Mix 9

vedic solutions

EKĀDHIKENA PŪRVEŅA

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

i) Squares of numbers ending in 5 :

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

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

Thus 252 = 2 X 3 / 25 = 625.

In the same way,

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

652= 6 X 7 / 25 = 4225;

1052= 10 X 11/25 = 11025;

1352= 13 X 14/25 = 18225;

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

Algebraic proof:

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

This identity for x = 10 and b = 5 becomes

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

= a2 . 102 + a. 102 + 52

= (a 2+ a ) . 102 + 52

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a) Division Method : Value of 1 / 19.

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

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

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

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

Step. 1 : Divide numerator 1 by 20.

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

Step. 2 : Divide 10 by 2

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

Step. 3 : Divide 5 by 2

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

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

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

Step. 5 : Divide 6 by 2

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

Step. 6 : Divide 3 by 2

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

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

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

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

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

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

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

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

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

Step. 11 : Divide 9 by 2

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

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

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

Step. 13 : Divide 7 by 2

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

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

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

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

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

Step. 16 : Divide 8 by 2

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

Step. 17 : Divide 4 by 2

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

Step. 18 : Divide 2 by 2

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

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

0 __________________ . .

1 / 19 = 0.052631578947368421 or 0.052631578947368421

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

b) Multiplication Method: Value of 1 / 19

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

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

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

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

Step. 1 : 1

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

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

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

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

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

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

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

= 7, put to left)

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

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

Step. 10 : 18947368421

Step. 11 : 178947368421

Step. 12 : 1578947368421

Step. 13 : 11578947368421

Step. 14 : 31578947368421

Step. 15 : 631578947368421

Step. 16 : 12631578947368421

Step. 17 : 52631578947368421

Step. 18 : 1052631578947368421

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

Thus 1 / 19 = 0.052631578947368421

It is interesting to note that we have

i) not at all used division process

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

Observations :

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

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

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

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

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

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

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

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

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

Now the complements of the numbers

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

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

i.e.,, 0.052631578947368421

Now taking the multiplication process we have

Step. 8 : 147368421

Step. 9 : 947368421

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

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

0.052631578947368421.

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

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

Algebraic proof :

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

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

= ________________________

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

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

( a + 1 ) x

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

( a + 1 ) x

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

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

This series explains the process of ekadhik.

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

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

( since a=1)

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

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

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

= 0.052631 - - - - - - -

Example1 :

1. Find 1 / 49 by ekadhikena process.

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

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

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

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

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

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

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

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

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

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

= .0204081613322615306111222244448 - -- - - - -

On completing 21 digits, we get 48

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

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

Thus 1 / 49 = 0.020408163265306122448

979591836734693877551

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

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

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

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

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

= ---- 483947294594118333617233446943383727551

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

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

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

= 0.020408163265306122448

979591836734693877551

Example 2: Find 1 / 39 by Ekadhika process.

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

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

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

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

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

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

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

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

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

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

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

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NIKHILAMitten using Rekhank (a bar on the number). Now observe the following table.

Number

Base

Number – Base

Deviation

14 - 10

8 - 10

-2 or 2

100

97 - 100

-03 or 03

112

100

112 - 100

___

993

1000

993 - 1000

-007 or 007

1011

1000

1011 - 1000

011

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

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

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

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

Eg : Multiply 7 by 8.

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

we write the numbers one below the other. 8

-----

Take the deviations of both the numbers from

the base and represent _

7 3

Rekhank or the minus sign before the deviations 8 2

------

or 7 -3

8 -2

-------

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

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

(or)

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

i.e., 7 3

8 2

_____________

/ (3x2) = 6

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

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

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

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

i.e., 8 - 3 = 5

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

i.e., (7 + 8) – 10 = 5

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

i.e., 10 – ( 3 + 2) = 5

Hence 5 is left hand side of the answer.

Thus 7 3

8 2

¯¯¯¯¯¯¯¯¯¯¯¯

5 /

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

7 3 7

8 2 i.e., X 8

¯¯¯¯¯¯¯ ¯¯¯¯¯¯

5 / 6 56

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

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

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

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

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

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

Ex. 2: 98 X 97 Base is 100.

Ex. 3: 75X95. Base is 100.

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

Ex. 5: 994X988. Base is 1000.

Ex. 6: 750X995.

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

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

Ex. 7: 13X12. Base is 10

Ex. 8: 18X14. Base is 10.

Ex. 9: 104X102. Base is 100.

104 04

102 02

¯¯¯¯¯¯¯¯¯¯¯¯

106 / 4x2 = 10608 ( rule - f )

¯¯¯¯¯¯¯¯¯¯¯¯

Ex. 10: 1275X1004. Base is 1000.

1275 275

1004 004

¯¯¯¯¯¯¯¯¯¯¯¯

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

____________ = 1280100

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

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

Ex.11: 13X7. Base is 10

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

Eg :

9 = 10 – 1 = 11

98 = 100 – 2 = 102

196 = 200 – 4 = 204

32 = 30 – 2 = 28

145 = 140 – 5 = 135

322 = 300 – 22 = 278. etc

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

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

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

Algebraic Proof:

Case ( i ):

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

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

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

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

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

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

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

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

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

Case ( ii ) :

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

Case ( iii ) :

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

Find the following products by Nikhilam formula.

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

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

7) 1234 X 1002 8) 118 X 105

Nikhilam in Division

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

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

since 9 ) 13 ( 1

____

ii) 34 ÷ 9, Q is 3, R is 7.

iii) 60 ÷ 9, Q is 6, R is 6.

iv) 80 ÷ 9, Q is 8, R is 8.

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

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

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

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

Eg. 1 / 3 3 / 4 8 / 0

______ , ______ , ______

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

Eg.

1 / 3 3 / 4 8 / 0

1 3 8

______ , ______ , ______

1 / 4 3 / 7 8 / 8

Now the problem is over. i.e.,

13 ÷ 9 gives Q = 1, R = 4

34 ÷ 9 gives Q = 3, R = 7

80 ÷ 9 gives Q = 8, R = 8

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

a) 21 ÷ 9 as

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

¯¯¯¯¯¯

2 / 3

b) 43 ÷ 9 as

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

¯¯¯¯¯¯

4 / 7

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

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

9 ) 104 ( 11 9 ) 10 / 4

99 1 / 1

¯¯¯¯¯¯ as ¯¯¯¯¯¯¯

5 11 / 5

ii)

9 ) 212 ( 23 9 ) 21 / 2

207 2 / 3

¯¯¯¯¯ as ¯¯¯¯¯¯¯

5 23 / 5

iii)

9 ) 401 ( 44 9 ) 40 / 1

396 4 / 4

¯¯¯¯¯ as ¯¯¯¯¯¯¯¯

5 44 / 5

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

Consider 511 ÷ 9

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

Thus

9 ) 51 / 1

5 / 6

¯¯¯¯¯¯¯

56 / 7

Q is 56, R is 7.

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

Eg : 1204 ÷ 9

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

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

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

In symbolic form 9 ) 120 / 4

13 / 3

¯¯¯¯¯¯¯¯

133 / 7

Another example.

9 ) 13210 / 1 132101 ÷ 9

gives

1467 / 7 Q = 14677, R = 8

¯¯¯¯¯¯¯¯¯¯

14677 / 8

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

Eg.

9 ) 3 / 6 9) 24 / 6

3 2 / 6

¯¯¯¯¯¯ or ¯¯¯¯¯¯¯¯

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

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

9 ) 3 / 6 9 ) 24 / 6

/ 3 2 / 6

¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯

3 / 9 26 / 12

¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯

4 / 0 27 / 3

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

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

9 ) 24 / 6

2 / 6

¯¯¯¯¯¯¯¯

26 /1 / 2

/ 1

¯¯¯¯¯¯¯¯

1 / 3

¯¯¯¯¯¯¯¯

27 / 3

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

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

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

Representing in the previous form of procedure, we have

89 ) 1 / 13 89 ) 100 / 15

/ 11 12 / 32

¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯

1 / 24 112 / 47

But how to get these? What is the procedure?

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

89 ) 1 / 13 89 ) 100 / 15

¯¯

/ 11 11 11 / first digit 1 x 11

¯¯¯¯¯¯¯¯

1 / 24 1 / 1 total second is 1x11

22 total of 3rd digit is 2 x 11

¯¯¯¯¯¯¯¯¯¯

112 / 47

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

Thus

98 ) 100 / 15

¯¯

02 02 / i.e., 10015 ÷ 98 gives

0 / 0 Q = 102, R = 19

/ 04

¯¯¯¯¯¯¯¯¯¯

102 / 19

In the same way

897 ) 11 / 422

¯¯¯

103 1 / 03

/ 206

¯¯¯¯¯¯¯¯¯

12 / 658

gives 11,422 ÷ 897, Q = 12, R=658.

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

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

* Obtain the Quotient and Remainder for the following problems.

1) 311 ÷ 9 2) 120012 ÷ 9 3) 1135 ÷ 97

4) 2342 ÷ 98 5) 113401 ÷ 997

6) 11199171 ÷ 99979

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

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PARĀVARTYA – YOJAYET

‘Paravartya – Yojayet’ means 'transpose and apply'

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

Example 1 : Divide 1225 by 12.

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

-2

¯¯¯¯

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

i.e.,, 12 122 5

- 2

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

i.e.,, 12 122 5

-2 -2

¯¯¯¯¯ ¯¯¯¯

Since 1 x –2 = -2 and 2 + (-2) = 0

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

12 122 5

- 2 -20

¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯

102 5

Step 5 : Continue the process to the last digit.

i.e., 12 122 5

- 2 -20 -4

¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯

102 1

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

Thus Q = 102 and R = 1

Example 2 : Divide 1697 by 14.

14 1 6 9 7

- 4 -4–8–4

¯¯¯¯ ¯¯¯¯¯¯¯

1 2 1 3

Q = 121, R = 3.

Example 3 : Divide 2598 by 123.

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

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

-2-3

¯¯¯¯¯ ¯¯¯¯¯¯¯¯

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

1 2 3 2 5 9 8

-2-3 -4 -6

¯¯¯¯¯ -2–3

¯¯¯¯¯¯¯¯¯¯

2 1 1 5

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

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

Hence Q = 21 and R = 15.

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

1 1 2 1 3 2 3 9 4 7 9

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

¯¯¯¯¯¯¯¯ -1-2-1-3 with 1

¯¯¯¯¯¯¯¯¯¯¯¯¯

2 1 4 0 0 6

Hence Q = 21, R = 4006.

Example 5 : Divide 13456 by 1123

1 1 2 3 1 3 4 5 6

-1–2–3 -1-2-3

¯¯¯¯¯¯¯ -2-4 –6

¯¯¯¯¯¯¯¯¯¯¯¯¯

1 2 0–2 0

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

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

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

1) 1234 ÷ 112 2) 11329 ÷ 1132

3) 12349 ÷ 133 4) 239479 ÷ 1203

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

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

X - 1 6x2 + 5x + 4

¯¯¯¯¯¯

1 6 + 11

¯¯¯¯¯¯¯¯¯¯¯¯

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

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

X - 5 x3 – 3x2 + 10x – 4

¯¯¯¯¯

5 5 + 10 100

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

x2 + 2x + 20, + 96

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

The procedure as a mental exercise comes as follows :

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

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

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

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

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

Example 3:

x4 – 3x3 + 7x2 + 5x + 7

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

x + 4

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

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

¯¯¯¯¯

-4 - 4 + 28 - 140 + 540

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

x3 - 7x2 + 35x - 135 547

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

or we proceed orally as follows:

x4 / x gives 1 as first coefficient.

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

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

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

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

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

Note :

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

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

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

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

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

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

x2 + 0 . x + 1 0 - 2

¯¯¯¯¯¯¯¯¯¯¯¯

0 - 1 0 + 3

0 + 2

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

2 - 3 - 2 0 4

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

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

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

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

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

2 0 - 3 4 0 - 6

-2 0 + 3

- 4 0 + 6

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

2 - 1 - 2 - 7 - 1 +13

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

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

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

1) (4x2 + 3x + 5) ÷ (x+1)

2) (x3 – 4x2 + 7x + 6) ÷ (x – 2)

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

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

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

Paravartya in solving simple equations :

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

Type ( i ) :

Consider the problem 7x – 5 = 5x + 1

7x – 5x = 1 + 5

i.e.,, 2x = 6 x = 6 ÷ 2 = 3.

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

d - b.

x = ¯¯¯¯¯¯¯¯

a - c

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

Hence 1 – (- 5) 1+5 6

x = _______ = ____ = __ = 3

7 – 5 7-5 2

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

d - b 6 - 4 2

x = _____ = _____ = __ = 2

a - c 3 - 2 1

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

cd - ab

x = ______________

(a + b) – (c + d)

It is trivial form the following steps

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

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

bx + ax – dx – cx = cd – ab

x( a + b – c – d) = cd – ab

cd – ab cd - ab

x = ____________ x = _________________

a + b – c – d ( a + b ) – (c + d.)

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

By paravartya

cd – ab 1 (2) – (-3) (-2)

x = __________ = ______________

a + b – c –d - 3 – 2 – 1 – 2

2 - 6 - 4 1

= _______ = ___ = __

- 8 - 8 2

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

Now cd - ab (3) (-4) – (7) (-6)

x = ___________ = ________________

a + b – c – d 7 + (-6) – 3 - (-4)

- 12 + 42 30

= ____________ = ___ = 15

7 – 6 – 3 + 4 2

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

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

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

Type ( iii) :

Consider the problems of the type ax + b m

______ = __

cx + d n

By cross – multiplication,

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

nax + nb = mcx + md

nax - mcx = md – nb

x( na – mc ) = md – nb

md - nb

x = ________

na - mc.

Now look at the problem once again

ax + b m

_____ = __

cx + d n

paravartya gives md - nb, na - mc and

md - nb

x = _______

na - mc

Example 1: 3x + 1 13

_______ = ___

4x + 3 19

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

x = ______ = ____________ = _______ = __

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

= 4

Example 2: 4x + 5 7

________ = __

3x + 13/2 8

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

x = _______________

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

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

= __________ = _________ = ______ = __

32 – 21 32 – 21 2 X 11 2

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

_____ + ____ = 0

x + a x + b

Take L.C.M and proceed.

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

______________ = 0

(x + a) (x +b)

mx + mb + nx + na

________________ = 0

(x + a)(x + b)

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

-mb - na

x = ________

(m + n)

Thus the problem m n

____ + ____ = 0, by paravartya process

x + a x + b

gives directly

-mb - na

x = ________

(m + n)

Example 1 : 3 4

____ + ____ = 0

x + 4 x – 6

gives -mb - na

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

(m + n)

-(3)(-6) – (4) (4) 18 - 16 2

= _______________ = ______ = __

( 3 + 4) 7 7

Example 2 :

5 6

____ + _____ = 0

x + 1 x – 21

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

x = ________________ = ______ = __ = 9

5 + 6 11 11

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

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

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

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

______ = __

3x - 5 8

4) x + 1 / 3

_______ = 1

3x - 1

5) 5 2

____ + ____ = 0

x + 3 x – 4

II)

1. Show that for the type of equations

m n p

____ + ____ + ____ = 0, the solution is

x + a x + b x + c

- mbc – nca – pab

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

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

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

Problem 3 2 5

____ + ____ - ____ = 0

x + 4 x + 6 x + 5

some more simple solutions :

m n m + n

____ + ____ = _____

x + a x + b x + c

Now this can be written as,

m n m n

____ + ____ = _____ + _____

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

m m n n

____ - ____ = _____ - _____

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

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

________________ = ________________

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

mx + mc – mx – ma nx + nb – nx – nc

________________ = _______________

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

m (c – a) n (b –c)

____________ = ___________

x +a x + b

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

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

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

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

x = ___________________

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

By paravartya rule we can easily remember the formula.

Example 1 : solve 3 4 7

____ + _____ = ____

x + 1 x + 2 x + 3

In the usual procedure, we proceed as follows.

3 4 7

____ + ____ = ____

x + 1 x + 2 x + 3

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

________________ = _____

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

3x + 6 + 4x + 4 7

_____________ = ____

x2 + 2x + x + 2 x + 3

7x + 10 7

_________ = ____

x2 + 3x + 2 x + 3

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

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

31x + 30 = 21x + 14

31 x – 21 x = 14 – 30

i.e.,, 10x = - 16

x = - 16 / 10 = - 8 / 5

Now by Paravartya process

3 4 7

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

x + 1 x + 2 x + 3

mb( a – c ) + na ( b – c )

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

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

3 . 2 ( 1 – 3 ) + 4 . 1 . ( 2 – 3)

= __________________________

3 ( 3 – 1 ) + 4 ( 3 – 2 )

6 ( -2)+ 4 (-1) - 12 – 4 - 16 - 8

= _____________ = _______ = ____ = ___

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

Example 2 :

3 5 8

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

x - 2 x – 6 x + 3

mb ( a – c ) + na ( b – c)

x = _____________________

m ( c – a ) + n ( c – b )

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

= __________________________________

3 ( 3 – ( -2 ) ) + 5 ( 3 – ( - 6 ) )

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

= ____________________________

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

90 + 90

= _______ = 180 / 60 = 3.

15 + 45

Solve the problems using the methods explained above.

1) 2 3 5

____ + ____ = ____

x + 2 x + 3 x + 5

2) 4 6 10

____ + ____ = ____

x + 1 x + 3 x + 4

3) 5 2 3

____ + ___ = ____

x - 2 3 - x x – 4

4) 4 9 15

_____ + _____ = _____

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

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

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

________ + ________ = _______ gives

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

12 18 30

_____ + _____ = _____ Now proceed.

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

Simultaneous simple equations:

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

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

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

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

4x + 5y = 23

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

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

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

Hence value of x = 4 ÷ 2 = 2.

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

13 x 4 – 23 x 2 = 52 – 46 = 6.

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

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

Example 2: 5x – 3y = 11

6x – 5y = 09

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

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

x = Nr ÷ Dr = 28 ÷ 7 = 4

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

Dr is 7

Hence y = 21 ÷ 7 = 3.

Example 3: solve 3x + y = 5

4x – y = 9

Now we can straight away write the values as follows:

(1)(9) – (-1)(5) 9 + 5 14

x = _____________ = _____ = ___ = 2

(1)(4) – (3)(-1) 4 + 3 7

(5)(4) – (9)(3) 20 – 27 -7

y = ____________ = _______ = ___ = -1

(1)(4) – (3)(-1) 4 + 3 7

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

Algebraic Proof:

ax + by = m ……… ( i )

cx + dy = n ………. ( ii )

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

adx + bdy = m.d

cbx + dby = n.b

____________________

( ad – cb ) .x = md – nb

md - nb bn - md

x = ______ = ______

ad - cb bc - ad

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

acx + bcy = m.c

cax + day = n.a

_____________________

( bc – ad ) . y = mc - na

mc - na

y = ______

bc - ad

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

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

1. 2x + y = 5 2. 3x – 4y = 7

3x – 4y = 2 5x + y = 4

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

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

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SŨNYAM SĀMYASAMUCCAYE

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

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

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

Otherwise we have to work like this:

7x + 3x = 4x + 5x

10x = 9x

10x – 9x = 0

x = 0

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

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

No need to proceed in the usual procedure like

5x + 5 = 3x + 3

5x – 3x = 3 – 5

2x = -2 or x = -2 ÷ 2 = -1

Simply think of the contextual meaning of ‘ Samuccaya ‘

Now Samuccaya is ( x + 1 )

x + 1 = 0 gives x = -1

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

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

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

Since it is same , we derive x = 0

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

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

Consider the example.

1 1

____ + ____ = 0

3x-2 2x-1

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

(2x-1)+(3x–2)

____________ = 0

(3x–2)(2x–1)

5x–3

__________ = 0

(3x–2)(2x–1)

5x – 3 = 0 5x = 3

x = __

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

i.e., 3x – 2 + 2x - 1 = 5x - 3 = 0

giving 5x = 3 x = 3 / 5

It is true and applicable for all problems of the type

m m

____ + _____ = 0

ax+b cx+d

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

- ( b + d )

x = _________

( a + c )

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

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

Consider examples of type

ax + b ax + c

_____ = ______

ax + c ax + b

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

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

2abx – 2acx = c2 – b2

x ( 2ab – 2ac ) = c2 – b2

c2–b2 (c+b)(c-b) -(c+b)

x = ______ = _________ = _____

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

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

2ax+b+c = 0

2ax = -b-c

-(c+b)

x = ______

2a Hence the statement.

Example 4:

3x + 4 3x + 5

______ = ______

3x + 5 3x + 4

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

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

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

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

6x = -9

-9 -3

x = __ = __

6 2

Example 5:

5x + 7 5x + 12

_____ = _______

5x +12 5x + 7

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

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

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

10x = -19

-19

x = ____

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

Example 6:

2x + 3 x + 1

_____ = ______

4x + 5 2x + 3

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

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

= 2 ( 3x + 4 )

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

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

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

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

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

Example 7:

3x + 2 2x + 5

_____ = ______

2x + 5 3x + 2

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

3x + 2 2x + 5

_____ = ______

2x + 5 3x + 2

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

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

9x2 + 12x + 4 - 4x2 - 20x – 25 = 0

5x2 – 8x – 21 = 0

5x2 – 15x + 7x – 21 = 0

5x ( x – 3 ) + 7 ( x – 3 ) = 0

(x – 3 ) ( 5x + 7 ) = 0

x – 3 = 0 or 5x + 7 = 0

x = 3 or - 7 / 5

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

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

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

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

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

Hence 5x + 7 = 0 , x – 3 = 0

5x = -7 , x = 3

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

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

Example 8:

3x + 4 5x + 6

______ = _____

6x + 7 2x + 3

Observe that

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

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

Further N1 ~ D1 = (3x + 4) – (6x + 7)

= 3x + 4 – 6x – 7

= -3x – 3 = -3 ( x + 1 )

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

By ‘Sunyam Samuccaye’ we have

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

8x = -10 x + 1 = 0

x = - 10 / 8 x = -1

= - 5 / 4

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

Example 9:

1 1 1 1

____ + _____ = ____ + ____

x - 4 x – 6 x - 2 x - 8

Usually we proceed as follows.

x–6+x-4 x–8+x-2

___________ = ___________

(x–4) (x–6) (x–2) (x-8)

2x-10 2x-10

_________ = _________

x2–10x+24 x2–10x+16

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

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

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

132x – 160 = 148x – 240

132x – 148x = 160 – 240

– 16x = - 80

x = - 80 / - 16 = 5

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

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

D3 + D4 = x – 2 + x – 8 = 2x – 10

By Samuccaya, 2x – 10 gives 2x = 10

x = __ = 5

Example 10:

1 1 1 1

____ + ____ = ____ + _____

x - 8 x – 9 x - 5 x – 12

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

D3 + D4 = x – 5 + x –12 = 2x – 17

Now 2x – 17 = 0 gives 2x = 17

x = __ = 8½

Example 11:

1 1 1 1

____ - _____ = ____ - _____

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

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

1 1 1 1

____ + ____ = ____ + _____

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

Now ‘Samuccaya’ sutra applies

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

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

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

x = - 16 / 2 = - 8.

Solve the following problems using Sunyam Samya-Samuccaye process.

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

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

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

1 1

4. ______ + ____ = 0

4 x - 3 x – 2

4 4

5. _____ + _____ = 0

3x + 1 5x + 7

2x + 11 2x+5

6. ______ = _____

2x+ 5 2x+11

3x + 4 x + 1

7. ______ = _____

6x + 7 2x + 3

4x - 3 x + 4

8. ______ = _____

2x+ 3 3x - 2

1 1 1 1

9. ____ + ____ = ____ + _____

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

1 1 1 1

10. ____ - ____ = _____ - _____

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

Sunyam Samya Samuccaye in Certain Cubes:

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

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

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

= 2 ( x3 – 15x2 + 75x – 125 )

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

156x – 280 = 150x – 250

156x – 150x = 280 – 250

6x = 30

x = 30 / 6 = 5

But once again observe the problem in the vedic sense

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

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

The traditional method will be horrible even to think of.

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

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

Algebraic Proof :

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

x – 2a + x – 2b = 2x – 2a – 2b

= 2 ( x – a – b )

Now the expression,

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

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

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

cancel the common terms on both sides

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

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

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

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

= ( a + b ) ( a2 + b2 – 2ab )

= ( a + b ) ( a – b )2

since x = a + b

Solve the following using “Sunyam Samuccaye” process :

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

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

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

Example :

(x + 2)3 x + 1

______ = _____

(x + 3)3 x + 4

with the text book procedures we proceed as follows

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

_______________ = _____

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

Now by cross multiplication,

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

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

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

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

56x + 32 = 54x + 27

56x – 54x = 27 – 32

2x = - 5

x = - 5 / 2

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

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

N2 + D2 on the right hand side

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

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

Solve the following by using vedic method :

(x + 3)3 x+1

______ = ____

(x + 5)3 x+7

(x - 5)3 x - 3

______ = ____

(x - 7)3 x - 9

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

2x2 + 2 = 5x

2x2 – 5x + 2 = 0

2x2 – 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

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:

________

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

x x + 1 1

_____ + _____ = 9 __

x + 1 x 9

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.

2x - 5

____________

(x – 2) (x – 3)

____________

(x + 1) (x – 2)

x – 13

__________

x2 - 2x - 15

3x + 4

__________

3x2 + 3x + 2Sanskrit तो

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What is Vedic Maths?

It is an ancient system of calculation found in the Vedas between 1911 and 1918 by Shri Bharati Krishna Tirathji. The place where he was born in 1884 is Puri in Orrisa. He had exceptional knowledge and skills in the subjects of Maths, Science, Sanskrit and Humanities. He had learnt these vedic techniques from “Rig Veda”.

Who is the father of Vedic Maths?

Shri Bharti Krishna Tirthaji is called the father of Vedic Maths. He took to self realization at Shringeri Matha with guidance of Shri Shankracharya, Shri Sachidananda Shiva Abhinava Narsimha Saraswati. He spent 7 years in deep meditation and study of Vedanta and lived the life of a Sadhu from 1911 to 1918. During this period , the 16 Sutras of Vedic Maths were made. He was called Shri Bharti Krishna Tirthaji after being initiated into Sanyas in July 1919 by Shri Trivikram Teerathaji of Varanasi.

What are the benefits of Vedic Maths?

Vedic Maths offers simpler and faster methods of calculation to students.

It develops their understanding of Maths.

Both the junior and Senior Vedic Maths helps children develop their mathematical ability to score high in school maths, and higher competitive exams.

It can be learnt and mastered with ease and in a very short time.

It also provides a set of independent cross-checking methods.

Problems are reduced to one-line answers.

Reduces dependence on calculators.

Games and maths activities in this program develops student’s interest and builds fun learning environment.

What is the duration of the program?

Vedic Maths Classes is relevant to students in every grade from Class 3 – 10th . Each student attends three sessions every week. However, the program would move as per the learning ability and pace of the child.

How many students are there in session?

Students are taught in small batch sizes to ensure personal attention to each and every student.

Are students given any assessments ?

Yes, periodic assessments are part of the program. Before joining the program, the child is assessed to know strength areas and areas of improvement of every child. Thereby after understanding the student’s current mathematical ability, students are guided to improve their logical skills. As per the approach guided in the curriculum, the child’s ability is re-assessed through continuous tests. This gives a clear progress chart of the child. Thereby teaming up with parents , the teacher and student work together for a common goal of student’s maths development.

How is it different from maths tuitions?

It is an (after-school) math-learning programme. As a result, while home-based tuitions focuses primarily on the school syllabus and preparing students for their tests and exams. This is a maths skill program which intends to develop calculative ability of the child. It thereby helps in school maths and also at maths olympiads.

How can Winaum Learning help in choosing the best Vedic Maths Training Centres near you?

Winaum Learning offers Vedic Maths OFFLINE training Centers all over the country and also classes can be taken ONLINE. To get the information about the course, just click on buttons for signup or link for free trial and fill in your details. The coordinating team shall contact you and give you information about the course , arrange a free trial for your child and share the fee details.

What Are The Benefits Of Vedic Maths Coaching Classes?

Most of the kids have struggled with math calculation at the different ages in school. Some children face math difficulty in early ages and others when they grow up. Sometimes it becomes a math phobia and children start hating math classes. But with Vedic maths coaching it can change. Under the banner of Maplemaths, Winaum Learning offer Vedic Maths coaching in India and around the world and maths becomes fun for the child. The engaging curriculum that it has , helps kids to eliminate their fear of maths and children start loving maths. These calculations have lot of tricks and short cuts which help children to calculate faster. Know more by clicking here.

Do Vedic maths coaching classes help you fall in love with maths ?

The most common problem that is found among children while doing maths is, that they are slow in calculations. Calculation is based on four main pillars which are addition, subtraction, multiplication and division and of course higher calculative ability also. Once the kids get comfortable with numbers and they know how to do it faster, their confidence keeps on growing from one class to another. The course offers a variety of short cut methods helping kids to solve difficult math calculations easily. Surely, once the kids are done with course , they love maths even more.

Can Vedic maths help you with time saving skills?

Many a time when we ask children, how would they feel if they get a little time during the exam. Almost all of them will say , they would feel great. Students with this extra time would be able to recheck their answers and revise the whole paper.Now since Vedic Maths helps children to do calculations faster with a lot tricks and techniques. Once the kids has understood the math question, knowing the faster method to calculate the answer , they would save a lot of time. Each second saved is extra time earned. So Vedic maths helps kids save a lot of time in while doing calculations with the variety of methods the course offers.

Will kids still depend on calculators if they join this classes?

When the kids grow up and join offices for work, they all use calculators and computers for calculation specially in financial and real estate field apart from other. But early dependence on calculators is not a good sign. The more we train our brain with mental calculations the brain also grows. With Vedic Maths you learn a lot of fast calculation techniques that help you to save time and thereby there is less dependency on calculators.

Does Vedic maths keep your brain active?

The brain becomes dull when it stops learning new things. With Vedic Maths, children learn to look at the numbers from a new angle. The logic of maths calculations is explained here in a very easy manner and it is easier to pick up for every child. With time the mind is able to deal with higher difficulty level and the brain grows. The neural network of brain create new pathways and you become more intelligent and it keeps your brain healthy for years.

What is the study material Vedic maths training centres provide ?

Students are given study material which includes textbooks, formula guides, and tips & tricks sheet.

What is the qualification of the faculty at Vedic maths training centres?

The faculty members are trained teachers and have years of experience in teaching Vedic math.

Will Vedic maths classes complement a student’s academic performance?

Yes, it will definitely help children in school maths and math olympiads. Vedic maths will help you improve your calculation speed which is a boon during examinations.

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हिन्दुओं के प्रमुख वंश, या पूर्वज

7. A रेखांक, रेखांकित अंक, ऋणात्मक अंक या विनकुलम्

संपूर्ण सुंदरकांड का प्रारंभ (शास्त्र सम्मत रूप से)

पुरुषोत्तम मास /अधिक मास / मल मास : 18 सितंबर से 16 अक्टूबर 2020 । अमिट पुण्य अर्जित करने का अवसर

सावन या श्रावण मास और भगवान शिव और उनका परिवार

3.1 A वैदिक गणित के 16 सूत्र व उनके अर्थ।

सूर्यग्रहण अर्थात् सूर्योपराग

4.4 A बीजांक या मूलांक:

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