VEDIC MATHS-34

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VEDIC MATHS

                           By OMKAR TENDOLKAR

Hello friends,

                      This is  post number 34 from the series of "Vedic maths" blogs. Here in this blog we will learn about "Example of Base method of multiplication Multiplying a number above the base with a number below the base".

The base method of multiplication is having large contribution in Vedic Mathematics. The name Base method is given by Vedic Mathematics in western countries. 

Actual Sanskrit sutra as given by Swamiji to define this system is:

‘Nikhilam Navatascaramam Dasatah.’ It means ‘all from 9 and the last from 10.’

For all practical purposes, we shall be calling the system elaborated in this chapter as the 'Nikhilam method' or simply the ‘Base Method'.

This method is used to multiply numbers. It is helpful in many  cases where traditional multiplication takes a long time to calculate the answer. Let us take the case of multiplying the number 9999999 by 9999991. 

If you go by the traditional method it will take a long time to multiply the numbers and calculate the product. However with the technique described in the Base Method one can find the answer in less than 5 seconds. The study of the Base Method is important to understand the other formulae of Vedic Mathematics.

There is a corollary of the Base Method which is called the Yavadunam Rule. This sutra is used in squaring numbers that we will discussed in the next post of our blogs.

Reference:

We had already learn about "Concept relating to Base" our previous blog. If you have missed my last blog then please visit "VEDIC MATHS-31".

We had already learn about "Base method of multiplication" our previous blog. If you have missed my last blog then please visit "VEDIC MATHS-32".

We had already learn about "Example of Base method of multiplication when the number of digits in RHS exceeds number of zeros in the Base" our previous blog. If you have missed my last blog then please visit "VEDIC MATHS-33"

We have dealt with examples where both the numbers are below a certain base or above it. We solved examples like 96 by 98 where both the numbers are below the base. We also solved examples like 1003 by 1050 where both the numbers are above the base. Now we will solve examples where one number is above a base and another number is below it.

                               N1    D1    B1
                   *          N2    D2    B2
             --------------------------------------
                 N1B2 + D2B1 | D1 * D2
                           or          
                 N2B1 + D1B2 |

N1 and N2  are the number to be multiplied D1 and D2 are deviation from base B1 and B2 are respective bases.

Examples:
  
1. Multiply 85 by 995

             8 5
       * 9 9 5
     --------------

Here, the number 85 is close to the base 100 and the number 995 is close to the base 1000

                               8 5   -1 5      1 0 0
                 *         9 9 5      -5   1 0 0 0 
(strikethrough common no of zeros in base)
                   -----------------------------------------
                      85*10 + -5*1 | -15 *-5
                                or        |                    
                  995*1 + -15*10 |
                                  8 4 5 | 7 5     
(number of digit in RHS is equal to common no of zeroes   in base)

Final answer: 
85 * 995 = 84575

2. Multiply 102 by 999

          1 0 2
       * 9 9 9
     ------------

Here, the number 102 is close to the base 100 and the number 999 is close to the base 1000

                            1 0 2       2      1 0 0
                   *       9 9 9      -1   1 0 0 0        
(strikethrough common no of zeros in base)
                  -----------------------------------------
                    102*10 - 1*1 | 2 * -1  
                             or         |
                   999*1 + 2*10 |         
                             1 0 1 9 | - 0 2                  

(number of digit in RHS is equal to common no of zeroes  in base)

                               1 0 1 8| 9 8                         

(-02 is viniculam number so take it's complement as 98 and subtract 1 from first  positive number while moving right to left )

Final answer: 
102 * 999 = 101898

Find multiplication of following:

  1. 111*1012=112332
  2. 999*10001=9990999
  3. 982*10078=9896596
The simplicity of this method can be vouched from examples given above.


You may try following example:

Find Multiplication of followings

1. 120 * 1020

2. 123 * 1003

3. 1020 * 10500

You may answer this in comment box. You may ask your any query or doubt in comment box. I will try to resolve as early as possible.

In next blog we will discuss about "Example of Base method of multiplication when base is not power of 10".     


Are you excited for this?...
Then, please wait for it.
I will post my new blog in next week.

We will meet very soon through our next  blog. Till that stay connected, stay healthy and stay safe.

Thanks

for giving your valuable time.

Good day😊.

Location: Kudal, Maharashtra, India

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