You can find all the previous posts about below:
Introduction to Vedic Mathematics
A Spectacular Illustration of Vedic Mathematics
Multiplication Part 1
Multiplication Part 2
Multiplication Part 3
Multiplication Part 4
Multiplication Part 5
Multiplication Special Case 1
Multiplication Special Case 2
Multiplication Special Case 3
Vertically And Crosswise I
Vertically And Crosswise II
Squaring, Cubing, Etc.
Division By The Nikhilam Method I
Division By The Nikhilam Method II
We have already seen the beginnings of how the process works in the previous two lessons. The key to the process is in the step where we multiplied the sum under each column by the 10's complement of the denominator. To illustrate the method further, we will take a few examples, starting with a simple case.
Let us work out 123/8. As before, our figure starts out looking like the one below:
As always, the first line consists of the denominator followed by its 10's complement (2 is 8's 10's complement). The numerator has been divided by a "|" such that there are as many digits to the right of the "|" as there are digits in the denominator. We then put a zero under the first digit of the numerator.
Now add up the digits in that column of the numerator to get a sum of 1 (1 + 0 = 1). Multiply it by the 10's complement to get 2 (1 x 2 = 2). Put that under the second digit of the numerator. The figure now looks as below:
The sum of the digits under the second column is 4. Multiplying this by the 10's complement gives us 8. Put the 8 under the third digit of the numerator, right of the "|". Now we add up the numbers under the columns to get the figure below:
Note that there is no carryover from the right of the "|" to the left of it. Following the rules on how to deal with a remainder greater than the denominator, we divide the remainder by the denominator and add the new quotient to the original quotient and retain the new remainder as the final remainder. Our final figure looks like this:
It is easy to verify that this is indeed the right answer to the problem.
Now, let us work out a larger problem such as 894378/7. We get the figure below:
On the first line, we have 7 and its 10's complement, 3. Then we have the numerator, with one digit behind the "|". We have a zero below the first digit of the numerator. The sum of those is 8 and the product of this sum with the 10's complement is 24. We have followed the usual rules of carryover by putting the 4 under the second digit of the numerator and the 2 under the first digit. Our next sum becomes 33 (24 + 9 = 33). Multiplying that by 3 gives us 99. Once again, we put the first 9 under the third digit of the numerator and the second 9 under the 2nd digit of the numerator to give us the figure below:
Now, we move on to the third digit of the numerator. The sum we have is now 99 + 4 = 103. Multiplying 103 by the 10's complement, we get 309. Now, we put the 9 below the 4th digit of the numerator, the 0 below the 3rd digit and the 3 below the 2nd digit of the numerator. We get the figure below:
Under the 4th digit of the numerator, we now have 309 + 3 = 312. Multiplying 312 by 3 gives us 936. We write it as below following normal carryover rules:
Coming to the last digit of the numerator before the "|", we get 936 + 7 = 943. Multiplying 943 by 3 gives us 2829. This leads to the figure below and the intermediate answer as below:
Deriving the final answer from this intermediate answer, we get:
To illustrate division by bigger denominators, let us start with 1123/88.
We start out by writing the problem as below:
As always, 88 and its 10's complement, 12, are on the first line. We have written the numerator on the next line, separating the last two digits behind the "|" because the denominator contains 2 digits. Now we write a 0 below the first digit of the numerator. We then add up the numbers under that digit, giving us 1. Now we have to multiply this number by the digits of the 10's complement. The product of 1 with the left digit of the 10's complement is 1 (1 x 1 = 1), and this goes under the second digit of the numerator. The product of 1 with the right digit of the 10's complement is 2 (1 x 2 = 2) and this goes under the 3rd digit of the numerator, to the right of the "|". The figure below reflects this:
Now add up the digits under the second digit of the numerator to get 1 + 1 = 2. We now multiply this sum by the digits of the 10's complement. The product of 2 with the left digit of the 10's complement is 2 (2 x 1 = 2), and this goes under the 3rd digit of the numerator. The product of 2 with the right digit of the 10's complement is 4 (2 x 2 = 4), and this goes under the 4th digit of the numerator. Since we have now dealt with all the digits of the numerator to the left of the "|", we are done. Add up the numbers under the columns to get the intermediate answer as below:
Since the intermediate remainder is less than the denominator, the intermediate answer above is also the final answer, and this can be verified using a calculator.
Let us try a couple more problems to convince ourselves that the method indeed works. We will start with 49857/79.
Note that 4 + 0 = 4, and the product of 4 with 2 and 1 gives rise to the 8 and 4 under the 2nd and 3rd digits of the numerator. Now, 9 + 8 = 17. The product of 17 with 2 and 1 gives rise to 34 under the 3rd digit of the numerator and 17 under the 4th digit. Since the 4th digit is the first digit of the numerator beyond the "|", and there is one more digit of the numerator to its right, we write the 17 as 170, with the addition of a zero to account for this extra digit. We get the figure below:
Since the 3 in the last row, from the 34 obtained as the product of 17 with 2, has never been dealt with in the problem before (and it is placed under the 2nd digit of the numerator, which we have already finished dealing with), we have to include it when we find the sum under the 3rd digit of the numerator. This gives us the sum of 46 (8 + 4 + 34). Multiplying 46 by 2 and 1 gives us 92 and 46. They are written as below (note the zero after the 92 to account for the fact that the 92 goes under the 4th digit of the numerator and there is one other digit of the numerator to the right of the "|"):
We then divide 1193 by 79 to get a final remainder of 8 and a new quotient of 15. Adding the original quotient to the new quotient gives us a final answer as below:
Let us now extend this method further by working out a few more problems. Just remember some simple rules and the rest becomes very easy:
- Write the 10's complement with as many digits as the denominator, padding to the left with zeroes as necessary
- Put a zero under the first digit of the numerator
- Add them (you get the first digit of the numerator once again), and multiply the sum by the individual digits of the 10's complement
- The product of the left-most digit of the 10's complement with the sum goes under the next digit of the numerator, while the product of the digit to its right with the sum goes under the next digit to the right and so on
- If some number from the product has not been dealt with when finding the sum of digits under a particular digit of the numerator (because of carryover as in the previous example), then it has to be dealt with when finding the sum under the next digit of the numerator
- There is no carryover from the right of the "|" to the left
- If a number goes under a digit to the right of the "|", pad it with as many zeroes to the right as there are digits in the numerator to its right
- Stop when you have dealt with the sum of digits under the last digit of the numerator to the left of the "|"
- Add up the digits to the right of the "|" with no carryover to the left
- Normal carryover rules apply when the carryover is entirely to the left of or right of the "|"
- If the intermediate remainder is larger than the denominator, perform division of this remainder by the denominator once again (this process is recursive, so remember this rule when you divide the remainder by the denominator!)
- The new quotient is added to the intermediate quotient to get the final quotient
- The new remainder is the final remainder
Let us now apply these rules to a few more problems that look difficult, but actually turn out to be quite easy once we start tackling them.
As you can see from the examples worked out above, this method works very well when the denominator is made up of large digits so that its 10's complement contains small numbers (mostly 0's, 1's etc.). The method becomes more cumbersome when the products are larger because of large numbers in the 10's complement. We will deal with such problems using other methods in future lessons.
This method is valuable though precisely because it is very useful when the denominator is composed of large numbers. It is in such division problems that most people face problems because trial and error multiplication becomes more difficult when the denominator consists of large numbers. Moreover, such problems also involve more difficult subtractions, especially when the numerator contains small numbers for its digits. Since the method illustrated here consists of very simple multiplications (mostly single digit by single digit, though occasionally one may need to multiply larger numbers) and additions, this method is ideal for such difficult division problems. Even though the method may seem complicated at first glance, it is very easy to master with practice. Good luck, and happy computing!