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
Division By The Nikhilam Method III
But, what if the denominator is not just below a power of 10? What if the denominator is close to a power of 10, but just above it? In this lesson we will learn a method that is very similar to the Nikhilam method, but which is much better suited for division by denominators that are larger than a power of 10. This method follows from the Paravartya Sutra which says Paravartya Yojayet. Literally, this means Transpose And Adjust. Division is just one of the many varied applications of this sutra.
We will illustrate by taking a simple example such as 256/11. Notice that 11 is just above 10. Its 10's complement is 89, which is very large and will make division using the Nikhilam method quite cumbersome. So, we take a different tack to perform this division.
First we write the denominator, then write a modified 10's complement next to it. In this case, the 10's complement is made of negative numbers. Just as 1 is the 10's complement of 9, we can consider -1 to be the 10's complement of 11 with respect to 10. Similarly, 112 will have 2 digits in its 10's complement, -1 and -2. Another way to think about this is to consider 112 as 1x100 + 1x10 + 2x1. To find the 10's complement of this number with respect to 100, for instance, we take the transpose of the two coefficients following the 100's digit. That is the origin of the "transpose" in transpose and adjust.
Going back to 256/11, let us write it down as below:
The first line, as usual, consists of the denominator and in this case, its modified 10's complement. Notice that we have written the numerator with just one digit to the right of the "|". The rule in this method is to put as many digits to the right of the "|" as there are digits in the modified 10's complement of the denominator. Now, we put a zero below the first digit of the numerator, as before. Adding up the first digit of the numerator with zero gives us the first digit itself, 2 in this case. Multiplying 2 by the 10's complement, we get 2 x -1 = -2. Put -2 below the second digit of the numerator. In the figure below, we designate a negative number by bold red characters (I want to fit it within the space of 1 character, hence this compromise).
Now add the numbers under the second digit of the numerator to get 5 + -2 = 3. Multiply this sum by the 10's complement to get 3 x -1 = -3. Put this -3 under the next digit of the numerator to get the figure below:
Now add up the numbers under each digit of the numerator as before to get the intermediate quotient and remainder:
In this case, the remainder is less than the denominator, so this is the final quotient and remainder. We can verify that this indeed is the correct answer to the problem.
Let us deal with another example. This time we will tackle 256/12.
We wrote 12 and its modified 10's complement (-2) on the first line. We wrote the numerator with one digit to the right of the "|". We put a zero below the first digit of the numerator and added to get 2. Multiplying 2 by -2 gives us -4. Put -4 below the second digit of the numerator and add them to get 1. Multiplying 1 by -2 gives us -2 which went under the third digit of the numerator. Now, add up the digits under each column of the numerator to get the answer. 21 is the quotient and 4 is the remainder. This can be verified using a calculator.
Now, let us deal with a special case by tackling 256/13. We get the figure below after a few steps of the process:
We are confronted with the job of adding 5 and -6. Obviously this will give us a negative number (-1) as the answer. This is a perfectly valid outcome from the process. Multiplying -1 by -3 (the 10's complement) gives us 3, which goes under the 3rd digit of the numerator. We get the figure below:
Obviously, when we add up the digits under the columns of the numerator, we have to make the appropriate adjustments for the negative number that results under the second digit of the numerator. Using normal carryover and borrowing rules of subtraction, we arrive at the answer below:
It can be verified that this is indeed the correct answer to the problem.
Now, let us tackle a different kind of special case by trying to solve 251/11.
We notice that in this case, adding up the numbers under each digit of the numerator leads to a negative answer in the remainder part of the answer. We follow a modified borrowing methodology to deal with this predicament. Figure out how many 11's have to be added to the negative remainder to make it positive. For instance if the intermediate remainder is R (R is negative), the denominator is D, and R + n*D = F where F is positive. In this case n is 1 (since -2 + 1*11 = 9). Subtract n from the intermediate quotient and make F the final remainder. Applying this procedure to the above problem we get:
The procedure for extending this to 3-digit and higher denominators is quite straight-forward. Note that the modified 10's complement will consist of more than 1 digit in this case. But the procedure is the same as how we dealt with multiple 10's complements digits like in the previous lesson. To illustrate, we will deal with a few examples below.
112 -1 -2
The problem is to find the solution to 589569/112. Notice that the 10's complement of 112 is written as -1 -2 next to 112. The sum under the first digit of the numerator is 5. Multiplying 5 by the left digit of the 10's complement (-1), we get -5 which goes under the second digit of the numerator. Multiplying 5 by the right digit of the 10's complement (-2) gives us -10. Since this is 2 digits long, the units digit goes under the 3rd digit of the numerator and the tens digit goes under the 2nd digit of the numerator.
Next we deal with the second digit of the numerator. We have 8 -5 -1 = 2 as the sum under that digit. 2 x -1 = -2 and this goes under the 3rd digit of the numerator. 2 x -2 = -4 and this goes under the 4th digit of the numerator.
Next we deal with the third digit of the numerator. Adding up the numbers under that digit gives us 9 - 0 -2 = 7. 7 x -1 = -7, and this goes under the 4th digit of the numerator. 7 x -2 = -14 and this goes under the 5th digit of the numerator. Since this is the first digit of the numerator after the "|" and there is one more numerator digit to its right, we pad the -14 to -140.
Now, we come to the 4th digit of the numerator, the last digit we have to deal with. Adding up the numbers under this digit gives us 5 -4 -7 = -6. -6 x -1 = 6. This goes under the 5th digit of the numerator. Again, since that is the first digit to the right of the "|", and there is one more numerator digit to its right, we pad it with one zero to make it 60. Then we get -6 x -2 = 12, which goes under the 6th digit of the numerator.
Now, we add up the numbers under each of the columns of the numerator to get the final answer. It can be verified that what we got is indeed the correct answer.
A couple more examples are worked out below to familiarize ourselves with the method so that we can apply it to other problems without any confusion.
1023 0 -2 -3
1123 -1 -2 -3
As you can see from the examples above, the method is simple to apply but one has to take care to keep the signs of the sums and products correct, otherwise the final answer will not be correct. Abundant practice is the key to doing this correctly and quickly.
In a future chapter, we will deal with yet another way of doing division that does not rely on the 10's complement or the modified 10's complement being composed of small numbers. This will enable us to deal with denominators that are neither just below nor just above a power of 10. Until then, good luck, and happy computing!