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 14 – 10 4

_

8 10 8 – 10 -2 or 2

__

97 100 97 – 100 -03 or 03

112 100 112 – 100 12

___

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,

7

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

_____________

/ (3×2) = 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 / 4×2 = 10608 ( rule – f )

¯¯¯¯¯¯¯¯¯¯¯¯

Ex. 10: 1275X1004. Base is 1000.

1275 275

1004 004

¯¯¯¯¯¯¯¯¯¯¯¯

1279 / 275×4 = 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

9

____

4

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

¯¯¯¯¯¯

2 / 3

b) 43 ÷ 9 as

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

4

¯¯¯¯¯¯

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.

i)

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 1×11

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.

ŨRDHVA TIRYAGBHYĀM

Urdhva – tiryagbhyam is the general formula applicable to all cases of multiplication and also in the division of a large number by another large number. It means “Vertically and cross wise.”

(a) Multiplication of two 2 digit numbers.

Ex.1: Find the product 14 X 12

i) The right hand most digit of the multiplicand, the first number (14) i.e., 4 is multiplied by the right hand most digit of the multiplier, the second number (12) i.e., 2. The product 4 X 2 = 8 forms the right hand most part of the answer.

ii) Now, diagonally multiply the first digit of the multiplicand (14) i.e., 4 and second digit of the multiplier (12) i.e., 1 (answer 4 X 1=4); then multiply the second digit of the multiplicand i.e., 1 and first digit of the multiplier i.e., 2 (answer 1 X 2 = 2); add these two i.e., 4 + 2 = 6. It gives the next, i.e., second digit of the answer. Hence second digit of the answer is 6.

iii) Now, multiply the second digit of the multiplicand i.e., 1 and second digit of the multiplier i.e., 1 vertically, i.e., 1 X 1 = 1. It gives the left hand most part of the answer.

Thus the answer is 16 8.

Symbolically we can represent the process as follows :

The symbols are operated from right to left .

Step i) :

Step ii) :

Step iii) :

Now in the same process, answer can be written as

23

13

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

2 : 6 + 3 : 9 = 299 (Recall the 3 steps)

Ex.3

41

X 41

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

16 : 4 + 4 : 1 = 1681.

What happens when one of the results i.e., either in the last digit or in the middle digit of the result, contains more than 1 digit ? Answer is simple. The right – hand – most digit there of is to be put down there and the preceding, i.e., left –hand –side digit or digits should be carried over to the left and placed under the previous digit or digits of the upper row. The digits carried over may be written as in Ex. 4.

Ex.4: 32 X 24

Step (i) : 2 X 4 = 8

Step (ii) : 3 X 4 = 12; 2 X 2 = 4; 12 + 4 = 16.

Here 6 is to be retained. 1 is to be carried out to left side.

Step (iii) : 3 X 2 = 6. Now the carried over digit 1 of 16 is to be added.

i.e., 6 + 1 = 7.

Thus 32 X 24 = 768

We can write it as follows

32

24

¯¯¯¯

668

1

¯¯¯¯

768.

Note that the carried over digit from the result (3X4) + (2X2) = 12+4 = 16 i.e., 1 is placed under the previous digit 3 X 2 = 6 and added.

After sufficient practice, you feel no necessity of writing in this way and simply operate or perform mentally.

Ex.5 28 X 35.

Step (i) : 8 X 5 = 40. 0 is retained as the first digit of the answer and 4 is carried over.

Step (ii) : 2 X 5 = 10; 8 X 3 = 24; 10 + 24 = 34; add the carried over 4 to 34. Now the result is 34 + 4 = 38. Now 8 is retained as the second digit of the answer and 3 is carried over.

Step (iii) : 2 X 3 = 6; add the carried over 3 to 6. The result 6 + 3 = 9 is the third or final digit from right to left of the answer.

Thus 28 X 35 = 980.

Ex.6

48

47

¯¯¯¯¯¯

1606

65

¯¯¯¯¯¯¯

2256

Step (i): 8 X 7 = 56; 5, the carried over digit is placed below the second digit.

Step (ii): ( 4 X 7) + (8 X 4) = 28 + 32 = 60; 6, the carried over digit is placed below the third digit.

Step (iii): Respective digits are added.

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 100th place) is obtained by vertical multiplication of a and c i.e., the digits in 10th place (coefficient of x) of both the numbers;

ii) The middle term, i.e., the coefficient of x (i.e., digit in the 10th 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 (3×2 + 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., 38×2

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

v) 3 X 0 = 0

Hence the product is 15×3 + 38×2 + 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) (6×2 + 5x + 2 ) (3×2 + 4x +7) 10) (4×2 + 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 + 5×2 + 3x + 7) by (x – 2) process by converse of urdhva – tiryak :

i) x3 divided by x gives x2 . x3 + 5×2 + 3x + 7

It is the first term of the Quotient. ___________________

x – 2

Q = x2 + – – – – – – – – – – –

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

x3 + 5×2 + 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 + 5×2 + 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 + 5×2 + 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) 3×2 – x – 6 2) 16×2 + 24x +9

¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯

3x – 7 4x+3

3) x3+ 2×2 +3x + 5 4) 12×4 – 3×2 – 3x + 12

¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

x – 3 x2 + 1

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.

12

-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

¯¯¯¯¯ ¯¯¯¯

10

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 6×2 + 5x + 4 by x – 1

X – 1 6×2 + 5x + 4

¯¯¯¯¯¯

1 6 + 11

¯¯¯¯¯¯¯¯¯¯¯¯

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

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

X – 5 x3 – 3×2 + 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 – 3×3 + 7×2 + 5x + 7

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

x + 4

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

x + 4 x4 – 3×3 + 7×2 + 5x + 7

¯¯¯¯¯

-4 – 4 + 28 – 140 + 540

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

x3 – 7×2 + 35x – 135 547

Thus Q = x3 – 7×2 + 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 – 7×2 + 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 2×4 – 3×3 – 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 2×4 – 3×3 + 0 . x2 – 3x + 2

x2 + 0 . x + 1 0 – 2

¯¯¯¯¯¯¯¯¯¯¯¯

0 – 1 0 + 3

0 + 2

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

2 – 3 – 2 0 4

Thus Q = 2×2 – 3x – 2 and R = 0 . x + 4 = 4.

Example 5 : 2×5 – 5×4 + 3×2 – 4x + 7 by x3 – 2×2 + 3.

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

x3 – 2×2 + 0 . x + 3 2×5 – 5×4 + 0.×3 + 3×2 – 4x + 7

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

2 0 – 3 4 0 – 6

-2 0 + 3

– 4 0 + 6

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

2 – 1 – 2 – 7 – 1 +13

Thus Q = 2×2 – 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) (4×2 + 3x + 5) ÷ (x+1)

2) (x3 – 4×2 + 7x + 6) ÷ (x – 2)

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

4) (2×5 + x3 – 3x + 7) ÷ (x3 + 2x – 3)

5) (7×6 + 6×5 – 5×4 + 4×3 – 3×2 + 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)

7×2 + 21x + 10x + 30 = 7×2 + 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

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 )

9×2 + 12x + 4 = 4×2 + 20x + 25

9×2 + 12x + 4 – 4×2 – 20x – 25 = 0

5×2 – 8x – 21 = 0

5×2 – 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)

2×3–20×2+32x–10×2+100x–160 = 2×3–20×2+48x–10×2+100x-240

2×3 – 30×2 + 132x – 160 = 2×3 – 30×2 + 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 – 12×2 + 48x – 64 + x3 – 18×2 + 108x – 216

= 2 ( x3 – 15×2 + 75x – 125 )

2×3 – 30×2 + 156x – 280 = 2×3 – 30×2 + 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)

= 2×3–6x2a–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

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 + 6×2 + 12x +8 x + 1

_______________ = _____

x3 + 9×2 + 27x +27 x + 4

Now by cross multiplication,

( x + 4 ) ( x3 + 6×2 + 12x + 8 ) = ( x + 1 ) ( x3 + 9×2 + 27x + 27 )

x4 + 6×3 + 12×2 + 8x + 4×3 + 24×2 + 48x + 32 =

x4 + 9×3 + 27×2 + 27x + x3 + 9×2 + 27x + 27

x4 + 10×3 + 36×2 + 56x + 32 = x4 + 10×3 + 36×2 + 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 – 9

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.

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

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 + 6×2 + 11 x + 6 = 0.

Since (x + 2 )3 = x3 + 6×2 + 12x + 8

Add ( x + 2 ) to both sides

We get x3 + 6×2 + 11x + 6 + x + 2 = x + 2

i.e.,, x3 + 6×2 + 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 + 8×2 + 17x + 10 = 0

We know ( x + 3 )3 = x3 + 9×2 + 27x + 27

So adding on the both sides, the term ( x2 + 10x + 17 ), we get

x3 + 8×2 + 17x + x2 + 10x + 17 = x2 + 10x + 17

i.e.,, x3 + 9×2 + 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 – 6×2 + 11x – 6 = 0

2. x3 + 9×2 + 23x + 15 = 0

3. x2 + 2x – 3 = 0

4. x4 + 4×3 + 6×2 + 4x – 15 = 0

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 1000th place is x, 100th place is ( y – 1 ) and the two digit number which makes up the 10th and unit place is the number obtained by subtracting the multiplicand from 100.(or apply nikhilam).

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

100th 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.

ĀNURŨPYENA

The upa-Sutra ‘anurupyena’ means ‘proportionality’. This Sutra is highly useful to find products of two numbers when both of them are near the Common bases i.e powers of base 10 . It is very clear that in such cases the expected ‘Simplicity ‘ in doing problems is absent.

Example 1: 46 X 43

As per the previous methods, if we select 100 as base we get

46 -54 This is much more difficult and of no use.

43 -57

¯¯¯¯¯¯¯¯

Now by ‘anurupyena’ we consider a working base In three ways. We can solve the problem.

Method 1: Take the nearest higher multiple of 10. In this case it is 50.

Treat it as 100 / 2 = 50. Now the steps are as follows:

i) Choose the working base near to the numbers under consideration.

i.e., working base is 100 / 2 = 50

ii) Write the numbers one below the other

i.e. 4 6

4 3

¯¯¯¯¯¯¯

iii) Write the differences of the two numbers respectively from 50 against each number on right side

i.e. 46 -04

43 -07

¯¯¯¯¯¯¯¯¯

iv) Write cross-subtraction or cross- addition as the case may be under the line drawn.

v) Multiply the differences and write the product in the left side of the answer.

46 -04

43 -07

____________

39 / -4 x –7

= 28

vi) Since base is 100 / 2 = 50 , 39 in the answer represents 39X50.

Hence divide 39 by 2 because 50 = 100 / 2

Thus 39 ÷ 2 gives 19½ where 19 is quotient and 1 is remainder . This 1 as Reminder gives one 50 making the L.H.S of the answer 28 + 50 = 78(or Remainder ½ x 100 + 28 )

i.e. R.H.S 19 and L.H.S 78 together give the answer 1978 We represent it as

46 -04

43 -07

¯¯¯¯¯¯¯¯¯

2) 39 / 28

¯¯¯¯¯¯¯¯¯

19½ / 28

= 19 / 78 = 1978

Example 2: 42 X 48.

With 100 / 2 = 50 as working base, the problem is as follows:

42 -08

48 -02

¯¯¯¯¯¯¯¯¯

2) 40 / 16

¯¯¯¯¯¯¯¯¯

20 / 16

42 x 48 = 2016

Method 2: For the example 1: 46X43. We take the same working base 50. We treat it as 50=5X10. i.e. we operate with 10 but not with 100 as in method

now

(195 + 2) / 8 = 1978

[Since we operate with 10, the R.H.S portion shall have only unit place .Hence out of the product 28, 2 is carried over to left side. The L.H.S portion of the answer shall be multiplied by 5, since we have taken 50 = 5 X 10.]

Now in the example 2: 42 x 48 we can carry as follows by treating 50 = 5 x 10

Method 3: We take the nearest lower multiple of 10 since the numbers are 46 and 43 as in the first example, We consider 40 as working base and treat it as 4 X 10.

Since 10 is in operation 1 is carried out digit in 18.

Since 4 X 10 is working base we consider 49 X 4 on L.H.S of answer i.e. 196 and 1 carried over the left side, giving L.H.S. of answer as 1978. Hence the answer is 1978.

We proceed in the same method for 42 X 48

Let us see the all the three methods for a problem at a glance

Example 3: 24 X 23

Method – 1: Working base = 100 / 5 = 20

24 04

23 03

¯¯¯¯¯¯¯¯

5) 27 / 12

¯¯¯¯¯¯¯¯

5 2/5 / 12 = 5 / 52 = 552

[Since 2 / 5 of 100 is 2 / 5 x 100 = 40 and 40 + 12 = 52]

Method – 2: Working base 2 X 10 = 20

Now as 20 itself is nearest lower multiple of 10 for the problem under consideration, the case of method – 3 shall not arise.

Let us take another example and try all the three methods.

Example 4: 492 X 404

Method – 1 : working base = 1000 / 2 = 500

492 -008

404 -096

¯¯¯¯¯¯¯¯¯¯¯

2) 396 / 768 since 1000 is in operation

¯¯¯¯¯¯¯¯¯¯¯

198 / 768 = 198768

Method 2: working base = 5 x 100 = 500

Method – 3.

Since 400 can also be taken as working base, treat 400 = 4 X 100 as working base.

Thus

No need to repeat that practice in these methods finally takes us to work out all these mentally and getting the answers straight away in a single line.

Example 5: 3998 X 4998

Working base = 10000 / 2 = 5000

3998 -1002

4998 -0002

¯¯¯¯¯¯¯¯¯¯¯¯

2) 3996 / 2004 since 10,000 is in operation

1998 / 2004 = 19982004

or taking working base = 5 x 1000 = 5,000 and

What happens if we take 4000 i.e. 4 X 1000 as working base?

_____

3998 0002

4998 0998 Since 1000 is an operation

¯¯¯¯¯¯¯¯¯¯¯¯

4996 / 1996

___ ___

As 1000 is in operation, 1996 has to be written as 1996 and 4000 as base, the L.H.S portion 5000 has to be multiplied by 4. i. e. the answer is

A simpler example for better understanding.

Example 6: 58 x 48

Working base 50 = 5 x 10 gives

Since 10 is in operation.

Use anurupyena by selecting appropriate working base and method.

Find the following product.

1. 46 x 46 2. 57 x 57 3. 54 x 45

4. 18 x 18 5. 62 x 48 6. 229 x 230

7. 47 x 96 8. 87965 x 99996 9. 49×499

10. 389 x 512

ĀDYAMĀDYENĀNTYA – MANTYENA

The Sutra ‘ adyamadyenantya-mantyena’ means ‘the first by the first and the last by the last’.

Suppose we are asked to find out the area of a rectangular card board whose length and breadth are respectively 6ft . 4 inches and 5 ft. 8 inches. Generally we continue the problem like this.

Area = Length X Breath

= 6’ 4″ X 5’ 8″ Since 1’ = 12″, conversion

= ( 6 X 12 + 4) ( 5 X 12 + 8) in to single unit

= 76″ 68″ = 5168 Sq. inches.

Since 1 sq. ft. =12 X 12 = 144sq.inches we have area

5168 = 144) 5168 (35

¯¯¯¯

144 432

¯¯¯¯

848

720 i.e., 35 Sq. ft 128 Sq. inches

¯¯¯¯¯

128

By Vedic principles we proceed in the way “the first by first and the last by last”

i.e. 6’ 4″ can be treated as 6x + 4 and 5’ 8″ as 5x + 8,

Where x= 1ft. = 12 in; x2 is sq. ft.

Now ( 6x + 4 )(5x + 8 )

= 30×2 + 6.8.x + 4.5.x + 32

= 30×2 + 48x + 20x + 32

= 30×2 + 68. x + 32

= 30×2 + ( 5x + 8 ). x + 32 Writing 68 = 5 x 12 + 8

= 35×2 + 8. x + 32

= 35 Sq. ft. + 8 x 12 Sq. in + 32 Sq. in

= 35 Sq. ft. + 96 Sq. in + 32 Sq. in

= 35 Sq. ft. + 128 Sq. in

It is interesting to know that a mathematically untrained and even uneducated carpenter simply works in this way by mental argumentation. It goes in his mind like this

6’ 4″

5’ 8″

First by first i.e. 6’ X 5’ = 30 sq. ft.

Last by last i.e. 4″ X 8″ = 32 sq. in.

Now cross wise 6 X 8 + 5 x 4 = 48 +20 = 68.

Adjust as many ’12’ s as possible towards left as ‘units’ i.e. 68 = 5 X 12 +8 , 5 twelve’s as 5 square feet make the first 30+5 = 35 sq. ft ; 8 left becomes 8 x 12 square inches and go towards right i.e. 8 x 12 = 96 sq. in. towards right ives 96+32 = 128sq.in.

Thus he got area in some sort of 35 squints and another sort of 128 sq. units. i.e. 35 sq. ft 128 sq. in

Another Example:

Now 12 + 2 = 14, 10 x 12 + 24 = 120 + 24 = 144

Thus 4′ 6″ x 3′ 4″ = 14 Sq. ft. 144 Sq. inches.

Since 144 sq. in = 12 X 12 = 1 sq. ft The answer is 15 sq. ft.

We can extend the same principle to find volumes of parallelepiped also.

I. Find the area of the rectangles in each of the following situations.

1). l = 3’ 8″ , b = 2’ 4 ” 2). l = 12’ 5″ , b = 5’ 7″

3). l = 4 yard 3 ft. b = 2 yards 5 ft.(1yard =3ft)

4). l = 6 yard 6 ft. b = 5 yards 2 ft.

II. Find the area of the trapezium in each of the following cases. Recall area = ½ h (a + b) where a, b are parallel sides and h is the distance between them.

1). a = 3’ 7″, b = 2’ 4″, h = 1’ 5″

2). a = 5’ 6″, b = 4’ 4″, h = 3’ 2″

3). a = 8’ 4″, b = 4’ 6″, h = 5’ 1″.

Factorization of quadratics:

The usual procedure of factorizing a quadratic is as follows:

3×2 + 8x + 4

= 3×2 + 6x + 2x + 4

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

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

But by mental process, we can get the result immediately. The steps are as follows.

i). Split the middle coefficient in to two such parts that the ratio of the first coefficient to the first part is the same as the ratio of the second part to the last coefficient. Thus we split the coefficient of middle term of 3×2 + 8x + 4 i.e. 8 in to two such parts 6 and 2 such that the ratio of the first coefficient to the first part of the middle coefficient i.e. 3:6 and the ratio of the second pat to the last coefficient, i.e. 2: 4 are the same. It is clear that 3:6 = 2:4. Hence such split is valid. Now the ratio 3: 6 = 2: 4 = 1:2 gives one factor x+2.

ii). Second factor is obtained by dividing the first coefficient of the quadratic by the fist coefficient of the factor already found and the last coefficient of the quadratic by the last coefficient of the factor.

i.e. the second factor is

3×2 4

____ + ___ = 3x + 2

x 2

Hence 3×2 + 8x + 4 = ( x + 2 ) ( 3x + 2 )

Eg.1: 4×2 + 12x + 5

i) Split 12 into 2 and 10 so that as per rule 4 : 2 = 10 : 5 = 2 : 1 i.e.,, 2x + 1 is first factor.

ii) Now

4×2 5

___ + __ = 2x + 5 is second factor.

2x 1

Eg.2: 15×2 – 14xy – 8y2

i) Split –14 into –20, 6 so that 15 : – 20 = 3 : – 4 and 6 : – 8 = 3 : – 4. Both are same. i.e., ( 3x – 4y ) is one factor.

ii) Now

15×2 8y2

____ + ___ = 5x + 2y is second factor.

3x -4y

Thus 15×2 – 14xy – 8y2 = ( 3x – 4y ) ( 5x + 2y ).

It is evident that we have applied two sub-sutras ‘anurupyena’ i.e.‘proportionality’ and ‘adyamadyenantyamantyena’ i.e. ‘the first by the first and the last by the last’ to obtain the above results.

Factorise the following quadratics applying appropriate vedic maths sutras:

1). 3×2 + 14x + 15

2). 6×2 – 23x + 7

3). 8×2 – 22x + 5

4). 12×2 – 23xy + 10y2

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): 1122

Base = 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.

ANTYAYOR DAŚAKE′PI

The Sutra signifies numbers of which the last digits added up give 10. i.e. the Sutra works in multiplication of numbers for example: 25 and 25, 47 and 43, 62 and 68, 116 and 114. Note that in each case the sum of the last digit of first number to the last digit of second number is 10. Further the portion of digits or numbers left wards to the last digits remain the same. At that instant use Ekadhikena on left hand side digits. Multiplication of the last digits gives the right hand part of the answer.

Example 1 : 47 X 43

See the end digits sum 7 + 3 = 10 ; then by the sutras antyayor dasakepi and ekadhikena we have the answer.

47 x 43 = ( 4 + 1 ) x 4 / 7 x 3

= 20 / 21

= 2021.

Example 2: 62 x 68

2 + 8 = 10, L.H.S. portion remains the same i.e.,, 6.

Ekadhikena of 6 gives 7

62 x 68 = ( 6 x 7 ) / ( 2 x 8 )

= 42 / 16

= 4216.

Example 3: 127 x 123

As antyayor dasakepi works, we apply ekadhikena

127 x 123 = 12 x 13 / 7 x 3

= 156 / 21

= 15621.

Example 4: 65 x 65

We have already worked on this type. As the present sutra is applicable.

We have 65 x 65 = 6 x 7 / 5 x 5

= 4225.

Example 5: 3952

3952 = 395 x 395

= 39 x 40 / 5 x 5

= 1560 / 25

= 156025.

Use Vedic sutras to find the products

1. 125 x 125 2. 34 x 36 3. 98 x 92

4. 401 x 409 5. 693 x 697 6. 1404 x 1406

It is further interesting to note that the same rule works when the sum of the last 2, last 3, last 4 – – – digits added respectively equal to 100, 1000, 10000 — – – . The simple point to remember is to multiply each product by 10, 100, 1000, – – as the case may be . Your can observe that this is more convenient while working with the product of 3 digit numbers.

Eg. 1: 292 x 208

Here 92 + 08 = 100, L.H.S portion is same i.e. 2

292 x 208 = ( 2 x 3 ) / 92 x 8

60 / =736 ( for 100 raise the L.H.S. product by 0 )

= 60736.

Eg. 2: 848 X 852

Here 48 + 52 = 100, L.H.S portion is 8 and its ‘ekhadhikena’ is 9.

Now R.H.S product 48 X 52 can be obtained by ‘anurupyena’ mentally.

_

48 2

52 2

_______

_

2) 50 4 = 24 / ( 100 – 4 )

¯¯

25 = 96

= 2496

and write 848 x 852 = 8 x 9 / 48 x 52

720 = 2496

= 722496.

[Since L.H.S product is to be multiplied by 10 and 2 to be carried over as the base is 100].

Eg. 3: 693 x 607

693 x 607 = 6 x 7 / 93 x 7

= 420 / 651

= 420651.

Find the following products using ‘antyayordasakepi’

1. 318 x 312 2. 425 x 475 3. 796 x 744

4. 902 x 998 5. 397 x 393 6. 551 x 549

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 + 2×2 + 7x + 3×2 + 6x + 21 = x3 + 3×2 + 5x + 2×2 + 6x + 10

x3 + 5×2 + 13x + 21 = x3 + 5×2 + 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

2×2 + 3x + 10 2x + 3

___________ = _____

3×2 + 4x + 14 3x + 4

Since

2×2 + 3x x (2x + 3) 2x+3

_______ = ________ = ____

3×2 + 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.

3×2 + 5x + 8 3x + 5

__________ = ______

5×2 + 6x +12 5x + 6

2.

4×2 + 5x + 3 4x + 5

__________ = ______

3×2 + 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.

2×2 + 3x + 9 2x + 3

__________ = ______

4×2 +5x+17 4x + 5

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: 3×2 + 7xy + 2y2 + 11xz + 7yz + 6z2.

Step (i): Eliminate z and retain x, y; factorize

3×2 + 7xy + 2y2 = (3x + y) (x + 2y)

Step (ii): Eliminate y and retain x, z; factorize

3×2 + 11xz + 6z2 = (3x + 2z) (x + 3z)

Step (iii): Fill the gaps, the given expression

= (3x + y + 2z) (x + 2y + 3z)

Example 2: 12×2 + 11xy + 2y2 – 13xz – 7yz + 3z2.

Step (i): Eliminate z i.e., z = 0; factorize

12×2 + 11xy + 2y2 = (3x + 2y) (4x + y)

Step (ii): Eliminate y i.e., y = 0; factorize

12×2 – 13xz + 3z2 = (4x -3z) (3x – z)

Step (iii): Fill in the gaps; the given expression

= (4x + y – 3z) (3x + 2y – z)

Example 3: 3×2+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 = 3×2 + 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. 3×2 + 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. 3×2 + 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 2×2 – x – 3 and 2×2 + x – 6

Example 3: x3 – 7x – 6 and x3 + 8×2 + 17x + 10.

Now by Lopana – Sthapana and Sankalana – Vyavakalanabhyam

Example 4: x3 + 6×2 + 5x – 12 and x3 + 8×2 + 19x + 12.

(or)

Example 5: 2×3 + x2 – 9 and x4 + 2×2 + 9.

By Vedic sutras:

Add: (2×3 + x2 – 9) + (x4 + 2×2 + 9)

= x4 + 2×3 + 3×2.

÷ x2 gives x2 + 2x + 3 —— (i)

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

Thus (2×4 + x3 – 9x) – (2×4 + 4×2 + 18)

= x3 – 4×2 – 9x – 18 —— ( ii )

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

x3 – 4×2 – 9x – 18 – (x3 + 2×2 + 3x)

= – 6×2 – 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 – 3×2 – 4x + 12, x3 – 7×2 + 16x – 12

3. x3 + 6×2 + 11x + 6, x3 – x2 – 10x – 8

4. 6×4 – 11×3 + 16×2 – 22x + 8,

6×4 – 11×3 – 8×2 + 22x – 8.

VILOKANAM

The Sutra ‘Vilokanam’ means ‘Observation’. Generally we come across problems which can be solved by mere observation. But we follow the same conventional procedure and obtain the solution. But the hint behind the Sutra enables us to observe the problem completely and find the pattern and finally solve the problem by just observation.

Let us take the equation x + ( 1/x ) = 5/2 Without noticing the logic in the problem, the conventional process tends us to solve the problem in the following way.

1 5

x + __ = __

x 2

x2 + 1 5

_____ = __

x 2

2×2 + 2 = 5x

2×2 – 5x + 2 = 0

2×2 – 4x – x + 2 = 0

2x (x – 2) – (x – 2) = 0

(x – 2) (2x – 1) = 0

x – 2 = 0 gives x = 2

2x – 1 = 0 gives x = ½

But by Vilokanam i.e.,, observation

1 5

x + __ = __ can be viewed as

x 2

1 1

x + __ = 2 + __ giving x = 2 or ½.

x 2

Consider some examples.

Example 1 :

x x + 2 34

____ + _____ = ___

x + 2 x 15

In the conventional process, we have to take L.C.M, cross-multiplication. simplification and factorization. But Vilokanam gives

34 9 + 25 3 5

__ = _____ = __ + __

15 5 x 3 5 3

x x + 2 3 5

____ + _____ = __ + __

x + 2 x 5 3

gives

x 3 5

_____ = __ or __

x + 2 5 3

5x = 3x + 6 or 3x = 5x + 10

2x = 6 or -2x = 10

x = 3 or x = -5

Example 2 :

x + 5 x + 6 113

____ + _____ = ___

x + 6 x + 5 56

Now,

113 49 + 64 7 8

___ = _______ = ___ + ___

56 7 x 8 8 7

x + 5 7 x+5 8

____ = __ or ____ = __

x + 6 8 x+6 7

8x + 40 = 7x + 42 7x + 35 = 8x + 48

or

x = 42 – 40 = 2 -x = 48 – 35 = 13

x = 2 or x = -13.

Example 3:

5x + 9 5x – 9 82

_____ + _____ = 2 ___

5x – 9 5x + 9 319

At first sight it seems to a difficult problem.

But careful observation gives

82 720 841 – 121 29 11

2 ___ = ___ = ________ = ___ – __

319 319 11 x 29 11 29

(Note: 292 = 841, 112 = 121)

5x + 9 29 -11

_____ = __ or ___

5x – 9 11 29

(Note: 29 = 20 + 9 = 5 x 4 + 9 ; 11 = 20 – 9 = 5 x 4 – 9 )

i.e.,

x = 4 or

5x + 9 -11

_____ = ___

5x – 9 29

145x + 261 = -55x + 99

145x + 55x = 99 – 261

200x = -162

-162 -81

x = ____ = ____

200 100

Simultaneous Quadratic Equations:

Example 1: x + y = 9 and xy = 14.

We follow in the conventional way that

(x – y)2 = (x + y)2 – 4xy = 92 – 4 (14) = 81 – 56 = 25

x – y = √ 25 = ± 5

x + y = 9 gives 7 + y = 9

y = 9 – 7 = 2.

Thus the solution is x = 7, y = 2 or x = 2, y = 7.

But by Vilokanam, xy = 14 gives x = 2, y = 7 or x = 7, y = 2 and these two sets satisfy x + y = 9 since 2 + 7 = 9 or 7 + 2 = 9. Hence the solution.

Example 2: 5x – y = 7 and xy = 6.

xy = 6 gives x = 6, y = 1; x = 1, y = 6;

x = 2, y = 3; x = 3, y = 2 and of course negatives of all these.

Observe that x = 6, y = 1; x = 1, y = 6: are not solutions because they do not satisfy the equation 5x – y = 7.

But for x = 2, y = 3; 5x – y = 5 (2) – 3 = 10 – 3 = 7 we have 5(3)–2≠7.

Hence x = 2, y = 3 is a solution.

For x = 3, y = 2 we get 5 (3) – 2 = 15 – 2 ≠ 7.

Hence it is not a solution.

Negative values of the above are also not the solutions. Thus one set of the solutions i.e., x = 2, y = 3 can be found. Of course the other will be obtained from solving 5x – y = 7 and 5x + y = -13.

i.e., x = -3 / 5, y = -10.

Partial Fractions:

Example 1: Resolve

2x + 7

___________ into partial fractions.

(x + 3) (x + 4)

2x + 7 A B

We write ____________ = ______ + ______

(x + 3)(x + 4) (x + 3) (x + 4)

A (x + 4) + B (x + 3)

= __________________

(x + 3) (x + 4)

2x + 7 ≡ A (x + 4) + B (x + 3).

We proceed by comparing coefficients on either side

coefficient of x : A + B = 2 ……….(i) X 3

Independent of x : 4A + 3B = 7 ………….(ii)

Solving (ii) – (i) x 3 4A + 3B = 7

3A + 3B = 6

___________

A = 1

A = 1 in (i) gives, 1 + B = 2 i.e., B = 1

Or we proceed as

2x + 7 ≡ A (x + 4) + B (x + 3).

Put x = -3, 2 (-3) + 7 ≡ A (-3 + 4) + B (-3 + 3)

1 = A (1) … A = 1.

x = -4, 2 (- 4) + 7 = A (-4 + 4) + B (-4 + 3)

-1 = B(-1) … B = 1.

2x + 7 1 1

Thus ____________ = _____ + _____

(x + 3) (x + 4) (x + 3) (x + 4)

2x + 7

But by Vilokanam ____________ can be resolved as

(x + 3) (x + 4)

(x + 3) + (x + 4) =2x + 7, directly we write the answer.

Example 2:

3x + 13

____________

(x + 1) (x + 2)

from (x + 1),(x + 2) we can observe that

10 (x + 2) – 7(x + 1) = 10x + 20 – 7x – 7 = 3x + 13

3x + 13 10 7

Thus ____________ = _____ – _____

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

Example 3:

9

________

x2 + x – 2

As x2 + x – 2 = (x – 1) (x + 2) and

9 = 3 (x + 2) – 3 (x – 1)

(3x + 6 – 3x + 3 = 9)

9 3 3

We get by Vilokanam, ____________ = ____ – ____

x2 + x – 2 x – 1 x + 2

I. Solve the following by mere observation i.e. vilokanam

1. 2.

1 25 1 5

x + __ = __ x – __ = __

x 12 x 6

3.

x x + 1 1

_____ + _____ = 9 __

x + 1 x 9

4.

x + 7 x + 9 32

____ – ____ = ___

x + 9 x + 7 63

II. Solve the following simultaneous equations by vilokanam.

1. x – y = 1, xy = 6 2. x + y = 7, xy = 10

3. 2x + 3y = 19, xy = 15

4. x + y = 4, x2 + xy + 4x = 24.

III. Resolve the following into partial fractions.

1.

2x – 5

____________

(x – 2) (x – 3)

2.

9

____________

(x + 1) (x – 2)

3.

x – 13

__________

x2 – 2x – 15

4.

3x + 4

__________

3×2 + 3x + 2

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) = 2×2 – 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 + 9×2 + 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 + 10×2 + 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) = 2×2 – x – 6

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

3. 12×2 + 13x – 4 = ( 3x – 4 ) ( 4x + 1 )

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

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

So far we have considered a majority of the upa-sutras as mentioned in the Vedic mathematics book. Only a few Upa-Sutras are not dealt under a separate heading . They are

2) S’ISYATE S’ESASAMJ ÑAH

4) KEVALAIH SAPTAKAMGUNYAT

5) VESTANAM

6) YAVADŨNAM TAVADŨNAM and

10) SAMUCCAYAGUNITAH already find place in respective places.

Further in some other books developed on Vedic Mathematics DVANDAYOGA, SUDHA, DHVAJANKAM are also given as Sub-Sutras. They are mentioned in the Vedic Mathematics text also. But the list in the text (by the Editor) does not contain them. We shall also discuss them at appropriate places, with these three included, the total number of upa-Sutras comes to sixteen.

TERMS AND OPERATIONS

a) Ekadhika means ‘one more’

e.g: Ekadhika of 0 is 1

Ekadhika of 1 is 2

—————–

Ekadhika of 8 is 9

——————-

Ekadhika of 23 is 24

———————

Ekadhika of 364 is 365——

b) Ekanyuna means ‘one less’

e.g: Ekanyuna of 1 2 3 ….. 8 ….. 14 …..69 ……

is 0 1 2 ….. 7 ……13 …. 68 ……

c) Purak means ‘ complement’

e.g: purak of 1 2 3 ….. 8., 9 from 10

is 9 8 7 ….. 2 1

d) Rekhank means ‘a digit with a bar on its top’. In other words it is a negative number.

_

e.g: A bar on 7 is 7. It is called rekhank 7 or bar 7. We treat purak as a Rekhank.

_ _

e.g: 7 is 3 and 3 is 7

At some instances we write negative numbers also with a bar on the top of the numbers as

_

-4 can be shown as 4.

__

-21 can be shown as 21.

e) Addition and subtraction using Rekhank.

Adding a bar-digit i.e. Rekhank to a digit means the digit is subtracted.

_ _ _

e.g: 3 + 1 = 2, 5 + 2 = 3, 4 + 4 = 0

Subtracting a bar – digit i.e. Rekhank to a digit means the digit is added.

_ _ _

e.g: 4 – 1 = 5, 6 – 2 = 8, 3 – 3 = 6

1. Some more examples

e.g: 3 + 4 = 7

_ _ _

(-2) + (-5) = 2 + 5 = 7 or -7

f) Multiplication and Division using rekhank.

1. Product of two positive digits or two negative digits ( Rekhanks )

_ _

e.g: 2 X 4 = 8; 4 X 3 = 12 i.e. always positive

2. Product of one positive digit and one Rekhank

_ _ _ __

e.g: 3 x 2 = 6 or -6; 5 X 3 = 15 or -15 i.e. always Rekhank or negative.

3. Division of one positive by another or division of one Rekhank by another Rekhank.

_ _

e.g: 8 ÷ 2 = 4, 6 ÷ 3 = 2 i.e. always positive

4. Division of a positive by a Rekhank or vice versa.

__ _ _ _

e.g: 10 ÷ 5 = 2, 6 ÷ 2 = 3 i.e. always negative or Rekhank.

g) Beejank: The Sum of the digits of a number is called Beejank. If the addition is a two digit number, Then these two digits are also to be added up to get a single digit.

e.g: Beejank of 27 is 2 + 7 = 9.

Beejank of 348 is 3 + 4 + 8 = 15

Further 1 + 5 = 6. i.e. 6 is Beejank.

Beejank of 1567 1 + 5 + 6 + 7 19 1 + 9 1

i.e. Beejank of 1567 is 1.

ii) Easy way of finding Beejank:

Beejank is unaffected if 9 is added to or subtracted from the number. This nature of 9 helps in finding Beejank very quickly, by cancelling 9 or the digits adding to 9 from the number.

eg 1: Find the Beejank of 632174.

As above we have to follow

632174 6 + 3 + 2 + 1 + 7 + 4 23 2 + 3 5

But a quick look gives 6 & 3 ; 2 & 7 are to be ignored because 6+3=9,2+7=9. Hence remaining 1 + 4 5 is the beejank of 632174.

eg 2:

Beejank of 1256847 1+2+5+6+8+4+7 33 3+3 6.

But we can cancel 1& 8, 2& 7, 5 & 4 because in each case the sum is 9. Hence remaining 6 is the Beejank.

h) Check by Beejank method:

The Vedic sutra – Gunita Samuccayah gives ‘the whole product is same’. We apply this sutra in this context as follows. It means that the operations carried out with the numbers have same effect when the same operations are carried out with their Beejanks.

Observe the following examples.

i) 42 + 39

Beejanks of 42 and 39 are respectively 4 + 2 = 6 and 3 + 9 = 12 and 1+2=3

Now 6 + 3 = 9 is the Beejank of the sum of the two numbers

Further 42 + 39 = 81. Its Beejank is 8+ 1 = 9.

we have checked the correctness.

ii) 64 + 125.

64 6 + 4 10 1 + 0 1

125 1 + 2 + 5 8

Sum of these Beejanks 8 + 1 = 9

Note that

64 + 125 = 189 1 + 8 + 9 18 1 + 8 9

we have checked the correctness.

iii) 134 – 49

134 1 + 3 + 4 8

49 4+9 13 1 + 3 4

Difference of Beejanks 8 -4 4, note that 134 – 49 = 85

Beejanks of 85 is 8 + 5 85 8 + 5 13 1 + 3 4 verified.

iv) 376 – 284

376 7 ( 6 + 3 9)

284 2 + 8 + 4 14 1 + 4 5

Difference of Beejanks = 7 – 5 = 2

376 – 284 = 92

Beejank of 92 9 + 2 11 1 + 1 2

Hence verified.

v) 24 X 16 = 384

Multiplication of Beejanks of

24 and 16 is 6 X 7 = 42 4 + 2 6

Beejank of 384 3 + 8 + 4 15 1 + 5 6

Hence verified.

vi) 237 X 18 = 4266

Beejank of 237 2 + 3 + 7 12 1 + 2 3

Beejank of 18 1 + 8 9

Product of the Beejanks = 3 X 9 27 2 + 7 9

Beejank of 4266 4 + 2 + 6 + 6 18 1 + 8 9

Hence verified.

vii) 242 = 576

Beejank of 24 2 + 4 6

square of it 62 36 9

Beejank of result = 576 5 + 7 + 6 18 1 + 8 9

Hence verified.

viii) 3562 = 126736

Beejank of 356 3 + 5 + 6 5

Square of it = 52 = 25 2 + 5 7

Beejank of result 126736 1 + 2 + 6 + 7 + 3 + 6 1 + 6 7

( 7 + 2 = 9; 6 + 3 = 9) hence verified.

ix) Beejank in Division:

Let P, D, Q and R be respectively the dividend, the divisor, the quotient and the remainder.

Further the relationship between them is P = ( Q X D ) + R

eg 1: 187 ÷ 5

we know that 187 = ( 37 X 5 ) + 2 now the Beejank check.

We know that 187 = (37 X 5) +2 now the Beejank check.

187 1 + 8 + 7 7( 1 + 8 = 9)

(37 X 5) + 2 Beejank [(3 + 7) X 5] + 2

5 + 2 7

Hence verified.

eg 2: 7986 ÷ 143

7896 = (143 X 55) + 121

Beejank of 7986 7 + 9 + 8 + 6 21

( 9 is omitted) 2 + 1 3

Beejank of 143 X 55 (1 + 4 + 3) (5 + 5)

8 X 10 80 (8 + 0) 8

Beejank of (143 X 55) + 121 8 + (1 + 2 + 1)

8 + 4 12 1 + 2 3

hence verified.

Check the following results by Beejank method

1. 67 + 34 + 82 = 183 2. 4381 – 3216 = 1165

3. 632 = 3969 4. (1234)2 = 1522756

5. (86X17) + 34 = 1496 6. 2556 ÷ 127 gives Q =20, R = 16

i) Vinculum : The numbers which by presentation contains both positive and negative digits are called vinculum numbers.

ii) Conversion of general numbers into vinculum numbers.

We obtain them by converting the digits which are 5 and above 5 or less than 5 without changing the value of that number.

Consider a number say 8. (Note it is greater than 5). Use it complement (purak – rekhank) from 10. It is 2 in this case and add 1 to the left (i.e. tens place) of 8.

_

Thus 8 = 08 = 12.

The number 1 contains both positive and negative digits

_ _

i.e. 1 and 2 . Here 2 is in unit place hence it is -2 and value of 1 at tens place is 10.

_

Thus 12 = 10 – 2 = 8

Conveniently we can think and write in the following way

General Number Conversion Vinculum number

_

6 10 – 4 14

_

97 100 – 3 103

__

289 300 – 11 311 etc.,,

The sutras ‘Nikhilam Navatascharamam Dasatah’ and ‘Ekadhikena purvena’ are useful for conversion.

eg 1: 289, Edadhika of 2 is 3

_

Nikhilam from 9 : 8 – 9 = -1 or 1

_

Charmam from 10 :9 -10 = -1 or 1

__

i.e. 289 in vinculum form 311

eg 2: 47768

‘Ekadhika’ of 4 is 5

___

‘Nikhilam’ from 9 (of 776) 223

_

‘Charmam from 10 (of 8) 2

____

Vinculum of 47168 is 5 2232

eg 3: 11276.

Here digits 11 are smaller. We need not convert. Now apply for 276 the two sutras Ekadhika of 2 is 3

__

‘Nikhilam Navata’ for 76 is 24

__

11276 = 11324

__

i.e. 11324 = 11300 – 24 = 11276.

The conversion can also be done by the sutra sankalana vyavakalanabhyam as follows.

eg 4: 315.

sankalanam (addition) = 315+315 = 630.

_

Vyvakalanam (subtraction) = 630 – 315 = 325

Working steps : _

0 – 5 = 5

3 – 1 = 2

6 – 3 = 3

Let’s apply this sutra in the already taken example 47768.

Samkalanam = 47768 + 47768 = 95536

Vyavakalanam = 95536 – 47768.

Consider the convertion by sankalanavyavakalanabhyam and check it by Ekadhika and Nikhilam.

eg 5: 12637

1. Sankalana ……. gives, 12637 + 12637 = 25274

_ _

25274 – 12637 = (2 – 1) / (5 – 2) / (2 – 6) / (7 – 3) / (4 – 7) = 13443

2. Ekadhika and Nikhilam gives the following.

As in the number 1 2 6 3 7, the smaller and bigger digits (i.e. less than 5 and; 5, greater than 5) are mixed up, we split up in to groups and conversion is made up as given below.

Split 1 2 6 and 3 7

_ _

Now the sutra gives 1 2 6 as 134 and 37 as 43

_ _

Thus 12637 = 13443

_

Now for the number 315 we have already obtained vinculum as 325 by “sankalana … ” Now by ‘Ekadhika and Nikhilam …’ we also get the same answer.

315 Since digits of 31 are less than 5,

We apply the sutras on 15 only as

Ekadhika of 1 is 2 and Charman of 5 is 5 .

Consider another number which comes under the split process.

eg 6: 24173

As both bigger and smaller numbers are mixed up we split the number 24173 as 24 and 173 and write their vinculums by Ekadhika and Nikhilam sutras as

_ __

24 = 36 and 173 = 227

_ __

Thus 24173 = 36227

Convert the following numbers into viniculum number by

i. Ekadhika and Nikhilam sutras ii. Sankalana vyavakalana sutra. Observe whether in any case they give the same answer or not.

1. 64 2. 289 3. 791

4. 2879 5. 19182 6. 823672

7. 123456799 8. 65384738

ii) Conversion of vinculum number into general numbers.

The process of conversion is exactly reverse to the already done. Rekhanks are converted by Nikhilam where as other digits by ‘Ekanyunena’ sutra. thus:

_

i) 12 = (1 – 1) / (10 – 2) Ekanyunena 1 – 1

_

= 08 = 8 Nikhilam. 2 = 10 – 2

__

ii) 326 = (3 – 1) / (9 – 2) / (10 – 6)

= 274

_ _

iii) 3344 = (3 – 1) / (10 – 3) / (4 – 1) / (10 – 4)

= 2736 (note the split)

__ __

iv) 20340121 = 2/(0–1)/(9–3)/(10–4)/(0–1)/(9–1)/(10–2)/1

_ _

= 21661881

_ _

= 21 / 6 / 61 / 881. once again split

= (2 – 1) / (10 –1) / 6 / (6 –1) / (10 –1) / 881

= 19659881

___

v) 303212 = 3 / 0321 / 2

= 3 / (0-1) / (9-3) / (9-2) / (10-1) / 2

_

3 / 1 / 6792

(3 –1) / (10 –1) / 6792

= 296792.

iii) Single to many conversions.

It is interesting to observe that the conversions can be shown in many ways.

eg 1: 86 can be expressed in following many ways

__

86 = 90 – 4 =94

__

= 100 – 14 = 114

___

= 1000 – 914 = 1914

_ __ ___ ____

Thus 86 = 94 = 114 = 1914 = 19914 = ………….

eg 2 :

_ _

07 = -10 +3 = 13

__ _

36 = -100 + 64 = 164

___ _

978 = -1000 + 22 = 1022. etc.,

* Convert by Vedic process the following numbers into vinculum numbers.

1) 274 2) 4898 3) 60725 4) 876129.

* Convert the following vinculum numbers into general form of numbers ( normalized form)

1) 283 2) 3619 3) 27216

4) 364718 5) 60391874

(iv) Vedic check for conversion:

The vedic sutra “Gunita Samuctayah” can be applied for verification of the conversion by Beejank method.

Consider a number and find its Beejank. Find the vinculum number by the conversion and find its Beejank. If both are same the conversion is correct.

eg.

_

196 = 216 . Now Beejank of 196 1 + 6 7

Beejank of 216 2 + ( -1 ) + 6 7. Thus verified.

But there are instances at which, if beejank of vinculum number is rekhank i.e. negative. Then it is to be converted to +ve number by adding 9 to Rekhank ( already we have practised) and hence 9 is taken as zero, or vice versa in finding Beejank.

eg:

__

213 = 200 – 13 = 187

_

Now Beejank of 213 = 2 + ( -1 ) + (-3 ) = -2

Beejank of 187 = 1 + 8 + 7 = 16 1 + 6 = 7

The variation in answers can be easily understood as

_ _

2 = 2 + 9 – 2 + 9 = 7 Hence verified.

Use Vedic check method of the verification of the following result.

_ _ _

1. 24 = 36 2. 2736 = 3344.

__ _ _

3. 326 = 274 4. 23213 = 17187

Addition and subtraction using vinculum numbers.

eg 1: Add 7 and 6 i.e., 7+6.

i) Change the numbers as vinculum numbers as per rules already discussed.

_ _

{ 7 = 13 and 6 = 14 }

ii) Carry out the addition column by column in the normal process, moving from top to bottom or vice versa.

iii) add the digits of the next higher level i.e.,, 1 + 1 = 2

_

13

_

14

____

_

27

iv) the obtained answer is to be normalized as per rules already explained. rules already explained.

_

i.e., 27 = (2 – 1) (10- 7) = 13 Thus we get 7 + 6 = 13.

eg 2 : Add 973 and 866.

_ _

973 = 1 0 3 3 1 0 3 3

_ _ _ _ _ _

866 = 1 1 3 4 1 1 3 4

______

_ _ _

2 1 6 1

___

But 2161 = 2000 – 161 = 1839.

Thus 973+866 by vinculum method gives 1839 which is correct.

Observe that in this representation the need to carry over from the previous digit to the next higher level is almost not required.

eg 3 : Subtract 1828 from 4247.

i.e.,, 4247

-1828

______

____

Step (i) : write –1828 in Bar form i.e.,, 1828

____

(ii) : Now we can add 4247 and 1828 i.e.,,

4247

____

+1828

_______

_ _

3621

_ _ _ _ _ _

since 7 + 8 = 1, 4 + 2 = 2, 2 + 8 = 6, 4 + 1 = 3

_ _

(iii) Changing the answer 3621 into normal form using Nikhilam, we get

_ _ _ _

3621 = 36 / 21 split

= (3 –1) / (10 – 6) / (2 – 1) / (10 – 1) = 2419

4247 – 1828 = 2419

Find the following results using Vedic methods and check them

1) 284 + 257 2) 5224 + 6127

3) 582 – 464 4) 3804 – 2612

ADDITION AND SUBTRACTION

ADDITION:

In the convention process we perform the process as follows.

234 + 403 + 564 + 721

write as 234

403

564

721

Step (i): 4 + 3 + 4 + 1 = 12 2 retained and 1 is carried over to left.

Step (ii): 3 + 0 + 6 + 2 = 11 the carried ‘1’ is added

i.e., Now 2 retained as digit in the second place (from right to left) of the answer and 1 is carried over to left.

step (iii): 2 + 4 + 5 + 7 = 18 carried over ‘1’ is added

i.e., 18 + 1 = 19. As the addition process ends, the same 19 is retained in the left had most part of the answer.

thus 234

403

564

+721

_____

1922 is the answer

we follow sudhikaran process Recall ‘sudha’ i.e., dot (.) is taken as an upa-sutra (No: 15)

consider the same example

i) Carry out the addition column by column in the usual fashion, moving from bottom to top.

(a) 1 + 4 = 5, 5 + 3 = 8, 8 + 4 = 12 The final result is more than 9. The tenth place ‘1’ is dropped once number in the unit place i.e., 2 retained. We say at this stage sudha and a dot is above the top 4. Thus column (1) of addition (right to left)

.

4

3

4

1

__

2

b) Before coming to column (2) addition, the number of dots are to be counted, This shall be added to the bottom number of column (2) and we proceed as above.

Thus second column becomes

.

3 dot=1, 1 + 2 = 3

0 3 + 6 = 9

6 9 + 0 = 9

2 9 + 3 = 12

__

2

2 retained and ‘.’ is placed on top number 3

c) proceed as above for column (3)

2 i) dot = 1 ii) 1 + 7 = 8

4 iii) 8 + 5 = 13 iv) Sudha is said.

.

5 A dot is placed on 5 and proceed

7 with retained unit place 3.

__

9 v) 3+4=7,7+2=9 Retain 9 in 3rd digit i.e.,in 100th place.

d) Now the number of dots is counted. Here it is 1 only and the number is carried out left side ie. 1000th place

..

Thus 234

403

.

564

+721

_____

1922 is the answer.

Though it appears to follow the conventional procedure, a careful observation and practice gives its special use.

eg (1):

.

437

. .

624

.

586

+162

______

1809

Steps 1:

i) 2 + 6 = 8, 8 + 4 = 12 so a dot on 4 and 2 + 7 = 9 the answer retained under column (i)

ii) One dot from column (i) treated as 1, is carried over to column (ii),

thus 1 + 6 = 7, 7 + 8 = 15 A’ dot’; is placed on 8 for the 1 in 15 and the 5 in 15 is added to 2 above.

5 + 2 = 7, 7 + 3 = 10 i.e. 0 is written under column (ii) and a dot for the carried over 1 of 10 is placed on the top of 3.

(iii) The number of dots counted in column (iii) are 2.

Hence the number 2 is carried over to column (ii) Now in column (iii)

2 + 1 = 3, 3 + 5 = 8, 8 + 6 = 14 A dot for 1 on the number 6 and 4 is retained to be added 4 above to give 8. Thus 8 is placed under column (iii).

iv) Finally the number of dots in column (iii) are counted. It is ‘1’ only. So it carried over to 1000th place. As there is no fourth column 1 is the answer for 4th column. Thus the answer is 1809.

Example 3:

Check the result verify these steps with the procedure mentioned above.

The process of addition can also be done in the down-ward direction i.e., addition of numbers column wise from top to bottom

Example 1:

Step 1: 6 + 4 = 10, 1 dot ; 0 + 8 = 8; 8 + 4 = 12;

1 dot and 2 answer under first column – total 2 dots.

Step 2: 2+2 ( 2 dots) = 4; 4+9 = 13: 1 dot and 3+0 = 3; 3+8 = 11;

1 dot and 1 answer under second column – total 2 dots.

Step 3: 3+2 ( 2 dots ) = 5; 5+6 = 11:1 dot and 1+7 = 8; 8+7 = 15;

1 dot and 5 under third column as answer – total 2 dots.

Step 4: 4 + 2 ( 2 dots ) = 6; 6 + 5 =11:

1 dot and 1+3 = 4; 4+2 = 6. – total 1 dot in the fourth 6 column as answer.

Step 5: 1 dot in the fourth column carried over to 5th column (No digits in it) as 1

Thus answer is from Step5 to Step1; 16512

Example 2:

Steps

(i): 8 + 9 = 17; 7 + 4 = 11; 1 + 1 = (2) (2dots)

(ii): 7 + 2 = 9; 9 + 1 = 10; 0 + 8 = 8, 8 + 9 = 17, (7) (2dots)

(iii): 2 + 2 = 4; 4 + 6 = 10; 0 + 0 = 0; 0 + 7 = (7) (1 dot)

(iv): 3 + 1 = 4; 4 + 4 = 8; 8 + 3 = 11; 1 + 1 = (2) (1 dot)

(v): 1

Thus answer is 12772.

Add the following numbers use ‘Sudhikaran’ whereever applicable.

1. 2. 3.

486 5432 968763

395 3691 476509

721 4808 +584376

+609 +6787 ¯¯¯¯¯¯¯¯

¯¯¯¯¯ ¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯

¯¯¯¯¯ ¯¯¯¯¯¯

Check up whether ‘Sudhkaran’ is done correctly. If not write the correct process. In either case find the sums.

SUBTRACTION:

The ‘Sudha’ Sutra is applicable where the larger digit is to be subtracted from the smaller digit. Let us go to the process through the examples.

Procedure:

i) If the digit to be subtracted is larger, a dot ( sudha ) is given to its left.

ii) The purak of this lower digit is added to the upper digit or purak-rekhank of this lower digit is subtracted.

Example (i): 34 – 18

34

.

-18

_____

.

Steps: (i): Since 8>4, a dot is put on its left i.e. 1

(ii) Purak of 8 i.e. 2 is added to the upper digit i.e. 4

_

2 + 4 = 6. or Purak-rekhank of 8 i.e. 2 is

_

Subtracted from i.e. 4 – 2 =6.

Now at the tens place a dot (means1) makes the ‘1’ in the number into 1+1=2.This has to be subtracted from above digit. i.e. 3 – 2 = 1. Thus

34

.

-18

_____

16

Example 2:

63

.

-37

_____

.

Steps: (i) 7>3. Hence a dot on left of 7 i.e., 3

(ii) Purak of 7 i.e. 3 is added to upper digit 3 i.e. 3+3 = 6.

This is unit place of the answer.

Thus answer is 26.

Example (3) :

3274

..

-1892

_______

Steps:

(i) 2 7 sudha required. Hence a dot on left of 9 i.e. 8

(iii) purak of 9 i.e. 1, added to upper 7 gives 1 + 7 = 8 second digit

.

(iv) Now means 8 + 1 = 9.

.

(v) As 9 > 2, once again the same process: dot on left of i.e., 1

(vi) purak of 9 i.e. 1, added to upper 2 gives 1 + 2 = 3, the third digit.

.

(vii) Now 1 means 1+1 = 2

(viii) As 2

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