Vedic Math
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VEDIC MATHS

1.  Special method for multiplication by numbers from 11 to 19.

Multiplication by 11

Rule:   1. Prefix a zero to the multiplicand
          2. Write down the answer one figure at a time, from right to left as in  
              any multiplication.
                 The figures of the answer are obtained by adding to each successive 
              digit of the multiplicand. its right neighbour. Remember the right 
              neighbour is the right, (i.e., the correct) neighbour to be added.
                 (1) 123 X 11 = ?
Step1:
    Prefix a zero to the multiplicand so that it reads 0123.
Step2:
    To the right digit 3, add its right neighbour.
             There is no neighbour on the right; so add 0.  0123 X 11

             
3 + 0 =3.                                                     3
                                                        
                        3

             To the next digit 2, add its right neighbour 3.  0123 X 11
             2 + 3 = 5.                                                    53
             To the next digit 1, add the right neighbour 2. 0123 X 11
              1 + 2=3.                                                     353
             To 0, add the right neighbour 1.                     0123 X 11
             0 + 1 = 1.                                                  1353

Therefore, 123 X 11 = 1353 (which you can easily verify by a conventional multiplication).

MULTIPLICATION BY 12:

The method is exactly the same as in the case of 11 except that you double each number before adding the right neighbour.

(1)        13 X 12 =?

Step1:   Prefix a zero to the multiplicand so that it reads 013.
Step2:
   Double 3 and add the right neighbour                013 X 12
            neighbour;therefore  add0).                                6

            Double 1 and add the right neighbour 3.             013 X 12
    
        2 X 1 + 3 =5                                                   56

           Double 0 and add the right neighbour 1.               013 X 12
      
    2 X 0 + 1 =1                                                   156

Therefore, 13 X 12 = 156 (which you can again verify by a conventional multiplication).

 

MULTIPLICATION FROM 13 TO 19: 

The reason why the rule is different for multiplication by 11 and    by 12 is obviously because the right digits are different. The right digit, we could call the Parent Index Number (PIN). Thus in 11, the PIN is 1 and in 12 it is 2. (In 13, it is 3; in 14, it is 4 etc.) When the PIN is 1, we are simply taking each figure of the multiplicand (we could call this figure the Parent Figure – PF in short ) as such and adding the right neighbour. When the PIN is 2, we are doubling the PF and then adding the right neighbour.

Obviously, if the PIN is 3 (as in 3), we would treble the PF and then add the right neighbour. If the PIN is 4 (as in 14), we would quadruple (i.e. multiply by 4) the PF and then add the right neighbour. If PIN is 9 (as in 19), we would multiply the PF by 9 and then add the right neighbour, What is the advantage of the method? We need know the tables only upto 9 and still multiply by a simple process of addition.

(1)        39942 X 13 = ?                        (2) 43285 X 14 = ?
            2331                                           21132 
            039942 X 13                                 043285 X 14                
            519246                                        605990

(3)        58265 X 15 = ?                        (4) 36987 X 16 = ?
            
34132                                         25654
           
058265 X 15                                 036987 X 16
           
873975                                        591792 

(5)        69873 X 17 = ?                        (6) 96325 X 18 = ?
           
57652                                         85224
           
069873 X 17                                096325 X 18
            1187841                                      1733850

(7)        74125 X 19 = ?
         
  73124
         
  074125 X 19
         
  1408375

 

2.         Multiplication of two 2 digit numbers

Consider the conventional multiplication of two 2 digit numbers 12 and 23 
        shown below:

                                   
12
                               X    23       
                        
            36       
                                  
24
 
               Ans.            276                      

It is obvious from the above that

(1) the right digit 6 of the answer is the product obtained by the "vertical"  
     multiplication of the  right digit of the multiplicand and of the multiplier.

(2) the left digit 2 of the answer is the product obtained by the "vertical"  
     multiplication of the left digit of the multiplicand and of the multiplier

(3) the middle digit 7 of the answer is the sum of 3 and 4. The 3 is the product 
     of the left digit of the multiplicand and the right digit of the multiplier; the 4 
     is the product of the right digit of the multiplicand and the left digit of the 
     multiplier. This means that, to obtain the middle digit, one has to multiply 
     "across" and add the two products (in our example 1 X 3 + 2 X 2)

     The working in our above example can therefore be depicted as
                       
1 2
  
                      2 3
           
1 X 2 / 1 X 3 + 2 X 2 / 2 X 3

       and can be summarised as

                                    1 2
                                   2 3

                       
 2 / 3 + 4 / 6     = 276

 

3.   When the units figure is "one" in both the numbers being multiplied, the 
     process of  multiplication is simplified further. Consider the following 
     multiplication:      

                                            31
                                            21
                        
2 X 3 / 2 X 1 + 1 X 3 / 1 X 1

     You will notice that the middle digit of the answer is 2 X 1 + 1 X 3 i.e. 
     (2 + 3) X 1. So, instead of multiplying "across" for the middle term, you  
     could simply add the tens digit of the two numbers.
      Therefore, 31 X 21 = 6 / (2 + 3) / 1 = 651
      Similarly, in 81 X 91, you could obtain the middle term as 17,  by merely 
     adding 8 and 9.
4.
  Like wise, when the tens figure is "one" in both the numbers being multiplied, 
    you could obtain the middle term by simply adding the units digit of the two 
    numbers. For instance, the  middle term in 12 X 17 is 2 + 7, i.e. 9, in 8 X 12 
    it is 8 + 2 i.e. 10 etc.
5. If the units figure or the tens figure is the same in the two numbers, the 
    process of multiplication could be simplified as shown in the following 
    examples:

    (1)              8 3  The middle term is obtained by multiplying 3 by 17, 17 being he sum of 8 9 and 93
               8      3
           
9     3     and 93
             72519   = 7719

     (2)   2      8  In this case, the middle term is 2 multiplied by 11, 11 being the sum of 8 2      3    and 3
               42 22 4   = 644

6.   Multiplication of 2 three-digit numbers  

        Let us consider the multiplicand to be ABC and the multiplier to be DEF, as   
      shown below:

                            
A B C
                 
    X     D E F
                          Answer

1.    The extreme right digit of the answer is obtained (as before) by 
               vertical multiplication as                         C X F

        2.    The extreme left digit is also obtained (as before) by vertical  
              multiplication as A X B

       3.    The "middle" digits are obtained (as before) by multiplying across
              Progressing one step at a time to the left, the "middle" digits are 
              successively

       
                      B X F + C X E
                            
A X F + C X D + B X E
                                  A X E + B X D
            The process is set out in detail for following examples below .
     1.                                             1          2          3
                                                     4          5          6                             
           1 X 4 / 1 X 5 + 2 X 4 / 1 X 6 + 3 X 4 + 2 X 5 / 2 X 6 + 3 X 5 / 3 X 6
                                              = 4/13/ 28/ 27 / 18 = 56088

     2.                                             2          4          5
                                                   1          9          8                        
           2 X 1/ 2 X 9 + 4 X 1 / 2 X 8 + 5 X 1 + 4 X 9 / 4 X 8 + 5 X 9 / 5 X 8
                                              = 2/ 22/ 57 / 77/ 40 = 48510

7.         Multiplication of Numbers of Different Lengths

In the examples we saw above, both the multiplicand and the multiplier contained the same number of digits. But what if the two numbers were to contain a different number of digits; for instance, how would we multiply 286 and 78? Obviously we could prefix a zero to 78 (so that it becomes 078, a 3 digit number) and proceed as in any multiplication of two 3 digit numbers.
The following examples will clarify the procedure:

(1)        286 X 78 =?                                        (2)        998 X 98 =?
                2  8  6                                                              9  9  8
                0  7  8                                                              0  9  8
          0 1472 106 48  = 22308                                0 81 153 144 64  = 97804

8.      Special case where the units figures of the multiplicand and the multiplier together total 10 and  the other figures are the same.
Consider             (i)         23 X 27
                        (ii)        94 X 96
                       
(iii)       982 X 988

In all of them, the units figures together total 10 [3 + 7] in (i), 4 + 6 in (ii) and 2 + 8 in (iii)], also the other figures of the multiplicand and the multiplier are the same [2 in (i) 9 in (ii) and 98 in (iii)]

In such cases, a special method can be used. But before we see the new method, let us consider the multiplication of 23 by 27 by the method we had learnt. 23 X 27 would give us the answer as 621. Do you notice anything special about the answer? The right part is the product of the unit figures 3 and 7 of the 2 numbers and the left part 6 of the answer is the product of 2 (the tens figure) and the next higher number 3.

            This gives us the special rule:

(1)    To obtain the right part of the answer, multiply the units (i.e. of the 
        extreme right) digit of the 2 numbers

(2)    To obtain the left part of the answer, multiply the other (i.e. the left), 
        digit/s by one more than itself/themselves. For example, if the left digit/s 
        of the 2 numbers is/are

       3,        multiply 3 by     (3 + 1) i.e. by 4
               5,        multiply 5 by     6
               6,        multiply 6 by     7
              99,       multiply 99 by 100
               100,     multiply 100 by 101
               888,     multiply 888 by 889 etc

The only thing you have to be careful about is to ensure that the right part of the answer always has 2 digits. For example in 29 X 21, the left part of the answer is 2 (2 + 1) and the right part is 9 X 1 i.e. 9; however the right part will be written as 09 and not as 9 because the right part has to contain 2 digits. Therefore, 29 X 21 = 609.

9.         Special Method for Squaring numbers ending in 5

An obvious extension of the above method will be in finding the square of numbers ending in 5. Consider 15 X 15. Here the sum of the units digit of the 2 numbers is the same. Hence, the above method will apply. Infact it will apply in squaring any number ending in 5. Since in all these cases the right digit of the multiplicand and the multiplier is 5, the right part of the answer is always 25 and therefore, we can mechanically set down 25 as the right part of the answer without any calculation – all that is needed is to find out the left part of the answer and this done exactly as in the previous section.

152 = 225            252 = 625             8752 = 765625 89952 = 80910025

TO MULTIPLY ANY TWO NUMBERS A AND B CLOSE TO A POWER OF 10

a.    Take as base for the calculations that power of 10 which is nearest to the 
       numbers to be
multiplied
b.    Put the two numbers A and B above and below on the left hand side
c.    Subtract each of them from the base (nearest power of 10) and write 
      down the remainders r1and r2 on the right hand side either with a 
      connecting minus sign between A & r1 and B & r2 if the numbers A and B  
      are less than the closest power of 10. Otherwise, use a connecting plus 
      sign between the numbers and the remainders.
d.   The final answer will have two parts. One on the left hand side and the 
      other on the right hand    side. The right hand side is the multiplication of 
      the two remainders and the left hand side is  either the difference of A and 
      r2 or B and r1 if the numbers are less than the closest power of 10. 
     Otherwise, it is the sum of A and r2 or B and r

     Few examples, which make the procedure clear, are:

    e.q. 1. Multiplication of 9 and 7
     The closest base to the two numbers in this case is 10
     Therefore       9 - 1 (The remainder after subtracting the number from 10)
                        7 – 3 (The remainder after subtracting the number from 10)
The right hand side of the answer will be 1 X 3 = 3
The left hand side can be computed either by subtracting 3 from 9 or 1 from 7 which is 6. Therefore, the answer is 63.
e.q. 2. A more difficult example will be the product of 94 and 87
The closest base in this case will be 100
Therefore                     94 – 6
                                  87 – 13  
                                  81 . 78  à 8178
Here, 6 in the first row is the difference 100 and 94 and the 13 in the second row is the difference between 100 and 87. The right hand side of the answer is obtained by the multiplication of 6 and 13 which is 78 and the left hand side is obtained by the difference between either 87 and 6 or 94 and 13, both of which give the answer 81.

e.q. 3. Taking numbers which are greater than the closest power of 10

Find the product of 108 and 112
The closest base is 100 in this case as well
Therefore                     108 + 8
                                  112 + 12  
                                  120,   96  
The procedure is the same with only difference being that instead of subtracting the remainder of one number from the other number, we add in this case as the number were marginally larger than the nearest power of 10
e.q. 4. When the number of digits of the product of the remainders is greater than the power of
10 closest to the two numbers
e.g. Product 84 and 92
            84 – 16
            92 – 8  
            76. 128  à (76 + 1), 28 à 7728
As the product of 16 and 8 is 128 which is a three digit number as against 2 being the power of 10 in100, we carry forward the digits on the left more than 2 digits (in this case) and add to 76, the left hand side of the answer

e.q. 5, When one of the number is lesser than the closest power of 10 and the other greater than the closest power of 10
Product of 88 and 106
            88 – 12
            108 + 8  
            96 .  96           

The operation is similar, excepting that as the right hand side of the answer is obtained by the multiplication of  a positive and a negative number the answer has to be subtracted from 100 by reducing the left hand side number by 1
Multiplication of numbers which are not close to the nearest power of 10  
Let us take the case of multiplication of 41 by 43. Going by the earlier method we have the nearest power of 10 as 100 or 10. In the former case. The remainders are 59 and 57, multiplication of which will be as tedious as the multiplication of these two numbers. In the latter case. the remainders will be 31 and 33 which will be equally difficult Therefore, we need to look at an alternative method. In this case,  we can take 50, which is a sub multiple of 100 or a multiple of 10 and proceed

Method 1: Take 50 as the base which is half of 100

            41 – 9
            43 – 7 
           
34 .63 à Since 50 = 100/2. we divide the left hand side number also by 2 while retaining the right hand side. Therefore the answer will be 1763.

Method 2  
We can start off instead of using 50 as the base, we can use 40 as the base  
           41 + 1
           43 + 3  
           44,   3 à and since 40 is 4 times 10, we multiply 44 by 4 to yield by 176 and join the right hand side to yield 1763 – the same answer.

To Calculate the squares from 51 to 59:- There is one general formula 
(5x)2 = (25 + x)/x2
Where x is a digit in units place

Note: Slash here does not represent division, It is just to differentiate between the two value. x2 should be a two digit no. If it is a single digit no, make it a two digit no by placing a prefix 0 before x2  
e.g.  32 = 9,      Make it of two digits i.e. 09
Let us calculate the squares of 54
In this case our x is 4
(54)2 = (25 + 4)/42 = 29/16 = 2916 is our answer

TO CALCULATE THE PRODUCT OF NOs. (Ending with 5)  WHICH DIFFER BY 10

            e.g. 45 x 35
Step 1:   First of all write 75 in the last two places of the product 45 x 
             35 =    __75
Step2:   Then Multiply 4 with 3 and add the smallest of these to the product 
             i.e. (4 x 3 + 3) = 15
Step3:   Now Place these two digit before 75 to get our required product i.e. 
               45 x 35 = 1575

   

 

 
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