4 ) ( x³ + 6x² + 12x + 8 ) = ( x + 1 ) ( x³ + 9x² + 27x + 27 )
       x⁴ + 6x³ + 12x² + 8x + 4x³ + 24x² + 48x + 32 =
                               x⁴ + 9x³ + 27x² + 27x + x³ + 9x² + 27x + 27
        x⁴ + 10x³ + 36x² + 56x + 32 = x⁴ + 10x³ + 36x² + 54x + 27
                                  56x + 32 = 54x + 27
                                 56x – 54x = 27 – 32
                                           2x = - 5
                                             x = - 5 / 2
Observe that ( N₁ + D₁ ) with in the cubes on
                             L.H.S. is x + 2 + x + 3 = 2x + 5 and

                          N₂ + D₂ on the right hand side
                               is x + 1 + x + 4 = 2x + 5.

      By vedic formula we have 2x + 5 = 0         x = - 5 / 2.

Solve the following by using vedic method :

                1.
                                (x + 3)³        x+1
                                ______   =   ____
                                (x + 5)³        x+7

                2.
                                (x - 5)³         x - 3
                                ______    =   ____
                                (x - 7)³         x -

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