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Thursday, September 3, 2009

Vedic Mathematics Lesson 16: Division By The Nikhilam Method II

In the previous lesson, we went over the basics of division using the Nikhilam Sutra. We dealt with examples that involve division by 9. In this lesson, we will continue our exploration by taking on problems where the denominator consists of several 9's strung together. We are talking about denominators such as 99, 999, etc.

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
Division By The Nikhilam Method I

Notice that numbers such as 99, 999, etc., all have 1 as their 10's complement. To use the technique illustrated in this lesson, we will consider the 10's complement of 99 to be 01, that of 999 to be 001, etc. The reason for this will become apparent when we explain the method and work out a few problems. Remember that, as in the previous lesson, the divisions in this lesson are going to be performed with trivial multiplications, basic additions, and no subtractions!

Let us start with a simple example. Let us try to find the quotient and remainder of the simple division problem, 800/99. First we write the problem as below:

99 01
8|00

We have written the denominator and its 10's complement (including the leading zero) on the first line. Then we have written the numerator on the second line. Since the denominator contains 2 digits, we have separated the right-most two digits of the numerator from the rest of the numerator using a "|".

As in the previous lesson, the first step is to write a 0 under the first digit of the numerator. Then add the zero to the first digit. In our case, we get 8.

99 01
8|00
0

Now, multiply the 8 by each digit of the 10's complement, and write the answers under the next digits of the numerator. In our case, we have to multiply 8 by 01. We first multiply 8 by 1 and get 8. Leave a gap in the third row for inserting the product of 8 and 0, and write the 8 we got as the product of 8 and 1 under the next digit of the numerator. We therefore get:

99 01
8|00
0|8

The "" in the third row shows the gap we have reserved for the product of 8 by the left digit of the 10's complement. That product is 0 (0 x 8 = 0). Now fill that gap with this product. We get the figure as below:

99 01
8|00
0|08
---------

Make sure the third row contains the "|" at the appropriate location. Now add up the 2nd and 3rd rows as in the previous lesson to get the figure below:

99 01
8|00
0|08
---------
8|08

It is easy to see that the last row contains the quotient of 8 before the "|" and the remainder of 8 after the "|".

Let us now expand this method to tackle larger numerators. Along the way, we will deal with and explain some special cases that may arise. First, let us try to find 12332/99. First we write the numbers as below:

99 01
123|32
0
----------

We have put a zero down below the first digit of the numerator. We add the first digit of the numerator to the zero to get 1. Now multiply 1 by each digit of the 10's complement and put the answers down under the appropriate columns of the numerator. Remember that the product of 1 with the left digit of the 10's complement goes to the left of the product of 1 with the right digit of the 10's complement. We then get the figure below:

99 01
123|32
001
----------

The product of 1 with 1 (the right digit of the 10's complement) went under the 3rd digit of the numerator, while the product of 1 with 0 (the left digit of the 10's complement) went under the 2nd digit of the numerator.

Now, add up the numbers under the second column of the figure above (obviously not including the first row which contains the denominator and its 10's complement). We get 2 + 0 = 2. Now, multiply this number by each digit of the 10's complement and put the answers under the appropriate digits of the numerator (product of 2 with the left digit of the 10's complement goes to the left of the product of 2 with right digit of the 10's complement). We get the figure below:

99 01
123|32
001|2
000|
----------

Notice that we have added another row to the figure. The product of 2 with 1 (the right digit of the 10's complement) went under the 4th digit of the numerator. Since it is beyond the "|", we draw a "|" in the third row and put the 2 to its right. The product of 2 with 0 (the left digit of the 10's complement) should go under the 3rd digit of the numerator (so that it is to the left of the product of 2 with the right digit of the 10's complement). Since there is already a number under the 3rd digit of the numerator in the previous row, we decided to add an extra row to the figure to hold this zero. This is a good idea in general when we are practicing, but since the number is zero and does not make a difference in addition, we can choose to ignore it and not create the extra row.

Now, there is one more important rule to follow in this case. The product of the right digit of the 10's complement with 2 went under the first digit of the numerator that is to the right of the "|". The rule to remember is that when this happens, pad that number with as many zeroes to the right as there are additional numerator digits to the right of the column where we put our product. In our case, the numerator has 1 additional digit beyond the "|", so we need to pad the 2 with 1 zero to abide by this rule. The figure now becomes:

99 01
123|32
001|20
000|
-----------

Now, add the numbers under the next digit of the numerator. We have already dealt with the second digit of the numerator, so we move on to the 3rd digit of the numerator. The numbers under that column are 3, 1 and 0. the sum of them is 4. Now multiply this sum by each of the digits of the 10's complement and put them under the 4th and 5th digits of the numerator. Make sure the product with the right digit of the 10's complement goes under the 5th digit and the product with the left digit goes under the 4th digit so that they maintain their relative positions to the right and left of each other. We get the figure below:

99 01
123|32
001|20
000|04
-------------

We have dealt with all the digits of the numerator that are to the left of the "|". That signals an end to the process of adding and multiplying by the 10's complement. All we have to do now is add up the digits under each column to find the answer. We get:

99 01
123|32
001|20
000|04
----------
124|56

It is easy to verify with a calculator that we have indeed ended up with the correct answer to the problem we started with: a quotient of 124 and a remainder of 56. Simple enough, isn't it?

Let us get a little more adventurous and explore a problem like 390879/99. We first start with the figure below:

99 01
3908|79
0
------------

We get the sum of the first digit of the numerator and 0 as the first digit of the numerator itself (3). We multiply this 3 by the 10's complement digit by digit to get the figure below:

99 01
3908|79
003
-----------

The sum of the second digit of the numerator with the number under it is 9. We repeat the procedure of multiplying this sum by the digits of the 10's complement and putting the answers in the appropriate columns. In the figure below, we have omitted writing the zero that results from the product of 9 with the left digit of the 10's complement under the 3rd digit of the numerator because it makes no difference to the sum of that column.

99 01
3908|79
0039
-----------

Now, we move to the 3rd digit of the numerator. The sum of numbers in that column is 3. Multiply this by the digits of the 10's complement and put them under the appropriate columns of the numerator. Note that in this case, the product of 3 with the right digit of the 10's complement goes under the first digit of the numerator to the right of the "|", so make sure it is padded with the appropriate number of zeroes. We get the figure below:

99 01
3908|79
0039|30
-----------

Once again, in the figure above, we have omitted the zero that should go below the 4th digit of the numerator (resulting from the product of 3 with the left digit of the 10's complement) because it does not make a material difference to the outcome. Now, add up the numbers under the 4th digit of the numerator to get 17. We now multiply this number by each digit of the 10's complement. Multiplying 17 by 0 (the left digit of the 10's complement) presents no problems since the answer is zero. Multiplying 17 by 1 (which is the right digit of the 10's complement) creates a problem because the answer is 2 digits long and therefore, will not fit under a single digit of the numerator. This answer is supposed to go under the 6th digit of the numerator. Since both the 5th and 6th digits of the numerator are to the right of the "|", normal carryover rules apply, and we just end up putting the right-most digit of the answer under the 6th digit of the numerator, and the left digit of the answer under the 5th digit of the numerator. We get the figure below:

99 01
3908|79
0039|30
0000|17
------------

Now add up the numbers in the columns to get the intermediate answer below:

99 01
3908|79
0039|30
0000|17
------------
3947|126

Obviously, this an intermediate answer because the remainder is larger than the denominator. As in the previous lesson, we resolve this by dividing the intermediate remainder by the denominator and add the resulting quotient to the intermediate quotient to get the final quotient. The resulting remainder becomes the final remainder to the problem. Performing this operation, we get the following final answer:

99 01
3908|79
0039|30
0000|17
------------
3947|126
3948|27

It is easy to verify that this is indeed the correct answer to the problem.

Let us work out a couple more problems with 99 as the denominator before we move on to larger denominators such as 999. We will first work out 797883/99 to illustrate the technique further. The following figure should be self-explanatory since we have just followed the steps explained at length above to derive it.

99 01
7978|83
0079|140 -----> See the notes below
0000|17
-----------
8057|240
8059|42

Note that the sum of 7 and 7 under the 3rd column of the numerator is 14. When 14 is multiplied by the right digit of the 10's complement (1), we get 14, which is supposed to go under the 5th digit of the numerator (first digit of the numerator beyond the "|"). Because of the rule we explained earlier, this 14 has to be padded with one zero to account for the fact that there is one more digit of the numerator after the "|", to the right of this 5th digit (note that the 14 is not subject to normal carryover rules because it is supposed to go under the first digit of the numerator to the right of the "|"). The intermediate remainder thus becomes 83 + 140 + 17, which is 240. We then divide 240 by 99 to get an adjustment to the intermediate quotient and the final remainder. One can verify that the final answer above is indeed correct.

Next, let us work out 9798474379/99.

99 01
97984743|79
00978634|80
00011110|07
----------------
98974487|166
98974488|67

The figure above should be self-explanatory. Notice that the third digit of the numerator, we encounter a 2-digit sum (9 + 9 = 18). Multiplying 18 by the right digit of the 10's complement gives a 2-digit product, 18. This is written with the 8 under the 5th digit of the numerator and the 1 under the 4th digit of the numerator (normal carryover to the left applies since the numbers are all to one side of the "|"). Now, when we add up the numbers under the 4th digit of the numerator for the next step, the sum becomes 8 + 7 + 1 = 16. Don't forget the carryover digit from the previous step when you are adding up the numbers under a particular column. The 16 that results from multiplying this sum by the right digit of the 10's complement then goes under two columns because of the carryover (the 6 under the 6th column and the 1 under the 5th column). The sum of the 5th column then includes this carryover 1.

When we get to the 7th digit of the numerator, the sum of digits under it is 8. Multiplying this by the right digit of the 10's complement results in 8 which needs to go under the 9th digit of the numerator. Since this is the first digit of the numerator to the right of the "|", we pad the 8 with one zero (to account for the fact that there is one more digit in the numerator beyond the "|").

Extending this method to denominators with more 9's should be quite easy to work out by now. Make sure that you write out the 10's complement with leading zeroes and multiply the sum by each digit of the 10's complement one by one. Let us illustrate by taking some examples.

999 001
1234133|321
0001235|300
0000000|060
0000000|008
----------------
1235368|689

Notice that the sum of the 1 and 0 in the first column of the numerator gives 1. Multiplying 1 by each digit of the 10's complement (001) produces the 1 under the 4th column of the numerator and the zeroes under the 2nd and 3rd columns. Similarly, under the second column of the numerator we have a 2 and a 0, whose sum is 2. Multiplying 2 by each digit of the 10's complement produces zeroes under the 3rd and 4th columns (which need not be written out) and a 2 under the 5th column.

When we come to the 5th column of the numerator, we find a 1 and a 2 in that column. Their sum is 3. Multiplying 3 by each digit of the 10's complement produces a 3 which needs to go under the 8th column of the numerator. This is the first digit of the numerator to the right of the "|", so we write the 3 there and then pad it with 2 zeroes because the numerator has 2 more digits to the right of the "|". Similarly, when we deal with the 6th column of the numerator, we obtain a sum of 6 which results in a product of 6 that needs to go under the 9th digit of the numerator. That 6 is then padded with a zero to the right because there is one more digit in the numerator to the right of the "|" beyond the 9th digit. Once this rule is remembered and applied, the problems become quite easy to work out correctly.

Extending this method to deal with denominators like 9999 is quite trivial, given how we have worked out several examples using 99 as a denominator and at least one problem involving 999 as the denominator. We will proceed to working with denominators that contain digits other than 9 in the next lesson. Until then, practice is the key to success in using this method, so good luck, and happy computing!

6 comments:

Anonymous said...

I have a query.if you see the division by 9 in your last post where you have shown the example of 497853/9 there in 2nd step the the total comes out to be 13 in which 3 comes under 7 while 1 under 9 but in the next step you do 13+7 to get 20 and then nxt digit is added to 20
In division by 99 this is not the case we take the sum of the digits which are in the column and we donot take the total.why is it so.I am little confused can you please clarify more on this

Blogannath said...

The reason for that is that in the previous lesson, when we wrote 13 with the left digit under the 9 and 4, we had already taken the sum of that column, and we did not take this 1 into account during the summing. So, we have to take that into account when summing the next column by considering it to be 13. In this lesson, if we get a sum such that one digit goes under a column that has already been summed up, we need to follow that kind of procedure and consider the place value of that carryover digit when calculating the sum under the column where its right most digit is placed.

So, the rule is, if the carryover digit is placed under a column which has not yet been summed up, everything is fine, just follow the procedure in this lesson. If the carryover digit is placed under a column which has already been summed up, then account for it by considering the place value of this digit during the addition of the next column.

I hope that explains things a little better.

Anonymous said...

Thanks for the explanation

Blogannath said...

You are welcome. Good luck!

Anonymous said...

wht is the derivation for this method sir?

Blogannath said...

I will try to come up with an explanation of the derivation and post it as a separate post. Thank you for your inquiry.

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