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Multiplication

Here, again addition is the key. However, it is advisable to know the basics of Vedic maths for multiplication.

Some of the methods are given below:

Base Method

In this method of multiplication, we use a number as a base, like 10, 50, 100 etc. We should try to assume the base that is closer to both the numbers.
 
Example-1
105 × 107
 
Solution
Both the numbers are close to 100, so let us assume 100 as the base. We will now find the deficit/surplus from the base.
 
Base = 100, Surplus = 5 and 7
 

Right part (after slash) is the product of the surplus. Since base = 100 and surplus are 5 and 7, so product would be 5 × 7 = 35.
 
Left part (before slash) could be either number plus surplus of the other multiplicand. Hence left part would be either (105 + 7) or (107 + 5) = 112 (both will always be the same) i.e., 112.
 
Left part would be the equivalent to Number × 100. In this case, 112 × 100 = 11200
 
Now we will add both the right part and the left part = 11200 + 35 = 11235
 
Hence, the result of the multiplication would be 11235.
 
 
Example-2
97 × 102
Solution

97 × 102
 
Base = 100, Deficit = 97 – 100 = – 3,
 
Surplus = 102 – 100 = 2
 
Right part will now be (–3) x 2 i.e., –06. To take care of the negative we will borrow 1 from the left part, which is equivalent to borrowing 100 (because we are borrowing from the hundred digits of the left part). Thus this part will be 100 – 06 = 94.
 
So, the answer = 9894
 

Place value Method

In this method of multiplication, every digit is assigned a place value and we do the multiplication by equating the place values of multiplicands with the place value of the product.
 
Example
Solution
Conventionally, the unit digit is assigned a place value of 0, tens place digit is assigned a place value of 1, hundreds place digit is assigned a place value of 2, thousands place digits is assigned a place value of 4 and so on.
 
 

Now, this multiplication is a two-step process:

Step 1

Add the place values of digits of the numbers given (1254 × 3321) to obtain the place value of the digits of the product.

 

For example, using the place values of the multiplicands i.e., using 0, 1, 2 and 3 of the number 1254 and the same place values 0, 1, 2 and 3 of the other multiplicand 3321, we can get 0 place value in the product in just one way i.e., adding 0 and 0.
 

Place value 1 in the product can be obtained in two ways:

 

 
Place value 2 can be obtained in three ways:
 

Place value 3 can be obtained in four ways:


Place value 4 can be obtained in three ways:


 
Place value 5 can be obtained in two ways:
 

 
Place value 6 can be obtained in one way:
 

And this is the maximum place value that can be obtained.

Step 2

Now multiply the corresponding numbers one by one.
 

 
And so on, we can find the product = 4164534
This method is most useful in case of multiplications of 2 digits × 2 digits or 2 digits × 3 digits or 3 digits × 3 digits multiplication.
 
Example
ab × cd
 
Solution

Similarly, we can have a proper mechanism for multiplication of 2 digits × 3 digits or 3 digits × 3 digits using the above given place value method.
 





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