Monday, November 16, 2020

Multiplication of numbers with a series of 9's

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 Multiplication of numbers with a series of 9's

     In my seminars , I often have an audience challenge round . In this round , the audience members ask me to perform various mental calculations and give them the correct answer . They generally ask me to multiply numbers which involve a lot of 9's in them . The general perception is that the higher the number of 9's the tougher it will be for me to calculate . However , the truth is exactly the opposite - the higher the number of 9's in the question , the easier it is for me to calculate the correct answer . I use two methods for this . The first method is given below and the second method is explained in the chapter ' Base Method of Multiplication ' .

       Using the method given below , we can multiply any given number with a series of nines . In other words , we can instantly multiply any number with 99 , 999,9999 , 99999 , etc.

        The technique is divided into three cases . In the first case , we will be multiplying a given number with an equal number of nines . In the second case we will be multiplying a number with a higher number of nines . In the third case , we will be multiplying a number with a lower number of nines , 

Case 1 ( Multiplying a number with an equal number of nines ) 

( Q ) Multiply 654 by 999

 • 654 

x 999

-------

=653 346 

• We subtract 1 from 654 and write half the answer as 653 . Answer at this stage is 653 . . 

• Now we will be dealing with 653. Subtract each of the digits six , five and three from nine and write them in the answer one by one ,

•  Nine minus six is 3. Nine minus five is 4. Nine minus three is 6 . 

• The answer already obtained was 653 and now we suffix to it the digits 3 , 4 and 6. The complete answer is 653346 

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( Q ) Multiply 9994 by 9999 

• 9094

    X9099

----------- 

=9993 0006

         We subtract one from 9994 and write it as 9993. This becomes our left half of the answer . Next , we . , 447 x 900 subtract each of the digits of 9993 from 9 and write the answer as 0006. This becomes the right half of the answer . The complete answer is 99930006

(Q) Multiply 456789 by 999999

•  156789

 x 99999

------------------

=456789 543211

 

       We subtract 1 from 456789 and get the answer 456788. We write this down on the left hand side . Next , we subtract each of the digits of 456788 ( left hand side ) from 9 and get 543211 which becomes the right hand part of our answer . The complete answer is 456788543211

More examples :

 •  7777

X 9999

----------

=77762223

 •  65432

X99999

-----------

=6543134568

 • 447

X999

----------

=446553

 • 90909

X99999

-----------

=9090809091

          The simplicity of this method can be vouched from the examples given above . Now we move toCase 2. In this case , we will multiply a given number with a higher number of nines . 

 Case 2 

( Multiplying a number with a higher number of nines ) 

 ( Q ) Multiply 45 with 999 

  45             045X999=044955

X 999

 --------

  There are three nines in the multiplier . However , the multiplicand 45 has only two digits . So we add a zero and convert 45 to 045 and make it a three digit number . After having done so , we can carry on with the procedure explained in Case 1 

  . First we subtract 1 from 045 and write it down as 044. Next , we subtract each of the digits of 044 from 9 and write the answer as 955. The complete answer is 044955 ,

 ( Q ) Multiply 888 with 9999

888             0888X9999 = 8879112

X9999

      We convert 888 to 0888 and make the digits equal to the number of nines in the multiplier . Next , we subtract 1 from 0888 and write the answer as 0887. Finally , we subtract each digit of 0887 from 9 and write the answer as 9112. The final answer is 08879112 which is 8879112 

 ( Q ) Multiply  123 by 99999    

  123              00123 X99999 =

x 99999          00122 / 99877

The multiplicand is a three - digit number and the multiplier is a five - digit number . Therefore  we

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    add two zeros in the multiplicand so that the digits are equal in the multiplicand and the multiplier . 

       We now subtract 1 from 00123 and write the left hand part of the answer as 00122. Next , we subtract each of the digits of the left hand part of the answer from 9 and write it down as 99877 as the right hand part of the answer . The complete answer is 12299877 

Other examples : 

•  162

       X9999

-----------

0161/9838

 •  5555

X99999

-----------

05554/94445

•  363

X999999

-------------

000362/999637

   We can see that this technique is not only simple and easy to follow , but it also enables one to calculate the answer in the mind itself . This is the uniqueness of these systems . As you read the chapters of this book , you will realize how simple and easy it is to find the answer to virtually any problem of mathematics that one encounters in daily life and especially in the exams . And the approach is so different from the traditional methods of calculation that it makes the whole process enjoyable

 Case 3 of this technique deals with multiplying a number with a lower number of nines . There is a separate technique for this in Vedic Mathematics and requires the knowledge of the Nikhilam Sutra ( explained later in this book ) . However , at this point of time , we can solve such problems using our normal practices of instant multiplication .

 ( Q ) Multiply 654 by 99 : In this case the number of digits are more than the number of nines in the multiplier . Instead of multiplying the number 654 with 99 we will multiply it with ( 100-1 ) . First we will multiply 654 with 100 and then we will subtract from it 654 multiplied by 

1.

654

X 99

     

       65400 

-       654

--------------

      = 64746 

 ( Q ) Multiply 80020 by 999

 We will multiply 80020 with ( 1000 - 1 ) . 

   80020000 

-       800203

---------------

   79939980

    This method is so obvious that it needs no further elaboration .

 

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