The Ultimate and Twice the Penultimate also known as 'Sopantyadvayamantyam" is the Thirteenth sutra in the Vedic Maths and it helps in solving problems immediately.Let us take any multiplication of 12 .

Example: 132 * 12 =?

Step 1 :
In this example, we are going to make a zero sandwich for the number 132 as follows
o 132 o

Step 2:
The above answer is our starter, from where we keep adding the last digit  and twice the second last digit .That means, we proceed for  o132o as follows

Last digit is "o" + Twice the second last digit " 2 "
= 0 + (2*2)
= 4      ------ (1)

Step 3:
Now we add the last digit 2 plus twice the second last digit again for the o132o

Last digit "2" + Twice the second last digit "3"
= 2 + ( 2 * 3)
= 8     ------- (2)

Step 4 :
Repeating the same steps as explained above,

Last digit "3" +Twice the second last digit "1"
= 3 + ( 2 * 1)
= 5   ...........(3)

Step 5:
Again,repeating the same steps

Last digit "1" + Twice the second last digit "0"
= 1   .........(4)

Step 6 :
Accumulating the derived answer in order as we move from left to right we , get

132 * 12 = (4)(3)(2)(1)

Therefore , 132  * 12 =1584

You can try it similarly for any other multiplication problems concerning 12 :).Visit this blog again , more on Vedic Maths to be blogged with more interesting sutra's .

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VERTICALLY AND CROSSWIRE (Urdhva-Tiryagbyham)

Example 1


 22 x 31 = ?

Follow the general steps :
First part : 2 x 3 = 6 (multiplying the left side )
Last part : 2 x 1 = 2 (multiplying the right side)
Middle part :  (2 x 1)  + (2 x 3) =  2  +  6 = 8 (cross-multiplying both sides)
The answer is (first part | middle part  | last part ) = 682

Therefore 22 x 31 = 682




Example :2

68 x 35 = ?

First part | Middle part | Last part
(6 x 3) | ( 6 x 5 )+ (8 x 3)  | (8 x 5)
18 | (30 + 24 ) | 40
18 | 54 | 40
1 8 | 5 4 | 4 0 (carry over and add the left side of the digits)
 = 2380

Therefore 68 x 35 = 2380

There is yet another technique known as the Grid Method ,good for multiplication

GRID METHOD  or Sum of the Products

Example : 1
15 x 17 =?

For this method rewrite 15 as (10 + 5) and 17 as (10 + 7)  and then fill in the grid as shown

Finally add all the numbers inside the grid for the final answer ( 100 + 50 + 70 + 35) = 255

Therefore 15 x 17 = 255

Example 2:

78 x 25 = ?

Rewriting the numbers and forming the grid , we get

Again adding all the digits inside the grids ,(1400+160+350+40) =1950

Therefore 78 x 25 = 1950

You can try with different two digit numbers for both the above methods.In case of a single digit number, say 35 x 6 ,it becomes 35 x 06 and calculate using the same rules as described above.

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The Remainders by the Last Digit or Shesanyankena Charamena is used to express a fraction as a decimal to all its decimal places.It is similar to Ekadhikina Purvena .

Example 1:


a) Express 1/7 as a decimal

1. Add 'zero' to the numerator which makes 1 as 10
2. If the numerator is less than the denominator, add another 'zero' , else proceed as follows ...
3. Divide : (10)/7 = 1 remainder 3
4. Adding zero to the remainder (since it is less than the denominator), we divide as (30)/7= 4 remainder 2
5. We continue taking the remainder the following the steps from the start as follows :

 (20) /7 = 2 remainder 6
 (60) /7 = 8 remainder 4
 (40) /7 = 5 remainder 5
 (50) /7 = 7 remainder 1

6. Now note that the remainder is '1' which is same as the numerator '1'.This means we would get the repetition of the answers again and again.So we will stop here.

7. Using the numbers(remainders) 3,2,6,4,5,1 got above , we multiply them with the denominator '7',

7 x 3 = 2 1
7 x 2 = 1 4
7 x 6 = 4 2
7 x 4 = 2 8
7 x 5 = 3 5
7 x 1 = 7

8. Now we take the red shaded numbers from the above steps as sequence ,142857 (which would be our final result)

Therefore, Result (1/7) = 0.142857142857

PS: Similarly we can find the answers for other fractions using the same procedure as the above illustrated one .

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Definition : Whatever the deficiency subtract that deficit from the number and write along side the square of that deficit.It is also called as Yaavadunam Sutra

This sutra is helpful in finding squares and cubes of any two digit-numbers .It can extend up to any number of digits for finding the square and cubes of  that particular number.But here I'm going use this sutra with different methods for finding squares for any two-digit  and three -digit numbers .

Method 1 :Using the Yaavadunam Sutra

1. 98 x 98 =9604
Steps:

1. The nearest power of 10 to 98 is 100.So we take our base value as 100.(Note: Base values should    always be powers of 10)
2. We can see clearly that 98 is 2 less than 100.So we call this 2 as deficiency.
3. Decrease the given number further by an amount equal to the deciency , meaning ( 98 -2 = 96).Now,this 96 becomes the left side of our answer.
4. On the right hand side, put the square of the difiency .i.e; 2 x 2= 04
5. Now for the final answer add(append), the results from Step 3 and Step 4.

Therefore , 98 x 98 =9604

Method 2 : Using Duplex Formula


Duplex Method Formulas:

D(X)= X^2
D(XY)=2 XY
D(XYZ)=(2XZ)+Y^2
D(ABCD)=(2AD) +(2BC)
D(ABCDE)=(2AE)+(2BD)+C^2

Following below are few examples :

\

3. Squaring any Three-digit Numbers :

Vedic Maths is a vast resource of mathematical knowledge.Try them and you get to like it.For further Vedic Maths ,check the Vedic Maths link below .

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Type 1 : Consider divisors more than one digit and the divisors are slightly greater than 10

Steps and Clues :

Follow the rules from left to right

1. Set the last digit of your dividend your main number as in red above and change the sign (-ve).
2. Carry the one forward and multiply with your main number as in step 2(pic)
3. Put you answer back into the table in step 1 (pic) and add to get your next answer.In our case it is -2 +2 =0
4. Repeat the steps for all other numbers as in step 2(pic).
5. Now the last digit becomes your remainder and the others becomes your quotient

As we go to for larger numbers following the same rules ,we can easily handle them and get our final result as soon as one minute.

Type 2 : The divisors is a three digit number 

Steps and Clues :

Follow the rules from left to right 

1. Set the last two digit of your dividend your main number as in red above and change their sign (-ve).
2. Carry the number 2 forward and multiply with your main number as in step 2(pic)
3. Put you answer back into the table in step 1 (pic) and add to get your second digit answer.In our case it is 5-4 = 1
4. With that answer from step 3(above) multiply again with the main number and put it in table (step 1 pic) and add them to get the results
5. Now the last two digit becomes your remainder and the others becomes your quotient

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As the name says ,this sutra has endings totaling to 10 (last totaling to 10).We have seen some in our first sutra By one more than the previous too.

1.Squaring Numbers Ending In 5

 85 ²

First two digits = the ten’s digit times one more than the ten’s digit.

Here ,the last two digits are always 25

8(8+1)  / 25

= 72/ 25
= 7525

2.Consecutive Decades 65 x 75

First two digits = the small ten’s digit times one more than the large ten’s digit.

Here, the last two digits are always 75

6(7+1)  / 75

= 48 / 75
= 4875

3.Ten’s Digits Both Even and ending in 5 :
25 x 65 

First two digits = the product of the ten’s digits plus ½ the sum of the ten’s digits.

Last two digits are always 25

2(6) + ½ (2+6) / 25

= (12 + 4) / 25
=16 /25
=1625

4.Ten’s Digits Both Odd and Ending in 5 :
15 x 75

First two digits = the product of the ten’s digits plus ½ the sum of the ten’s digits.

Last two digits are always 25

1(7) + ½ (1+7)  / 25

= (7+4) / 25
=12 / 25
= 1225

5.Ten’s Digits Odd&Even and Ending in 5:
 35 x 85

First two digits = the product of the ten’s digits plus ½ the sum of the ten’s digits. Always drop the remainder.

Last two digits are always 75

3(8) + ½ (3+8) / 75
=(24+5) / 75

= 29 / 75
= 2975

This can be extended further :


1.For same digits appearing on L.H.S and digits totaling 10 on R.H.S 
27 x 23 
becomes
= 2(2+1) / (7x3)
=6 / 21
=621


86 x 84
= 8 (8+1) / (6x4)
=8x9 / 24
=72 / 24
=7224


Similarly for any last two or three digit number totaling 100,1000 etc,we can apply the same sutra ,


884 x 816
=8 (84) x 8 (16)
= 8x9 / (84x16)
=72/1344 ..add zero to left hand side and carry over 1
= 720/1344 carry over the nearest digit ..i.e;'1'
=721344 

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This sutra is helpful in case of factorization of quadratic equation.

It is very difficult to factorise the long quadratic (2x²+ 6y² + 3z² + 7xy + 11yz + 7zx). But "Lopana-Sthapana" or By Alternative Elimination and Retention removes the difficulty

Example 1

Lets assume E =(2x² + 6y² + 3z² + 7xy + 11yz + 7zx)

By "Lopana-Sthapana" we eliminate z by putting z = 0.

Hence the given expression becomes ,

                   E = 2x² + 6v+ 7xy = (x+2y) (2x+3y)

Similarly, if y=0, then,

                   E = 2x²+ 3z² + 7zx = (x+3z) (2x+z)

We get , E = (x+2y+3z) (2x+3y+z)

Now,

Factorizing we get 

2x² + 2y²+ 5xy + 2x- 5y –12 = (x+3) (2x-4) and (2y+3) (y-4)

Hence Result , E = (x+2y+3) (2x+y-4)

* This "Lopana-sthapana" method (of alternate elimination and retention) will be found highly useful in HCF, in Solid Geometry and in Co-ordinate Geometry of the straight line, the Hyperbola, the conjugate Hyperbola, the Asymptotes etc.

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

Example 1:

98^{2 =?}

Our nearest base is 100 and 98 is 2 less than 100.So '2 ' is called the deficiency.Subtract it from 98, we get 96 our left part of answer.Next,square it (02 * 02) to get the right part of answer.

98 * 98 = (98 - 2) | (2 *2)

             =  96|04

Result 98²  = 9604

Example 2:

104²=?

Our nearest base is 100 and 104 is 4 more  than 100.So '4 ' is called the surplus .Add it to 104, we get 108 our left part of answer.Next,square it (04 * 04) to get the right part of answer.

104 * 104 =(  104+4) | (4*4)

                = 108 | 16

Result 104²=10816

Example 3:

1006 ²=?
Calculating mentally as fast as we can,
(1006 +6) | 6 *6
=1012036
Result : 1006 ²=1012036

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Generally we come across problems which can be solved by mere observation.But we follow the conventional procedure to get our results.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.


Example 1:
Solve : x + (1/x) =3/2


Conventional method :


x +1/x =5/2


x² +1        5
------- = -----
x               2
2 x²+2 =5x
2x²-5x+2=0
2x²-4x-x+2=0
2x(x-2)-(x-2)=0
(x-2)(2x-1)=0
or x=2 ,(1/2)


Now by Vilokanam or by mere observation :


x + 1      5
    --- = ----
     x       2    can be viewed as ,


x + 1              1
     ----= 2 +-----
      x              2   where we get our results straight away as x= 2 ,(1/2)




Example 2 : 

x + 5      x + 6      113
____ + _____ = ___
x + 6      x + 5      56

Now  by Vilokanam or by mere observation :


113      49 + 64          7           8
___    = _______ = ___ +     ___
56            7 x 8          8           7

x + 5         7                x+5      8
____    = __          or ____ = __
x + 6        8                 x+6      7  by splitting the R.H.S and L.H.S on both sides

8x + 40 = 7x + 42  and 7x + 35 = 8x + 48
Solving for x,
x = 42 - 40 = 2 -x = 48 – 35 = 13
x = 2 or x = -13.

Example 3:

Simultaneous Quadratic Equations:

Solve:    x + y = 9 and xy = 14.

By mere observation,

xy = 14 gives x = 2, y = 7 or x = 7, y = 2

These two sets satisfy x + y = 9 since 2 + 7 = 9 or 7 + 2 = 9. 

Hence the solution.

Similarly, 

Solve : 5x – y = 7 and xy = 6.

By mere observation,

xy = 6 gives x = 2, y = 3 or x = 3, y = 2

These two sets satisfy 5x - y = 7 since 5(2) -3 = 7 but not 5(3)- 2 = 12.

Hence the solution is x=2 and y =3

Try them ,you will be surprised of Vedic Maths ! 

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