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
Division By The Paravartya Method
However, unfortunately, not all divisions we encounter have denominators that are conveniently close to powers of 10. Fortunately, there are other methods we can use to overcome this limitation of the Nikhilam and Paravartya methods. The method we will learn in this lesson is derived from a sutra which reads Dhvajanka. Literally, this means On Top Of The Flag. This method is also referred to as straight division.
Before we dive into the method let us first review the basics of division by simple single-digit denominators. This review will serve as a foundation for the division method we will move on to.
Let us consider 25/2 for instance. Let us write the problem out as below:
Now, we divide the first digit of the numerator, 2, by the denominator to get 2/2 = 1. Let us put down 1 as the first digit of the quotient. The reminder of the division of 2 by 2 is zero. Let us put this zero just before the 5 as in the figure below:
Now, divide the two-digit number obtained by placing the remainder and the next digit of the numerator together (in this case, 05), by the denominator, 2. This results in the second digit of the quotient being 2 with a remainder of 1. This is shown below:
•| 0 1
Since there are no more digits in the numerator, we are left with the final answer. We can write it by moving the last remainder to the answer line as below:
•| 0 1
This gives us a quotient of 12 and a remainder of 1 (which is obviously the correct answer).
We can also choose to proceed further by considering the numerator to be 25.0000... instead of just 25 with no digits after the decimal point. If we do, then, we have to follow the same procedure as before with the new remainder we have found. The two-digit number we get by placing the remainder next to the next digit of the numerator is 10 (place the remainder 1 before 0 in the numerator after the decimal point). Since the remainder has been attached to the first digit of the numerator beyond the decimal point, we also put a decimal point in the answer line. Then, we do the division by 2 to 10/2 = 5 as the next digit of the quotient. The remainder is zero. This is illustrated below:
•| 0 1
Since the remaining numbers are all uniformly zero, we know that there will be no further addition to the answer line except zeroes (and since these zeroes will be after the decimal point, they do not change the final answer). That completes this division.
To tackle a slightly more interesting case, let us consider 85/7. The figure below illustrates what happens when we stop with a quotient and remainder:
•| 1 1
The quotient of 12 and remainder of 1 can be verified to be correct. When we proceed with the division instead of stopping with a remainder we get the figure below:
7|8 5.0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
•| 1 1 3 2 6 4 5 1 3
•|1 2.1 4 2 8 5 7 1 4 . . .
We see that since the remainders repeat themselves, we know that the answer digits will also repeat themselves. Instead of continuing the calculations, we stop at that point.
These examples form the basic foundation on which straight division using the Dhvajanka sutra rests. To illustrate straight division, let us now take the example of 562/41. First we separate the denominator into two parts, the base and the flag digits. Usually, the base consists of just the left-most digit of the denominator. Sometimes, it helps to make more digits part of the base if division by the base will remain easy. The rest of the digits become flag digits (hence the name of the sutra: On Top Of The Flag).
The numerator is also divided into two parts. A quotient part and a remainder part. The remainder part consists of as many digits as there are on the flag. In this particular problem, we will consider 4 to be the base digit of the denominator and hoist 1 on top of the flag. Similarly, we will consider 56 to be the quotient part of the numerator and 2 to be the remainder digit. We now get the figure below:
We begin by trying to divide the first digit of the numerator by the base digit of the denominator. We get 5/4, which has a quotient of 1 and a remainder of 1. We write them as below:
4 |5 6|2
The difference between the single-digit division we reviewed earlier and straight division comes at this step: Instead of trying to divide the 2-digit number obtained by putting the remainder and the next digit of the numerator together (16 in this case), we do something slightly different. We take the 16 we obtain by putting the remainder digit together with the next digit of the numerator and subtract from it the product of the previous quotient digit and the flag digit. That product is 1 x 1 = 1 in this case. The subtraction gives us 15. This now becomes the next dividend for division by the base digit of the denominator. 15/4 is 3 with a remainder of 3. This is written in the figure below:
4 |5 6|2
••| 1 3
We now repeat the same procedure with this remainder digit as we did in the previous step. We take the 2-digit number we get by putting it together with the next digit of the numerator (32 in this case), and subtract from it the product of the previous quotient digit by the flag digit (3 x 1 = 3, in this case). We get 29. Since this 29 is obtained after taking into account the remainder digit of the numerator, we can stop here with a remainder of 29 if we want to, giving us the answer as below:
4 |5 6|2
••| 1 3
Thus, we find the answer to our problem to be a quotient of 13 with a remainder of 29. This can be verified to be true. The advantage with straight division using the Dhvajanka sutra is that we don't have to stop with a remainder. We can proceed as far as we want by finding digits beyond the decimal point using the same procedure as we followed above. This is illustrated below:
4 |5 6|2.0 0 0 0 0 0 0 0 0
••| 1 3 1 3 2 1 3 1 3 2 1
••|1 3.7 0 7 3 1 7 0 7 3 . . .
We see that the digits start repeating, so there is no point in continuing the division further. The general method can be summarized in the following steps:
- Split the denominator into one or more base digits and one or more flag digits (in most cases, it makes the most sense to have one base digit and all the rest of the digits raised on to the flag)
- Split the numerator into quotient digits and remainder digits. The number of remainder digits has to be the same as the number of flag digits
- Now perform division of the first digit of the numerator by the base digit of the denominator (if this is not possible, take the next digit of the numerator also to perform the division)
- Put the quotient on the answer line and put the remainder just before the next digit of the numerator in the remainder line
- Take the 2-digit line formed by the combination of this remainder digit with the next digit of the numerator and subtract from it the product of the flag digit(s) and the last quotient digit (on the answer line). Call this sum the intermediate numerator
- Perform division of this intermediate numerator by the base digit of the denominator and put the quotient on the answer line as the next digit of the quotient. Put the remainder just before the next digit of the numerator in the remainder line
- Now, perform the process in the above 2 steps over and over again until you have taken into account the first of the remainder digits of the numerator
- At this point, one can stop, and consider the number on the answer line to be the quotient, and the intermediate numerator to be the final remainder of the division problem
- If one wants to find a decimal answer to the problem, add a decimal point to the answer line as soon as one of the remainder digits of the numerator is involved in the calculation of the intermediate numerator
- Then, add a decimal point to the numerator, pad it with zeroes to the right and continue this process until either the intermediate numerators become zero, or the quotient digits and intermediate numerators start repeating
That is all there is to this method! There is no multiplication by multi-digit denominators, there is no complicated multi-digit subtractions, and most importantly, no trial-and-error in figuring out the next quotient digit. If you can do simple division by single-digit denominators, you can master this method and perform division by much higher denominators.
Let us now deal with a special case that one might encounter when using this method. At the step where one subtracts the product of the quotient digit and the flag digit from the two-digit number formed by the remainder and the next digit of the numerator, you may get a negative number as the intermediate numerator. We haven't made any provisions for such a contingency in the procedure above, so we need a method to deal with it if the situation arises.
The solution to this problem is actually quite simple: we simply have to remember that even though we are used to thinking that only one quotient and remainder exist for a given division problem, actually an infinite number of them exist. Take the case of 9/2 for instance. The traditional answer to this problem is a quotient of 4 and remainder of 1. However, technically, once can consider a quotient of 3 and a remainder of 3 to be a perfectly valid answer, as are a quotient of 2 and remainder of 5, and so on.
So, the solution to our special case is simply to reduce the quotient by 1 to get a higher remainder which will then make the negative answer go away. Let us illustrate this by finding the answer to 819/49. After following the procedure through the first few steps, we get the figure below:
4 |8 1|9
Now we find the 2-digit number formed by the remainder and the next digit of the numerator to be 01. When we try to subtract the product of the quotient digit and the flag digit (2 x 9 = 18) from 01, we get -17, which threatens to throw a wrench into the works here! As explained above though, the way out of this predicament is to consider the answer to the first division to be a quotient of 1 and remainder of 4 rather than a quotient of 2 and remainder of 0. When we do that, the answer flows naturally based on the steps outlined in the procedure:
4 |8 1|9
••| 4 0
Now, we find the 2-digit number to be 41, and it is easy to subtract 1 x 9 = 9 from 41. We get 32 as our intermediate numerator. At this step, if we consider 8 to be our quotient and 0 to be our remainder from the division of 32 by the flag digit, 4, we encounter the same problem as before. Our 2-digit number becomes 09, and the product of 8 and 9 is much higher. So, let us try the same method as before and reduce the quotient to 7 with a remainder of 4. We get the figure below:
4 |8 1|9
••| 4 4
But, we find that this small concession is not actually sufficient to get over the problem yet. We find our 2-digit number to be 49 and our product of the quotient digit with the flag digit to be 63. To get over the problem, we need to make a bigger concession. We finally solve the problem by considering the result of the second division to be a quotient of 6 with a remainder of 8. We illustrate the final solution below:
4 |8 1|9
••| 4 8
We can verify that a quotient of 16 and a remainder of 35 is indeed the correct (and most widely accepted!) solution to our problem.
In future lessons, we will learn how to extend this method to perform divisions by denominators that have more than just 2 digits. In the meantime, hope you will find time to practice this method so that you can do it rapidly without errors. Practice makes perfect, so good luck, and happy computing!