Vedic Maths Shortcut Techniques for Multiplication - Lesson 4

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Friends, here is the 4th lesson of our Vedic Math Multiplication shortcut techniques series (you can read remaining lessons from here). This method is called as "Eka Nyunena Purvena". That means "By one less than the previous one". This method works if one of the given numbers consists of all 9s. Lets try to understand this method with an example.

Example 1 : 68 x 99 = ?

Method 1 :

• Put four blanks in the result

Step 1 :

1. The number preceding 99 = 68
2. Complement of 68 = 32
3. Put this complement in the two blanks on RHS of the result

Present Status :

Step 2 :

1. The number preceding 99 = 68
2. A number that is less by 1 than the given number = 68-1 = 67
3. Put this 67 in the two blanks on LHS of the result

So 68 x 99 = 6732

Method 2 : 99 x 68 = ?

1. One of the given numbers is 99. It is having all 9s
2. The second number can be anything.
3. Put both the numbers as shown above.
4. Put 4 blanks for the result.
5. Reduce the second number by 1. It becomes 67 (68-1). Put this 67 exactly below the number 99 as shown below.
   
6. Subtract the digit 6 from the 9 above. 9-6 = 3
7. Put this 3 in the blank that belongs to 1.'s place of the result
• Present Status : 
  
8. Subtract the digit 7 from the 9 above. 9 - 7 = 2
9. Put this 2 in the blank that belongs to units place of the result
   

Result = 99x68 = 6732

Lets see another example.

Example 2 : 53 x 99 = ?

We shall solve this using the method 1.

Step 1 : 

1. The number preceding 99 = 53
2. Complement of 53 = 47
3. Put this complement in the two blanks on RHS of the result
   

Step 2 :

1. The number preceding 99 = 53
2. A number that is less by 1 than the given number is = 53-1 = 52
3. Put this 52 in the two blanks on LHS of the result
   

So, the answer is 53 x 99 = 5247

Note : There are two 9s' in the multiplier 99 in the given examples. Other number that is being multiplied is also of two digits. Hence the result may be of maximum of four digits. Hence we use four blanks for the result.

If the numbers are of three digits each, and one of them is 999, then we have to have 6 blanks for the result.

Example 3 : 123 x 999 = ?

Step 1 :

1. The number preceding 999 = 123
2. Complement of 123 = 877
3. Put this complement in the three blanks on RHS of the result
   

Step 2 :

1. The number preceding 999 = 123
2. A number that is less by 1 than the given number = 123-1 = 122
3. Put this 122 in the three blanks on LHS of the result
   

Example 4 : 15064 x 99999 = ?

Step 1 :

1. The number preceding 99999 = 15064
2. Complement of 15064 = 84936
3. Put this complement in the five blanks on RHS of the result

Step 2 :

1. The number preceding 99999 = 15064
2. A number that is less by 1 thank the given number = 15064-1 = 15063
3. Put this 15063 in the five blanks on LHS of the result.

15063 | 84936

So, 15064 x 99999 = 1506384936

Method 2 : 99999 x 15064 = ?

1. One of the given numbers is 99999. It is having all 9's
2. The second number can be anything.
3. Put both the numbers as shown above.
4. Put ten blanks for the result
5. Reduce the second number by 1. It becomes 15063 (15064-1). Put this 15063 exactly below the number 99999 as shown below.
   
6. Subtract each digit of 15063 from 99999
• 9-1 = 8
• 9-5 = 4
• 9-0 = 9
• 9-6 = 3
• 9-3 = 6

Put these digits in the remaining blanks (RHS of the result) from left to right

Result : 99999 x 15064 = 1506384936 

That's all for now friends. In our next post we shall discuss another useful method of Vedic Maths. Happy Reading :)

About The Author                                             

Rupali Shete (M.Sc., M.Phil., Ph.D) is a gold medalist from Osmania University (Hyderabad), one of the working members of Rananujan Mathematics Academy and Institute of Scientific Research on Vedas (ISERVE), Currently working in University of Pune.

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