VEDIC MATHS-33

-

 

VEDIC MATHS

                           By OMKAR TENDOLKAR

Hello friends,

                      This is  post number 33 from the series of "Vedic maths" blogs. Here in this blog we will learn about "Example of Base method of multiplication when the number of digits in RHS exceeds number of zeros in 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 have studied the four steps used in the base method of multiplication. Step 2 says that the RHS should be filled by as many digits as the number of zeros in the base. In some cases, the answer on the RHS was small and so we filled the remaining places by inserting zeros. An opposite situation may also arise when the number of digits on the RHS is more than the number of zeros in the base. In that case, we will have to carry over as we do in normal multiplication. Have a look at the following examples:

Examples:
  
1. Multiply 950 by 950

         9 5 0
       * 9 5 0
     ------------

STEP1: Find the Base and the Difference

Here in this example, both the number are closer to 1000 and we know that as a base we can take only power of 10's thus we will take 1000 as a base.
In this example, the difference between 1000 and 950 is 50.
After completing step 1, the question look as given below:

                   1000
             9 5 0  -  5 0
         *  9 5 0  -  5 0
         ----------------------


STEP 2: Number of digit on the RHS = Number of zeroes in the base

In the example, base 1000 has three zeros. Hence, the RHS will be filled in by three-digit number.

                   1000
             9 5 0  -  5 0
         *  9 5 0  -  5 0
         ----------------------
                       | _ _ _


STEP 3: Multiply the difference on The RHS

In this example, we multiply the difference , viz.. -50 by -50 and get the answer as 2500. Note that the RHS can be filled by a three-digit answer only but we have a four-digit number, namely 2500. We carry over the extra digit 2 to the LHS in next step.
                   1000
             9 5 0  -  5 0
         *  9 5 0  -  5 0
           ----------------------
                        | 5 0 0

STEP 4:  Put the cross answer in the LHS

In given example, the cross answer can be obtained by doing (950 - 50). in either case the answer will be 950. This 950 we will put on the LHS. We carry over extra digit2 to LHS  and add it to the number 900 and make it 902. But we already had 500 on the RHS.

Thus out multiplication process is complete.

                   1000
              9 5 0  -  5 0
          *  9 5 0  -  5 0
           ----------------------
              9 0 2 | 5 0 0

Hence complete answer is 902500.


3. Multiply 150 by 140

                    100
             1 5 0  +  5 0
          *  1 4 0  +  4 0
         ----------------------
               2 1 0 |  0 0

The base is 100 and the difference is 50 and 40 respectively. 
The number of digits on the RHS will be 2 Vertical multiplication is 2000, a four-digit number But the RHS can only be filled in with two digits. 
We carry over the two excess digits 20 to the LHS and add it to 190.The answer is 210. The LHS is 210 and the RHS is 00.

The final answer is 21000.

Examples:

Find multiplication of following:

  1. 112*100=11200
  2. 1300*1020=1326000
  3. 9200*9200=8464000
The simplicity of this method can be vouched from examples given above.


You may try following example:

Find Multiplication of followings

1. 1200 * 1020

2. 1230 * 1003

3. 10200 * 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 Multiplying a number above the base with a number below the base".     


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😊.

Comments

Post a Comment

Popular posts from this blog

VEDIC MATHEMATICS

-
Image
  VEDIC MATHS                             By OMKAR TENDOLKAR Hello friends,                        I am Omkar Tendolkar from Kudal. I am studying in S.Y.B.Sc at SRM College, Kudal. I always enjoy sharing my knowledge. This is one of my attempts.            Some people find mathematics challenging to understand, while others struggle with calculations. Certainly, the comprehension of the subject depend on IQ (intelligent quotient), but equally crucial is one's perspective towards it. Because mastering calculation doesn't solely rely on a sharpe IQ; it also demands some basic tricks that can greatly assist in handling complex calculations.                  Starting today, I want to launch a "VEDIC MATHS" blog. Here, you will find a series of blog's explaining simple tricks that will...
Read more

VEDIC MATHS-1

-
Image
VEDIC MATHS                             By OMKAR TENDOLKAR Hello friends,                          This is our first blog in the series of 'Vedic Maths' blogs. Here, we will learn about the basic principles required in Vedic Maths. This blog will cover all the fundamental concepts necessary for understanding our next few blogs.   Base: It is the first number in the series. For a two-digit number: 10. Starting with 10, 100, 1000, 10000, etc. Complement: . Adding two numbers results in the next nearest base. 48 + 52 = 100. Then, We can say that 48 is the complement of 52 or 52 is the complement of 48. How to find: Subtract each digit from 9 and the last digit from 10. If... The last digit is 0, then take the second-to-last digit as the last one. For example: 60 is the complement of 40. All these examples, ranging from 2 digits to 6 digits, w...
Read more

VEDIC MATHS-15

-
Image
  VEDIC MATHS                             By OMKAR TENDOLKA R Hello friends,                       This is our 15th blog from the series of "Vedic maths" blogs. Here in this blog we will learn about   "Criss-Cross System of Multiplication " . Criss-Cross System of Multiplication :                In Vedic Mathematics , we have similar system as like traditional way of multiplication it helps us to get the answer much faster. This system is also universal system and can be used for any combination of numbers of any length.                      This system of Multiplication is given by the " Urdhva-Tiryak Sutra. " It means ' vertically and cross-wise '. The application of this system are manifold, but in this blog we shall confine our study only to it's utility in...
Read more