Vedic Mathematics is that one field of knowledge which fulfils the purpose of education by developing the total creative genius of the individual, giving him/her the ability to be always spontaneously right and automatically precise so that his/her action, supported by Natural Law, is always effortlessly fulfilling.

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4 Paraavartya Yojayet

5 Shunyam Saamyasamuccaye

6 (Anurupye) Shunyamanyat

7 Sankalana-vyavakalanabhyam

8 Puranapuranabyham

9 Chalana-Kalanabyham

10 Yaavadunam

11 Vyashtisamanstih

12 Shesanyankena Charamena

13 Sopaantyadvayamantyam

14 Ekanyunena Purvena

15 Gunitasamuchyah

16 Gunakasamuchyah

The Sutra (formula) Ekādhikena Pūrvena means:

*“By one more than the previous one”.*

Now let us apply this sutra to the ‘**squaring of numbers ending in 5**’.

Consider the example 25^2.

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 5^2, that is, 25.

Thus 25^2 = 2 X 3 / 25 = 6/25=625.

In the same way,

35^2= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;

65^2= 6 X 7 / 25 = 4225;

105^2= 10 X 11/25 = 11025;

135^2= 13 X 14/25 = 18225;

Now try to find out the squares of the numbers 15, 45, 85, 125, 175 and verify the answers.

Now let us apply this sutra to the ‘

Consider the example 25^2.

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 5^2, that is, 25.

Thus 25^2 = 2 X 3 / 25 = 6/25=625.

In the same way,

35^2= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;

65^2= 6 X 7 / 25 = 4225;

105^2= 10 X 11/25 = 11025;

135^2= 13 X 14/25 = 18225;

Now try to find out the squares of the numbers 15, 45, 85, 125, 175 and verify the answers.

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

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 :*1*68421 (multiply 8 by 2="16, *1* carried over, 6 put to left)

Step. 6 :*1*368421 ( 6 X 2 =12,+1 = 13, *1* carried over, 3 put to left )

Step. 7 : 7368421 ( 3 X 2, = 6 +1 = 7, put to left)

Step. 8 :*1*47368421 (as in the same process)

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

Step. 10 :*1*8947368421

Step. 11 :*1*78947368421

Step. 12 :*1*578947368421

Step. 13 :*1*1578947368421

Step. 14 : 31578947368421

Step. 15 : 631578947368421

Step. 16 :*1*2631578947368421

Step. 17 : 52631578947368421

Step. 18 :*1*052631578947368421

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

Thus 1 / 19 = 0.052631578947368421

Find the recurring decimal form of the fractions 1 / 29, 1 / 59, 1 / 69, 1 / 79, 1 / 89 using Ekadhika process

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 :

Step. 6 :

Step. 7 : 7368421 ( 3 X 2, = 6 +1 = 7, put to left)

Step. 8 :

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

Step. 10 :

Step. 11 :

Step. 12 :

Step. 13 :

Step. 14 : 31578947368421

Step. 15 : 631578947368421

Step. 16 :

Step. 17 : 52631578947368421

Step. 18 :

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

Thus 1 / 19 = 0.052631578947368421

Find the recurring decimal form of the fractions 1 / 29, 1 / 59, 1 / 69, 1 / 79, 1 / 89 using Ekadhika process

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(eg: 96 x 98 or 102 x 104). 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.

**Case (i) : when both the numbers are lower than the base.**

Find 97 X 94. Here base is 100. Now following the rules, the working is as follows: 97 is 3 less than the nearest base 100. and 94 is 6 less than the same nearest base 100. Hence 3 and 6 are called deviations from the base. Always the base should be same for the two numbers.

=91/18=9118

*Note: Here '/' signifies just a seperation and has nothing to do with division.*

In genreal, 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 ( ii) : When 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.104 X 102. Base is 100.

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(eg: 96 x 98 or 102 x 104). 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.

=91/18=9118

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. 13 X 7. Base is 10

10/(-9) should be read as 'one zero, nine bar'. Here 'one' and 'zero' are in normal form. 'nine' is in complement form. so, when we bring a carry from normal form to complment form, '10' becomes '9' and '9-bar' becomes '1'( 10's complement of 9). Hence 10/(-9)-91.

Another example would be 94/(-3)=93/7=937.

5) 1003 X 997 6) 11112 X 9998 7) 1234 X 1002 8) 118 X 105

The upa-Sutra '*Anurupyena*' means '*proportionality*' or '*similarly*'. This Sutra is highly useful to find products of two numbers when both of them are near the Common bases like 50, 60, 200 etc(multiples of powers of 10).

**Example 1: 46 X 43**

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

This is much more difficult and of no use.

Now by ‘Anurupyena’ we consider a different working base through we can solve the problem.

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

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

iv) Write cross-subtraction or cross- addition as the case may be under the line drawn. Multiply the differences and write the product in the left side of the answer.

v) 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

**Solve 58 x 48 : **

Working base 50 = 5 x 10 gives

**Find the following products.**

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

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

This is much more difficult and of no use.

Now by ‘Anurupyena’ we consider a different working base through we can solve the problem.

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

Treat it as 100 / 2 = 50.

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

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

iv) Write cross-subtraction or cross- addition as the case may be under the line drawn. Multiply the differences and write the product in the left side of the answer.

v) 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

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

Multiplication of two 2 digit numbers.

Ex.1: Find the product 14 X 12

The symbols are operated from right to left .

Step i) :

Step ii) :

Step iii) :

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

Steps:

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.

General rule for a 3 digit by 3 digit multiplication:

The Sutra ' Adyamadyena-Antyamantyena' 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

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 )

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.

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

From Sankalana – vyavakalanabhyam

add them,

i.e., ( 45x – 23y ) + ( 23x – 45y ) = 113 + 91

i.e., 68x – 68y = 204 i.e., x – y = 3

subtract one from other,

i.e., ( 45x – 23y ) – ( 23x – 45y ) = 113 – 91

i.e., 22x + 22y = 22 i.e., x + y = 1

and repeat the same sutra, we get x = 2 and y = - 1

Example 2:

1955x – 476y = 2482

476x – 1955y = -4913

Oh ! what a problem ! And still

just add, 2431( x – y ) = - 2431 i.e., x – y = -1

subtract, 1479 ( x + y ) = 7395 i.e., 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

Example 1:

45x – 23y = 113

23x – 45y = 91

From Sankalana – vyavakalanabhyam

add them,

i.e., ( 45x – 23y ) + ( 23x – 45y ) = 113 + 91

i.e., 68x – 68y = 204 i.e., x – y = 3

subtract one from other,

i.e., ( 45x – 23y ) – ( 23x – 45y ) = 113 – 91

i.e., 22x + 22y = 22 i.e., x + y = 1

and repeat the same sutra, we get x = 2 and y = - 1

Example 2:

1955x – 476y = 2482

476x – 1955y = -4913

Oh ! what a problem ! And still

just add, 2431( x – y ) = - 2431 i.e., x – y = -1

subtract, 1479 ( x + y ) = 7395 i.e., 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

This sutra means whatever the extent of its deficiency, lessen it still further to that very extent; and also set up the square of that deficiency. This sutra is very handy in calculating squares of numbers near(lesser) to powers of 10

For instance in computing the square of 98 we go through the following steps:

1. The nearest power of 10 to 98 is 100.

2. Therefore, let us take 100 as our base.

3. Since 98 is 2 less than 100, we call 2 as the deficiency.

4. Decrease the given number further by an amount equal to the deficiency.

i.e., perform ( 98 -2 ) = 96. This is the left side of our answer!!.

5. On the right hand side put the square of the deficiency, that is square of 2 = 04.

6. Append the results from step 4 and 5 to get the result. Hence the answer is 9604.

For instance in computing the square of 98 we go through the following steps:

1. The nearest power of 10 to 98 is 100.

2. Therefore, let us take 100 as our base.

3. Since 98 is 2 less than 100, we call 2 as the deficiency.

4. Decrease the given number further by an amount equal to the deficiency.

i.e., perform ( 98 -2 ) = 96. This is the left side of our answer!!.

5. On the right hand side put the square of the deficiency, that is square of 2 = 04.

6. Append the results from step 4 and 5 to get the result. Hence the answer is 9604.

Note: while calculating step 5, the number of digits in the squared number (04) should be equal to number of zeroes in the base(100). Hence in our case, the base 100 has 2 zeros and hence sqaure of 2 is 04 and not just 4.

Now consider another example: Calculate the square of 94. ( following the steps given above...)

1. The neareset power of 10 to given number 94 is 100

2. Hence the base is 100

3. Deficiency = 6

4. Subtract the deficiency from the given number ( 94 - 6 ) = 88

5. Square of deficiency = 36 ( 2 digits)

6. Append the results from step 4 and 5 to get the result. i.e., 8836

1. The neareset power of 10 to given number 94 is 100

2. Hence the base is 100

3. Deficiency = 6

4. Subtract the deficiency from the given number ( 94 - 6 ) = 88

5. Square of deficiency = 36 ( 2 digits)

6. Append the results from step 4 and 5 to get the result. i.e., 8836

Try out calculating squares of 96, 994, 9999995

<br>108 x 109 can be computed in one second by understanding this Pattern Recogition:

I add the excess of "8" to the other number "109" and then tag on the multiplicatin of those two excesses: "8" and "9".

= 109 + 8 / 8 x 9

= 117 / 72

= 11,772

= 117 / 72

= 11,772

Sutra: By The Deficiency

The Squaring of Numbers Under a Base

To solve 98 Squared (98 x 98) we must first determine what Base we are in. It is close to 100, therefore we say Base 100. We must now choose one of the 16 Sutras to effectively solve the problem. It is called: "By The Deficiency":

"By whatever the deficiency, lessen it further by that much and set up the square thereof"

says one of the 16 Sutras. Sounds cryptic and meaningless yet it quickly solves the problem.

We get our answer by merely knowing how much is 100 less 98. Knowing that the deficiency is 2, we merely lessen 98 by 2 and then we tag on the squaring of that 2. As a one-line answer, the setting out would appear as thus:

98 Squared = 98 - 2 / 2x2. Simplifying it:

= 96 / _ 4

We almost have our answer. What we need to know is that since our Base is 100, it has 2 zeroes, therefore this fact governs the need for 2 spaces or 2 digits after the "forward slash" symbol ( / ). By inserting or inventing The Zero as a "Place Marker", the answer is achieved:

= 96 / 04

= 9,604.

Observe similar examples:

97

= 94 / 09

96

= 92 / 16.

What if we enlarged our numbers to 998 Squared?

It is close to 1,000 so we say Base 1,000 and know to have 3 zeroes on the right hand side of the ( / ).

<span style="font-weight:bold;">998<sup>2</sup></span> = 998 - 2 / 2x2

= 996 / _ _ 4

= 996 / 004.

= 996,004

Understanding this, you can be calculating digits in the millions:

9998

= 9996 / _ _ _ 4

(Since we are in Base 10,000 the 4 Zeroes determine the need for 4 digits after the ( / ).

= 9996 / 0004

= 99,960,004.

This sutra means whatever the extent of its surplus, increment it still further to that very extent; and also set up the square of that surplus. This sutra is very useful in calculating the sqaures of numbers nearer(greater) to powers of 10.

For instance: in computing the square of 103 we go through the following steps:

1. The nearest power of 10 to 103 is 100.

2. Therefore, let us take 100 as our base.

3. Since 103 is 3 more than 100(base), we call 3 as the surplus.

4. Increase the given number further by an amount equal to the surplus.

i.e., perform ( 103 + 3 ) = 106. This is the left side of our answer!!.

5. On the right hand side put the square of the surplus, that is square of 3 = 09.

6. Append the results from step 4 and 5 to get the result.

Hence the answer is 10609.

Note: while calculating step 5, the number of digits in the squared number (09) should be equal to number of zeroes in the base(100). Hence in our case, the base 100 has 2 zeros and hence sqaure of 3 is 09 and not just 9.

Similarly if we need to calculate the square of 1009,

1. The nearest power of 10 to 1009 is 1000.

2. Therefore, let us take 1000 as our base.

3. Since 1009 is 9 more than 1000(base), we call 9 as the surplus.

4. Increase the given number further by an amount equal to the surplus.

i.e., perform ( 1009 + 9 ) = 1018. This is the left side of our answer!!.

5. On the right hand side put the square of the surplus, that is square of 9 = 081.(not just 81 !)

6. Append the results from step 4 and 5 to get the result.

Hence the answer is 1018081.

Try to find the square of 106 and 19 using this sutra.

Now here's a twist. What is sqaure of 112 using this rule? will this rule apply?

Answer is Yes. But we need to stick to basics and use some common sense.

Here we go...

1. & 2. the base is 100.

3. Surplus = 12

4. Increase the given number by surplus i.e. ( 112 + 12 ) = 124

5. Square of surplus, 12 = 144 . Now what?

by rule we need to consider only 2 digits in step 5 ( as the base has 2 zeros )

6. We have to append 124 and 144. The '1' in 144 becomes 'carry' to 124.

Hence the result is (124 +1 ) appended with 44 i.e., 12544 is the result.

dont get confused. try this example to make yourself more confident.

Find the square of 111. Now try to find the square of 10101.

For instance: in computing the square of 103 we go through the following steps:

1. The nearest power of 10 to 103 is 100.

2. Therefore, let us take 100 as our base.

3. Since 103 is 3 more than 100(base), we call 3 as the surplus.

4. Increase the given number further by an amount equal to the surplus.

i.e., perform ( 103 + 3 ) = 106. This is the left side of our answer!!.

5. On the right hand side put the square of the surplus, that is square of 3 = 09.

6. Append the results from step 4 and 5 to get the result.

Hence the answer is 10609.

Note: while calculating step 5, the number of digits in the squared number (09) should be equal to number of zeroes in the base(100). Hence in our case, the base 100 has 2 zeros and hence sqaure of 3 is 09 and not just 9.

Similarly if we need to calculate the square of 1009,

1. The nearest power of 10 to 1009 is 1000.

2. Therefore, let us take 1000 as our base.

3. Since 1009 is 9 more than 1000(base), we call 9 as the surplus.

4. Increase the given number further by an amount equal to the surplus.

i.e., perform ( 1009 + 9 ) = 1018. This is the left side of our answer!!.

5. On the right hand side put the square of the surplus, that is square of 9 = 081.(not just 81 !)

6. Append the results from step 4 and 5 to get the result.

Hence the answer is 1018081.

Try to find the square of 106 and 19 using this sutra.

Now here's a twist. What is sqaure of 112 using this rule? will this rule apply?

Answer is Yes. But we need to stick to basics and use some common sense.

Here we go...

1. & 2. the base is 100.

3. Surplus = 12

4. Increase the given number by surplus i.e. ( 112 + 12 ) = 124

5. Square of surplus, 12 = 144 . Now what?

by rule we need to consider only 2 digits in step 5 ( as the base has 2 zeros )

6. We have to append 124 and 144. The '1' in 144 becomes 'carry' to 124.

Hence the result is (124 +1 ) appended with 44 i.e., 12544 is the result.

dont get confused. try this example to make yourself more confident.

Find the square of 111. Now try to find the square of 10101.

The Sutra simply means - numbers of which the last digits added up give 10.

This sutra is helpful in multiplying numbers whose last digits add up to 10(or powers of 10). The remaining digits of the numbers should be identical.

i.e. the Sutra works in multiplication of numbers for example:

25 and 25, 2 is common and 5 + 5 = 10

47 and 43, 4 is common and 7 + 3 = 10

62 and 68,

116 and 114.

and also for 425 and 475

The last example can be looked as digit 4 being common and 25 and 75 add upto 100.

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 : 57 X 53**

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

Ekadhikena to the remaining digits means, increment the remaining digits by 1 and multiply it with the same.

57 x 53 = ( 5 + 1 )x5 / 7x3*( the '/' is just a seperator and not a division mark)&*nbsp = 30 / 21

= 3021.

**Example 2: 62 x 68**

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

Ekadhikena(increment) of 6 gives 7

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

= 42 / 16

= 4216.

Use Vedic sutras to find the products

1. 125 x 125

2. 34 x 36

3. 98 x 92

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.

Find the following products using*‘Antyayordasakepi’*

1. 318 x 312

2. 425 x 475

This sutra is helpful in multiplying numbers whose last digits add up to 10(or powers of 10). The remaining digits of the numbers should be identical.

i.e. the Sutra works in multiplication of numbers for example:

25 and 25, 2 is common and 5 + 5 = 10

47 and 43, 4 is common and 7 + 3 = 10

62 and 68,

116 and 114.

and also for 425 and 475

The last example can be looked as digit 4 being common and 25 and 75 add upto 100.

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.

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

Ekadhikena to the remaining digits means, increment the remaining digits by 1 and multiply it with the same.

57 x 53 = ( 5 + 1 )x5 / 7x3

= 3021.

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

Ekadhikena(increment) of 6 gives 7

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

= 42 / 16

= 4216.

Use Vedic sutras to find the products

1. 125 x 125

2. 34 x 36

3. 98 x 92

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.

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.

Find the following products using

1. 318 x 312

2. 425 x 475

Lopana sthapanabhyam means 'by alternate elimination and retention'.

Consider the case of factorization of quadratic equation of type ax^{2} + by^{2} + cz^{2} + 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.

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

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.

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

3x

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

3x

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

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

Step (i) : Eliminate y and z, retain x and independent term

i.e., y = 0, z = 0 in the expression (E).

Then E = 3x

Step (ii) : Eliminate z and x, retain y and independent term

i.e., z = 0, x = 0 in the expression.

Then E = 6y

Step (iii): Eliminate x and y, retain z and independent term

i.e., x = 0, y = 0 in the expression.

Then E = 2z

Step (iv) : The expression has the factors

(think of independent terms: constants)

= (3x + 2y + z + 4) (x + 3y + 2z + 5).

Solve the following expressions into factors by using appropriate sutras:

1. x

2. 3x

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)/p>

It is intended for the purpose of verifying the correctness of obtained answers in multiplications, divisions and factorizations.

It means in this context:

Symbolically we represent as sc of the product = product of the sc (in the factors)/p>

Now ( x + 3 ) ( x + 2 ) = 4 x 3 = 12 : Thus verified.

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

i.e., 6 x 8 x -1 = 22 - 70

i.e., -48 = -48 Verified.

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) = 2x

2. 12x

3. 12x

4. ( x + 1 ) ( x + 2 ) ( x + 3 ) = x