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Vedic Math - Multiplication of numbers with a series of 9’s

Another special case of multiplication is, multiplication with numbers like 9, or 99, or 999, or 9999.....so on. It feels like if multiplier is a big number, the calculation will be tough. But, with the help of vedic math formulae, the multiplication is much easier for all '9' digits multiplier. By using the method given below, we can multiply any number with 99,999,9999, etc. very quickly.

Please note that the methods or the vedic formulae, that we use in this calculation, are "By one less than the one before"  and "All from 9 and the last from 10".

There are three cases for the multiplication of numbers with a series of 9's.
  • Case 1: Multiplying a number with a multiplier having equal number of 9’s digits                                              (like 587 x 999)
  • Case 2: Multiplying a number with a multiplier having more number of 9’s digits                                             (like 4678 x 999999)
  • Case 3: Multiplying a number with a multiplier having lesser number of 9’s digits                                             (like 1628 x 99)

The method to solve 'Case 1' and 'Case 2' is the same, but for 'Case 3', the method is different. Let us start with 'Case 1'.

Case 1: Multiplying a number with a multiplier having equal number of 9’s digits

Multiply 587 by 999

           587
       x  999
       ------------
        586 413
       
Solution is,
  •  Let us first do the calculation by conventional method to understand the solution. Result will be 586413.
  • Split the answer in two parts i.e. '586' and '413'.
  • Let's see the first part of the result, i.e. 586. It is reduced by 1 from the number being multiplied i.e. 587 - 1 = 586. {Vedic sutra "By one less than the one before"}
  • Now see the last part, i.e. 413. Subtract the multiplicand i.e. 587 from 1000 (multiplier + 1). Vedic Sutra applied here is "All from 9 and the last from 10", and hence we substract 587 from 1000. So the outcome will be (9 -5 = 4, 9 - 8 = 1, 10 - 7 = 3) , and result is 413. Refer to image below for more clarity:

Case 2: Multiplying a number with a multiplier having more number of 9’s digits
Multiply 4678 by 999999

           4678
     x   999999
   ----------------
    4677 995322   

Here, 4678 is 4-digit number and 999999 is 6-digit number. So, we can rewrite 4678 as 004678 to make 6-digit number. Now, it turns out to be same as case 1. So, same method is going to be applied.

The result has two parts:

  • The first part of the result, i.e. 4677, which is reduced by 1 from the number being multiplied      (4678 - 1). {Vedic sutra "By one less than the one before"}
  • For the last part of answer (995322), Subtract the multiplicand i.e. 004678 from 1000000   (1000000 - 004678). Here we are applying the vedic formula "All from 9 and the last from 10" on 004678. Calculation is like:
  •  9 - 0  =9
  •  9 - 0  =9
  •  9 - 4  =5
  •  9 - 6  =3
  •  9 - 7  =2
  • 10 - 8 =2
  • So the result is 995322
  • Now combine first and last part, and you get the final answer i.e. '4677995322'

Case 3: Multiplying a number with a multiplier having lesser number of 9’s digits

Multiply 1628 by 99

For this case, method is changed. In this, we rewrite 1628 x 99 as 1628 x (100 - 1). So,

                     1628(100 - 1)
                   = 162800 - 1628
                   = 161172

Here, calculation might get slow during substraction. But once you learn the vedic subtraction, it will become fast. In coming sessions, we shall discuss it.

OR  there is another approach given below, by spliting the solution in three parts.

                1628 x 99 = 16 : 11 : 72

Now, see how we get this result using Vedic Math:

                        16     : 28    : 100
                                 : 17    :   28
                      -----------------------  
                        16     : 11     :  72



                                                                                          
Steps to follow are given below:
  • Split the multiplicand in such a way that the right hand side number contains the digits equal to the number of digits in multiplier. Like here, multiplier is 99 (having 2 digits). So after splitting the multiplicand, the right hand side number will be 28. Now, there are two parts of multiplicand after split i.e. 16 and 28.
  • Create three different virtual columns, and place two parts of splited number in first and second column.
  • In third column, put (multiplier + 1) i.e. 100 in this case
  • Now add '1' to first part of multiplicand, and place it in middle column of second row. Number of digits should be equal to number existed in first row in same column. If it is not, shift the extra digits in first column of second row.
  • Place second part of multiplicand (28, from first row) in third column.
  • Now subtract numbers in second row from first row numbers, in their respective columns. In case of middle column, if first row number is lesser than second row number, we can borrow '1' from first column value.
  • If we follow this, the answer for each column will be
  • Third Column: 100 - 28 = 72
  • Second Column: 28 -17 = 11
  • First Column: 16 - 0 = 16
  • So the answer is, '16 11 72'


Example where number in middle column of first row is lesser than second row value (63 x 9):
                6    : 3    :  _
                _    : 7    :  3
               -----------------
                5    : 6    : 7

Here, 3 is less than 7 and subtraction is not possible. Hence we borrow '1' from first column value i.e. borrow 1 from 6 and 3 becomes 13 and left part is now 5.

One more example to make it more clear: 17119 x 99

                       171   : 19   : _ _
                           1   : 72   :  19
                    ----------------------   
                       169   : 47   : 81


Here, don't get confuse by seeing '1' in the first column of second row. It is just taken from '172' (which is one more than '171'). As per the method, middle column should have number having digits equal to multiplier. So we shift '1' to first column. Rest all is same.

Hope you understandand this method. If you have any question, please post in comments. In the next session, we shall discuss about multiplication of numbers with a series of 1's.

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