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Tuesday, September 1, 2009

Vedic Mathematics Lesson 15: Division By The Nikhilam Method I

I have been slacking off in my learning of Vedic Mathematics due to a confluence of various factors. The method I was trying to learn was a little complicated and the book I was learning it from did not have a good explanation of how and why the method works. I went off and started working on other things while pondering the method to find a good way to explain it to my kids. The basic point of my learning Vedic Mathematics was to teach it to my kids, so I had to make sure I understood the method well enough to make it easy for my kids to understand when I taught it to them.

Then, we got a steady stream of guests at home. First came my friend and his family from San Diego. Then, it was my brother-in-law and his family. This last weekend, I was visited by a cousin and his wife. It now looks like my hosting days are over for the next couple of weeks at least, so I decided I would take advantage of the down-time to get this method squared away and documented.

You can find all the previous posts about Vedic Mathematics below:


Introduction to Vedic Mathematics
A Spectacular Illustration of Vedic Mathematics
10's Complements
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.
Subtraction

In this lesson, we are going to explore the process of division. In particular, the method, as the name indicates, is based on finding and using the 10's complement of the number in the denominator (remember that the Nikhilam sutra goes as follows: Nikhilam Navatascaramam Dasatah, literally meaning All From 9 And The Last From 10).

To illustrate the use of this method, we will first perform some divisions by 9. Then we will extend the method to other denominators in future lessons. Remember that the 10's complement of 9 is 1. Let us first try to find the answer to the division problem of 33/9. To perform this division, write down the problem as below:

9 1
3|3

What we have done is write the denominator on the top line along with its 10's complement and the numerator on the next line. Next, we have divided the numerator into two parts with "|". The part to the right of | contains the same number of digits as the denominator (1 in this case) and the part to the left contains all the other digits of the numerator.

Now, put a zero below the first digit of the numerator. Take the sum of zero and this first digit (this will obviously be the first digit itself).

9 1
3|3
0
-----

Now, multiply this sum by the 10's complement of the denominator (1 in this case) and put it under the next digit of the numerator. So, we multiply 3 by 1 to get 3. Since the next digit of the numerator is after the "|", put down a similar "|" in the third row and write the 3 to the right of it.

9 1
3|3
0|3
-----

We then add up the numbers in the columns (ignore the row with the denominator and its 10's complement during this addition) and put them in the answer row. Now, in the answer line, we have two parts. We have carried down the dividing "|" from the rows above to the answer row also. We get the figure below:

9 1
3|3
0|3
------
3|6

It should be easy to recognize that the 3 to the left of the "|" is the quotient and the 6 to the right is the remainder. The fact of the matter is that that is all there is to division of any number by 9. Let us go ahead and test this with larger numbers to make sure!

Let us take the case of 1232/9. Let us start with the figure below:

9 1
123|2

Now, put down 0 under the first digit of the numerator. We get:

9 1
123|2
0
--------

Now, multiply the sum of 0 and 1 (this is just 1) the 10's complement and put it under the next digit of the numerator. We get:

9 1
123|2
01
-------

We get 2 + 1 = 3. Take the 3 and multiply it by the 10's complement of 9 to get 3. Put that under the next digit of the numerator. We get:

9 1
123|2
013
-------

Adding 3 to 3, we get 6. Multiply 6 by the 10's complement (1), we get 6. Put this under the next digit of the numerator. Since the next digit of the numerator is to the right of the "|", put a "|" in the third row before putting the 6 to its right. We get the following figure.

9 1
123|2
013|6
-------

Now, add up the columns to get the figure below:

9 1
123|2
013|6
-------
136|8

It is easy to verify with a calculator that the answer to our division problem is indeed a quotient of 136 and a remainder of 8.

I admit that it is hard to believe that division can be this easy. After all, we did not perform any trial and error multiplication. We did absolutely no subtraction. We did a little multiplication, but in the case of 9, that multiplication turned out to be trivial because the 10's complement of 9 is 1. We then did a little basic addition. That is all there is to division using the Nikhilam method!

Let us now deal with some special cases before we proceed to apply this method to more complicated problems. Consider the example below:

9 1
123|4
013|6
-------
136|10

In the first step, we put down 0 under the first digit. Then we multiplied the sum of 0 and the first digit by the 10's complement of 9 (1), and added it to the second digit. The procedure is the same as before all the way until the very end. But at the very end, we find that the sum to the right of the "|" is a 2-digit sum that can not really be the remainder because it is larger than the denominator.

The special procedure to deal with this kind of situation is to then perform division of the remainder by the same denominator to get a new quotient and remainder. Then add the new quotient to our already obtained quotient. The new remainder from this secondary division becomes the remainder to the original problem. We easily see that the quotient of dividing 10 by 9 is 1 and the remainder is 1. Adding the new quotient of 1 to the already obtained 136 results in a final quotient of 137 and a final remainder of 1. This can be verified to be the correct answer to the problem.

The second special case we will deal with is illustrated below:

9 1
47|6
04|11
-------
51|17

Note that in this case, when we added 4 with 7 following the regular procedure, we get a 2-digit sum (4 + 7 = 11). Since that sum needs to be added to the next digit of the numerator and this next digit is beyond the "|", we don't perform the normal carryover we are used to. Instead, the 11 is added directly to the 6 already to the right of the "|" to get 17 as the intermediate remainder to the problem. To the left of the "|", we follow the normal rules of addition (including carryover from right to left), to get 51.

Dividing 17 by 9 leads to a quotient of 1 and a remainder of 8. So, the final answer to our problem becomes a quotient of 51 + 1 = 52, and a remainder of 8. This can be verified to be the correct answer to the problem.

Let us now see how to deal with one more special case. For this, let us try to find the answer to 4975831/9. Let us first draw the figure below:

9 1
497583|1
0

Adding 0 to 4 gives us 4. Multiplying 4 by the 10's complement of 9 (1), gives us 4 which is then added to 9, the next digit in the numerator. This is shown in the figure below:

9 1
497583|1
04

Adding 9 to 4 though gives us a 2-digit answer (9 + 4 = 13). To accommodate this, we don't write 13 fully under the 7, instead we write the right-most digit (3) under the 7 and the remaining digits under the digits to the left of the 7 as below (this is similar to normal carryover rules when we encounter more than one digit where only one digit will fit):

9 1
497583|1
043
01

The 1 in the 4th row and the 3 in the 3rd row are from the sum of 9 and 4. Now, we add 13 to 7 to get 20. This again is a 2-digit number. The procedure is similar to what we did before, and we write it as below:

9 1
497583|1
0430
012

The 2 in the 4th row and the 0 at the right extreme of the 3rd row are the result of the 20 that resulted from adding 13 to 7. Now add 20 (after multiplying it by the 10's complement of 9) to 5 to get 25. Write it down as below:

9 1
497583|1
04305
0122

The second 2 in the 4th row and the 5 in the third row are obtained from the 25 that resulted from adding 20 to 5. The next step must be familiar by now: Add 25 and 8 (we omitted mentioning multiplying the 25 by the 10's complement of 9 because it always results in the same number, but this becomes important when we perform division by numbers other than 9, whose 10's complements are not just 1) to get 33. Write it down as below:

9 1
497583|1
043053
01223

Now, we add 33 to 3 to get 36. Since 36 is to be added to the 1 that is to the right of the "|", it is not subject to the normal carryover rules we have used so far. It is to be added to 1 as is to result in an intermediate remainder of 37. The interim quotient is obtained by adding up the numbers in the columns using regular addition rules. We get the figure below:

9 1
497583|1
043053|36
01223
-------------
552866|37

Dividing 37 by 9 gives us a quotient of 4 and remainder of 1. Adding this 4 to our intermediate quotient gives us a final quotient of 552870 and a final remainder of 1. This can be verified to be the correct answer to this problem.

9 1
497583|1
043053|36
01223
-------------
552866|37
552870|01

A couple more problems are worked out below to further illustrate this technique. Where necessary, the final quotient and remainder are written right below the intermediate quotient and remainder, as in the problem above.

9 1
347038|5
037447|25
00111
------------
385595|30
385598|3

9 1
9437583|6
0936386|39
011223
--------------
10486199|45
10486204|0

Note that since the intermediate remainder is divisible by 9 without a remainder, the final remainder becomes 0.

In the next lesson, we will deal with denominators other than 9. First we will extend the method to work with numbers such as 99, 999, etc., whose 10's complements are 1. Then, we will further extend the method to denominators whose 10's complements are not 1. In the meantime, hope you will find the time to practice so that you will be ready to move on without confusion! Good luck!!

7 comments:

Vedmath said...

Hi...nice article....just want to point out that the answer to the sum 4975831 / 9 is actually 552870 - with a remainder of 1. You have wrongly given the answer as 542870. - by Vedmath

Blogannath said...

I made a mistake in the addition and got 542870 instead of 552870 for the quotient, but I do mention that the final reminder is 1, not 0. Thank you for pointing out the error. I have corrected the lesson so that it reads correctly now.

Maverick said...

Can you kindly explain me 14525 divide by 98 by this method...seems to be not workin for me.

Blogannath said...

This lesson deals specifically only with division by 9 using the Nikhilam method. I suggest you read the next two lessons in which I talk about dividing by series of 9's, and with denominators that are close to a power of 10, but not a series of 9's. If you still have problems with your division, please post back a comment in the next lesson or the lesson after that with the nature of your problem, and I will be happy to help you out.

chandrakant said...

nice article, then show how can we divide by 8?

srikanth yadav said...

Hi,

Could any one please help me in solving 10015/89 & 10015/98 in the same nihalam method. I would be very glad if i get any response.

Blogannath said...

As explained in the last paragraph of this blog post, the method for extending the Nikhilam method to denominators other than 9 is explained in the next two posts: http://blogannath.blogspot.com/2009/09/vedic-mathematics-lesson-16-division-by.html, and http://blogannath.blogspot.com/2009/09/vedic-mathematics-lesson-17-division-by.html

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