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

Division By The Nikhilam Method II

Division By The Nikhilam Method III

Division By The Paravartya Method

Digital Roots

Straight Division I

In this lesson, we will extend the straight division method to deal with denominators that may have more than 2 digits. There are two main ways in which straight division can be performed when the denominator contains more than 2 digits.

The first method is very similar to the method we used for 2-digit denominators. We raise only the right-most digit of the denominator "on to the flag", and keep the rest of the denominator as part of the base. Since the number of flag digits is just 1, our method is not much changed from what we saw in the previous lesson. But since the base digits of the denominator now form a much larger number (at least a 2-digit number), we may have to make some adjustments to make sure our intermediate numerators are not negative. As explained in the previous lesson, this involves reducing the quotient and increasing the remainder so that the intermediate numerators are always positive.

To illustrate this method, let us tackle an example. We will start out by trying to find the answer to 8941/121. We divide the denominator into a base part of 12 and a flag part of 1. We then get the figure below:

••1

12 |894|1

•••|

------------

We then proceed with the method as outlined in the previous lesson. Our first division is of 89 by 12. This gives us a quotient of 7 and a remainder of 5. These are written as below:

••1

12 |89 4|1

•••|••5

-------------

•••| 7

Now we find our next intermediate numerator as 54 - 7 = 47 (54 is the number we obtain by putting the remainder of the previous division together with the next digit of the numerator, and 7 is obtained by multiplying our flag digit by the previous quotient digit). Dividing 47 by 12 gives us a quotient of 3 and a remainder of 11. These are written down as shown below:

••1

12 |89 4| 1

•••|••5 11

--------------

•••| 7 3

Since we have now dealt with all the quotient digits of the numerator, we can find the remainder and stop, or proceed further to find a decimal representation of the answer. If we do stop at this point, the answer we obtain is below:

••1

12 |89 4| 1

•••|••5 11

---------------

•••| 7 3|108

We can verify that the answer to the problem is indeed a quotient of 73 and a remainder of 108.

If we proceed further, we obtain the first few digits beyond the decimal point as below:

••1

12 |89 4| 1. 0 0 0 0 0••0••0

•••|••5 11 12 4 7 8 3 12 11

---------------------------

•••| 7 3. 8••9 2 5 6 1••9••8 . . .

One more example will further solidify this approach in our mind. With that mind, let us take the example of 8878867/1001. We know that it is very easy to deal with this example using the Paravartya method as illustrated below:

1001 0 0 -1

8878|867

0008|

0000|800

0000|070

-------------

8870|-3

8869|998

Now, doing it by the method of straight division, we get the figure below. Once again we have hoisted only 1 digit "on to the flag", and retained the rest of the denominator as base digits.

•••1

100 |887••8••8•••6|••7

••••|•••87•70•100•100

------------------

••••|••8••8••6•••9|998

We get the identical result. With straight division, we don't have to stop with a remainder though. We can carry on the division and find as many digits beyond the decimal point as we want for any division problem.

This method of dealing with multi-digit denominators (retaining all but one digit as base digits and hoisting just one digit on to the flag) works when the base digits are easy to divide by (small numbers like 11, 12, etc., or round numbers like 100, 50, etc.).

Now, let us look at another way of dealing with multi-digit denominators. In this method, we retain the part of the denominator that is easy to divide by as the base digits and hoist all the other digits on to the flag as flag digits. In many cases, this may involve just one base digit and 2 or more flag digits. The procedure for dealing with multiple flag digits is a little different from how we dealt with a single flag digit. We will now go over that procedure in detail as we apply it to a real problem. For the purpose of this illustration, let us take the example of 977995/212.

This time, instead of keeping 2 base digit and 1 flag digit, we will split the denominator into 1 base digit and 2 flag digits. This is illustrated in the figure below. Notice that because there are 2 flag digits, we set aside two digits of the numerator as remainder digits.

•12

2••|9779|95

•••|

-------------

Our first division involves dividing 9 by 2. The quotient is 4 and the remainder is 1. This is written below:

•12

2••|9 779|95

•••| 1

------------

•••|4

The next step in this procedure is not to multiply the first quotient digit by the numbers on the flag. The procedure is very similar to the procedure we use in multiplying vertically and cross-wise using the Urdhva-Tiryak sutra. In this case, the multiplicands are always the flag digits and the last n digits of the quotient, where n is the number of flag digits.

Since the quotient, at this time, contains only one digit, we pad it with as many zeroes to the left as is necessary to make sure that the quotient contains the same number of digits as the flag. In this case, we need to pad the quotient with one zero. In general, this will require padding with (n-1) zeroes where n is the number of flag digits. This is shown in the figure below:

•12

2••|9 779|95

•••| 1

---------------

••0|4

Now, we obtain the sum of the cross-products of 04 and 12 as explained in the lesson on Vertically and Cross-wise. This results in 0 x 2 + 4 x 1 = 4. Subtracting 4 from 17 gives us 13, which becomes our next intermediate numerator. Dividing 13 by 2 gives us a quotient of 6 and a remainder of 1. This is shown in the figure below:

•12

2••|9 7 79|95

•••| 1 1

---------------

••0|4 6

At this point, the 2-digit number we get by putting together the remainder and the next digit of the numerator is 17. From this 17, subtract the sum of the cross-wise products of the 2 flag digits and the last 2 quotient digits. This sum works out to 4 x 2 (first digit of the quotient x second flag digit) + 1 x 6 (second digit of the quotient x first flag digit) = 14 (we don't have to use the zero in front of the first quotient digit anymore since we now have enough real quotient digits to perform our computations with). 17 - 14 = 3. Dividing 3 by 2 gives us a quotient of 1 with a remainder of 1. This is illustrated in the figure below:

•12

2••|9 7 7 9|95

•••| 1 1 1

---------------

••0|4 6 1

Now, the 2-digit number we get by juxtaposing the remainder and the next digit of the numerator is 19. The number to be subtracted from this is the sum of the cross-products of the flag digits and the last 2 quotient digits found so far. The last 2 digits of the quotient are 61 and the flag digits are 12. So, the sum of the cross-products is 6 x 2 + 1 x 1 = 13. 19 - 13 = 6. Dividing 6 by 2 gives us a quotient of 3 and a remainder of 0. This is now added to our figure to give us the following:

•12

2••|9 7 7 9|95

•••| 1 1 1 0

----------------

••0|4 6 1 3

We have now dealt with all quotient digits of the numerator. So, we can now calculate a remainder or we can continue the division by adding a decimal point and proceeding as above.

The calculation of the remainder is also slightly different from what we are used to with a single flag-digit. To calculate the remainder, we now take the last remainder digit and add it to the front of the remainder digits of the numerator. This gives us 095.

Then we calculate the sum of the cross-products of the last 2 digits of the quotient and the 2 flag digits. The sum of the cross products in this case is 1 x 3 + 1 x 2 = 5. We multiply this by the power of 10 that has the same number of digits as the number of flag digits. In this case, that power of 10 is 10 itself. 095 - 5 x 10 = 45.

Now, as in the lesson on vertically and cross-wise, we subtract out from this, the vertical product of the last digit of the quotient and the last flag digit. This vertical product is 3 x 2 = 6. 45 - 6 = 39, and this becomes the remainder for the division problem. This is shown in the figure below:

•12

2••|9 7 7 9|95

•••| 1 1 1 0

------------------

••0|4 6 1 3|39

It is easy to verify that a quotient of 4613 and a remainder of 39 is indeed the correct solution to the divison problem we undertook to solve.

The following steps explain the process to be used when dividing with more than one flag digit:

- The numerator is divided into quotient and remainder digits such that the number of remainder digits is the same as the number of flag digits
- The first quotient digit is obtained by dividing the first digit (or digits if necessary) of the numerator by the base digit(s) of the denominator
- The remainder from the above division is placed in front of the next numerator digit to form a 2-digit number
- Pad the quotient with zeroes to the left such that the number of quotient digits is the same as the number of flag digits
- The last n (where n is the number of flag digits) quotient digits are then multiplied and added vertically and cross-wise with the flag digits and this sum of products is subtracted from the 2-digit number we obtained earlier by placing the remainder before the next digit of the numerator. This gives us the next intermediate numerator
- The intermediate numerator is divided by the base digit(s) of the denominator for the next quotient and remainder. The quotient goes on the quotient line and the remainder goes in front of the next digit of the numerator to form a 2-digit number
- We now proceed by calculating the sum of cross-products using the last n digits of the quotient (where n is the number of flag digits), and subtracting this from the 2-digit number obtained in the previous step. This gives us the next intermediate numerator
- This process is continued until we have dealt with all quotient digits of the numerator and our next remainder digit is placed in front of a remainder digit of the numerator
- If we want a decimal answer, we put a decimal point on the quotient line when our 2-digit number on the numerator line is formed by including one of the remainder digits of the numerator. The method proceeds as outlined in the previous steps, and can be continued to as many digits beyond the decimal place as needed by padding the numerator with zeroes to the right. Do not perform the steps below which are for calculation of a remainder
- If we want a remainder, then we stop as soon as any step above involves a remainder digit of the numerator. Perform the steps below to calculate the remainder
- Place the last remainder that was calculated by division of an intermediate numerator by the base digit(s) in front of the remainder digits of the numerator to get an (n+1)-digit number, where n is the number of flag digits (call this the intermediate remainder)
- Take the last n quotient digits where n is the number of flag digits and calculate cross-products of this number with all the flag digits as in vertically and cross-wise
- Multiply this cross-product by the power of 10 that has n digits and subtract from the intermediate remainder to get the new intermediate remainder
- Next, take the last (n-1) quotient digits where n is the number of flag digits and calculate cross-products with the last (n-1) flag digits
- Multiply this by the power of 10 that has (n-1) digits and subtract from the intermediate remainder to get the new intermediate remainder
- Proceed this way, winding down the number of digits in the cross-multiplication until you are left with just a vertical multiplication of the last quotient digit with the last flag digit. Subtract this final vertical product from the intermediate remainder calculated through the above steps to get the final remainder

The procedure above looks long and complicated, but is actually quite easy to perform once the basics of vertically and cross-wise are clear. In fact, an astute person would have easily seen that padding the quotient digits with zeroes to the left so that we can perform a full vertically and cross-wise cross-product with the flag digits is actually the same as starting with a vertical product of the first quotient digit with the first flag digit, then in the next step, expanding the cross-multiplication to 2 quotient digits and the first 2 flag digits, and so on until the cross-product expands to all the flag digits. This is the same procedure as is used in vertically and cross-wise to calculate products. And then, the process is wound down by eliminating flag digits from left to right (since we expanded from left to right, we have to wind down from left to right also) until we are left with a vertical product of the right-most quotient digit with the right-most flag digit, just as in vertically and cross-wise. If this is not clear, it is advisable to review vertically and cross-wise once more to be clear on the procedure, especially for multi-digit multiplications, as explained in this past lesson.

We will conclude this lesson with a couple more examples.

64794/423 worked out in decimal form gives us 153.17730496. . . as below (note that in several of the steps below, we have adjusted the division by reducing the quotient and increasing the remainder so that the subtraction step does not produce a negative answer):

•23

4••|6 4 7|9 4 0 0 0 0 0 0 0

•••| 2 2 2 4 5 5 3 3 5 6 6

--------------------------------

••0|1 5 3.1 7 7 3 0 4 9 6 . . .

64794/423 with a remainder gives us a quotient of 153 and remainder of 75 as below:

•23

4••|6 4 7|94

•••| 2 2 2

----------------

••0|1 5 3|75

943809843/2432 with remainder gives us a quotient of 388079 and a remainder of 1715. Note how the quotients and remainders of the individual divisions by the base digit (2) have been adjusted to avoid negative numbers after subtraction. Also note that ultimately, we obtain a negative remainder, which has to be adjusted by subtracting 1 from the quotient and adding the denominator, 2432, to the previous remainder.

•432

2•••|9 4 3 8 0 9|8 4 3

••••| 3 6 6 6 4 1

---------------------------

••00|3 8 8 0 8 0|-717

••00|3 8 8 0 7 9|1715

Hope you will take the time to practice this technique so that it becomes more and more of a habit or second nature rather than a difficult intellectual exercise that requires you to drop everything else to concentrate on this. The whole idea behind mental computation is to be able to perform them "on the side" while trying to think through bigger problems that require answers to these computations only as a part of the whole "bigger picture". After all, I don't think any of us is going to be doing these computations just for the sake of doing them, in real life! So, good luck, and happy computing!!

## 13 comments:

Hi.. Thanks for this informative article. I really liked it. I also came across this similar site with nice articles and videos on vedic prinicples - http://www.vichaar.tv

hey boss, this is some of the best stuff I've found on Vedic Math so far. Nice examples and explained well. I have run into some areas that aren't working on the first 2 of my own practice problems!Can you enlighten me where I'm going wrong? I love this technique but must be able to depend on it... and count on it to be fast...(I'm taking the GRE test soon)

787243/464

•64

4••|7 8 7 2|43

•••| 3 5

----------------

••0|1 7 2

I stopped here because it wasn't coming close to the answer (1696.644). I initially came up with 8 as the quotient for the second intermidate number but the remainder came out to zero so I backed it down to a quotient of 7 with a remainder of 5 but it still leads me down the wrong path. That's where I quit. (so I created another random problem and ran into other stopping points)

Problem #2 was...9812/382(answer should be 25.68)

•82

3••|9 8 |12

•••| 6

----------------

••0|2

68-(0+16)= 52 (huh? I'd have a quotient of 16 here) What am I doing wrong?

Let us take 787243/464 first. We draw the figure as below:

•64

4••|7 8 7 2|43

•••| 3

----------------

••0|1

The next step is to take 38 and subtract from it 0x4 + 1x6 (crosswise multiplication between 01 and 64). We get 38 - 6 = 32. Division of 32 by 4 gives us a quotient of 8 and remainder of 0 (which is not suitable since we would get a negative intermediate numerator). So, we can take the quotient to be 7 and get a remainder of 4 (note that 7x4 + 4 = 32, that is why 4 is the remainder when the quotient is reduced to 7. In general, everytime you reduce the quotient by 1, add the divisor to the remainder).

For some reason, you had a remainder of 5 at this second step, leading you astray. So, the second and third steps result in the figure below:

•64

4••|7 8 7 2|43

•••| 3 4 1

----------------

••0|1 7 0

Now, we see that this is a problem since the intermediate numerator will become 12 - 28 which is negative. Since we can not reduce the 0 any further, we have to go back a couple of steps to reduce the 7 to a 6 and make 8 the remainder from that step. After doing that, we get the final figure as below:

•64

4••|7 8 7••2|43

•••| 3 8 11 2

----------------

••0|1 6 9 8|-629

••0|1 6 9 6|299

The easier way to handle problem like this is to read my lesson on vinculums (http://blogannath.blogspot.com/2009/09/vedic-mathematics-lesson-22-vinculums.html) and then convert 464 to 5(4)4 (I denote a negative number by enclosing it in parentheses. Then the division problem becomes as below:

•(4)4

5••|7 8 7 2|43

•••| 2 2 2 4

----------------

••0|1 6 9 6|299

As you can see, much simpler!

9812/382 can similarly be simplified by changing the 382 to 4(2)2. I will leave 9812/4(2)2 for later. But to handle the original problem using straight division, we draw the figure below:

•82

3••|9 8 |12

•••| 3

----------------

••0|2

First of all instead of a quotient of 3 and a remainder of 0 in the first division, we get a quotient of 2 and a remainder of 3 (not 6). Note that 2x3 + 3 = 9. Again, when you reduce the quotient by 1, add the divisor to the remainder (in this case 3).

The rest of the division follows in a straightforward fashion after that as below:

•82

3••|9 8 |12

•••| 3 1

----------------

••0|2 7|-502

••0|2 5|262

Even if one of the quotients becomes larger than 9 (a 2-digit quotient like 10, 11, etc.), it is handled by carrying over digits to the left. This is also illustrated in the lesson on vinculums.

If you have more questions, please feel free to post back. Good luck!

Hey, can you solve this... 449 / 64

I wanted an answer up to last decimal place, but using this method I get 7.016 when the actual answer is 7.015625

Again tried using this 449 / 68 using the complement method i.e. instead of 68 used 7-2.

Here the answer gets solved to 6.60294 when the actual answer can goes beyond 5 decimal places... 6.6029411764705882352941176470588

Is there issues with these methods when decimals are to be found?

There should not be a problem with 449/64 or any other division problem with this method. I think the mistake you are making is that at one particular step, you have an intermediate numerator of 36, which is exactly divisible by 6. But that leaves one with a remainder of 0, and when you put that 0 in front of the next digit of the dividend, you get 00. At this point though, you can not subtract the product of the flag digit and the last digit on the quotient line from 00 (the answer will become negative). So, you have to put down 5 as the quotient digit and 6 as the remainder and continue until you don't have this problem. I have solved 449/64 below. The same procedure will apply for 449/68 also. Always make sure you start getting quotient digits as well as remainder digits of 0 before you consider the division as complete.

•4|

6•|4 4| 9 0 0 0 0 0 0 0 0

••| |2 1 4 6 4 4 2 0 0

-----------------------------

••| 7| 0 1 5 6 2 5 0 0

The answer is therefore 7.015625, which is exactly correct. Notice how both remainder and quotient digits are zero towards the end of the division in the above example. If you just stop when the remainder digit becomes zero, that is not correct.

Thanks a lot sir! This is of help!!!

Only a small suggestion, there are way to many things (ads, counters, flash etc) on one page of the blog and at times it hogs up the browser. It would certainly help if these non-core contents are minimized or split across different pages. Cleaner the better, THINK Google!!!

Hi, I was trying to divide 1375 by 175 and then something happened.

The rest of one step was 10, otherwise the result would be negative.

I proceeded with 10 as the rest and put it together with the number

Here's my final step

•75|

1• |1 3| 7 5 0 0 0 0 0 0 0

••| |1 6 10 9

-----------------------------

••| 007| 8 5

Is that correct?

Seems to be correct so far. And you can verify that you are on the right track by using a calculator to see what the first few digits of 1375/175 are supposed to be.

Thanks for the quick reply and for the great site. I am studying for a test that will have alternatives.

It would be great to calculate only 2 digits of the answer really fast.

For example, if the answers for a particular question involving math are:

a)12345

b)17623

c)12435

d)12965

e)12975

If I could rapidly discover the penultimate digit of any operation, I would promptly solve the test (because these digits are all different).

In general, the answers are very similar in the first few digits. At least to the 10³ order they're almost equal (to prevent any quick approximation)

When it's a sum or a subtraction, it's trivial to discover the digit, but when it's a multiplication or division, i don't know any methods.

Do you know if there's an specific technique for that?

I was thinking about using some sort of cross-wise, but i'm afraid the carrys in the final sum would ruin my approximation.

The method could be only for multiplication or only for division, because I can test all alternatives. e.g if the final calculation is in the form x/y i can multiply Y by the answers and see if the digit corresponds to X.

Other thing I would really bennefit from is a quick aproximation method for division and multiplication (especially cubing, squaring, etc). In one test, for instance, the final answer depended on the following calculation:

(1.06)^12 - 1

The answers were considered only until the 4th digit after the decimal point (that prevented me to check if the alternatives + 1 were divisible by 1.06.

But in this question, as opposed to the ones I described in the beginning, the answers were very distants from one another. Here are the options

(A) 51,2196%

(B)101,2196%(C) 151,5456%

(D) 201,2196%

(E) 251,5456%

The quickest way I found to solve this was using pascal triangle to cube it and then square it two times.

Well, sorry about that long post. In the end, I just wanted to know for specific ways to take advantage of having the possible answer at your disposal.

Thanks again for the great blog! I am following all the Vedic Course and loving it!

P.S: Sorry about my bad english, I am from Brazil.

I wish you all the best in your test! Sounds like an interesting test.

For multiplication and division problems, I would suggest that you review my blog post on Digital Roots (http://blogannath.blogspot.com/2009/09/vedic-mathematics-lesson-19-digital.html), and another post on the math behind using digital roots to check the results of mathematical operations (http://blogannath.blogspot.com/2011/04/math-behind-using-digital-roots-to.html). These can be useful in quickly discarding wrong answers from a bunch of options (especially if you practice digital roots in base 7 or 11 so that you can do them very quickly). Obviously, if the answers are rounded off, then it may not work, so you just have to come up with other ways to approximate the solution.

I am glad that you are finding my blog useful in your preparation for this test. Once again I wish you all the best, and if there is anything I can do on my blog that would help you and other people in your situation, please feel free to comment with your suggestions, and I will try my best.

And don't worry about your English. It is very good, and I would be ecstatic if my Portuguese were half as good!

I'm having a problem doing straight division with the following problem:

1411/29

9 is the flag digit, and 2 is the divisor. 2 goes into 14 7 times. Now I know that the first digit should be 4. Even if I say that 2 goes into 14 4 times with a remainder of 6, that doesn't work either.

What am I missing here?

Thanks.

I am not sure what the problem is. Your logic is exactly correct, and you do put down 4 as the first digit of the quotient, and have a remainder of 6. 61 - 36 is 25, which forms the next dividend. To make sure the next dividend is positive, you have to put down 8 as the next digit of the quotient, and carry over 9 as the remainder. 91 - 72 is 19, and that forms the final remainder of the division. So, the solution to 1411/29 is 48 with a remainder of 19, which can be verified with a calculator.

The flag division figure is given below.

*9

2 |14 1| 1

6 9

---------------

**|4 8 19

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