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

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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: