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Vedic Math Techniques in Multiplication

There are several techniques of multiplication. We will discuss them one by one.

Method 1: Base Method

In this method, one number is used as a base; for example, 10, 50, 100 etc. The number that is closer to both the numbers should be taken as the base.
 
Example-1
105 × 107
Solution
In this case, both the numbers are close to 100, so 100 is taken as the base. We will now find the deficit/surplus from the base.
 
Base = 100, Surplus = 5 and 7
Description: vm-1.tif 
 
The right part (after slash) ⇒ this is the product of the surplus. Since the base = 100 and the surpluses are 5 and 7, the product would be 5 × 7 = 35.
 
The left part (before slash) ⇒ It could be either of the numbers plus the surplus of the other multiplicand. Hence, the left part would be either (105 + 7) or (107 + 5) = 112 (both will always be the same) i.e., 112.
 
The left part would be equivalent to the number × 100. In this case, 112 × 100 = 11200
 
Now, we add both the right part and the left part = 11200 + 35 = 11235
 
Hence, the result of the multiplication would be 11235.
 

Example-2
108 × 104
Solution
Description: vm-2.tif
 
 
Example-3
111 × 112
Solution
Description: vm-3.tif
 
Here, it is 11 × 12 = 132. But it can have only two digits. Thus, 1 will be carried over to the left part and the right part will be only 32. Left part will be either 111 + 12 + 1 (1 for the carry over) or (112 + 11 + 1), i.e., 124. So, the result will be 12432.
 
For 102 × 104, the answer will be 10608. Please note that the right part will be 08 and not simply 8.
 
 
Example-4
97 × 95
Solution
Description: vm-4.tif
 
Base=100, Deficit = 97 − 100 = 3 and 95–100 =5
 
 
Example-5
97 × 102
Solution
Description: vm-5.tif
 
97 × 102
Base = 100, Deficit = 97 – 100 = −3,
Surplus = 102 – 100 = 2
 
The right part will now be (−3) × 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 = 98 94
 
 
Example-6
62 × 63
Solution
Description: vm-6.tif
 
We will assume here the base as 50 owing to the fact that the numbers are close to 50.
 
Base = 50, Surplus = 62 – 50 = 12,
Surplus = 63 – 50 = 13
 
The left hand side = 156 and the right hand side = 75. Since the base is assumed to be equal to 50, so the value of the right hand side = 75 × 50 = 3750. Besides, only two digits can be there on the right hand side, so 1(100) is transfered to the left hand side leaving 56 only on the left hand side.
 
So, the value on the right hand side = 3750 + 100 = 3850
 
Value on the left hand side = 56
 
Net value = 3850 + 56 = 3906
 
Let us do the same multiplication by assuming 60 as the base.
Description: vm-7.tif

Base = 60, Surplus = 62 – 60 = 2, Surplus = 63 – 60 = 3
 
Since the base is assumed to be equal to 60, the value of the right hand side = 65 × 60 = 130 × 30 = 3900
 
So, net value = 3906
 

Method 2: Place Value Method

In this method of multiplication, every digit is assigned a place value and the multiplication is done by equating the place values of multiplicands with the place value of the product.
 
Let us see this with some examples:
Description: vm-8.tif 

 

Conventionally, the unit digit is assigned a place value 0, the tens place digit is assigned a place value 1, the hundreds place digit is assigned a place value 2, the thousands place digits is assigned a place value 3 and so on.

This multiplication is a two-step process.
 
Step 1 Add the place values of the 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 another multiplicand 3321, we can get 0 place value in the product in just one way, i.e., adding 0 and 0.
Description: vm-9.tif 

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

 
Description: vm-10.tif 
 
Place value 2 can be obtained in three ways.
 
Description: vm-11.tif 

Place value 3 can be obtained in four ways.

 
Description: vm-12.tif 
 
Place value 4 can be obtained in three ways.
 
Description: vm-13.tif 
 
Place value 5 can be obtained in two ways.

 
Description: vm-14.tif 

Place value 6 can be obtained in one way.

 
Description: vm-15.tif 
 

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

Step 2 Multiply the corresponding numbers one by one.
 
Description: vm-17.tif
Description: vm-16.tif

In this manner, we can find the product = 4164534

This method is most useful in case of the multiplications of 2 digits × 2 digits or 2 digits × 3 digits or 3 digits × 3 digits multiplication.
 
E.g., ab × cd
Description: vm-18.tif

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

Method 3: Units Digit Method

This method of multiplication uses the sum of the units digit, provided all the other digits on the left hand side of the unit digit are the same.
 
Example-7
75 × 75
Solution
Description: vm-19.tif
 
The sum of the units digit = 10, so we add 1.0 in one of the digits on the left hand side.
 

Example-8
62 × 63
Solution
Description: vm-20.tif
 
The sum of the units digit = 5, so we add 0.5 in one of the digits on the left hand side.
 





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