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
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:
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:
12 |89 4|1
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:
12 |89 4| 1
•••| 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:
12 |89 4| 1
•••| 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:
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
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.
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.
Our first division involves dividing 9 by 2. The quotient is 4 and the remainder is 1. This is written below:
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:
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:
2••|9 7 79|95
•••| 1 1
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:
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:
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:
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):
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:
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.
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!!