You can find all my previous posts about Vedic Mathematics 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
Straight Division II
Solving Equations 1
Solving Equations 2
Solving Equations 3
Solving Equations 4
Simultaneous Equations 1
Simultaneous Equations 2
Quadratic Equations 1
Quadratic Equations 2
Quadratic Equations 3
Quadratic Equations 4
Polynomial Division 1
Polynomial Division 2
Polynomial Division 3
Square Roots 1
Square Roots 2
Square Roots 3
Most of us are familiar with the fact that certain fractions (such as 1/3) have a recurring decimal representation. 1/3 is 0.333333. . ., with the series of digits continuing on forever with no end. On the other hand, some fractions (such as 1/2) have a decimal representation that has no recurring part. Thus, 1/2 is exactly 0.5, and the decimal representation does not continue on forever.
The rules that determine whether a fraction has recurring decimals or not are really quite simple. First represent the fraction in its simplest form, by dividing both numerator and denominator by common factors. Now, look at the denominator. If the prime factorization of the denominator contains only the factors 2 and 5, then the decimal fraction of that fraction will not have recurring digits. If the prime factorization yields factors like 3, 7, 11 or other primes (other than 2 and 5), then that fraction will have a decimal representation that includes recurring digits.
Moreover, if the denominator's prime factors include 2 and/or 5 in addition to other prime factors like 3, 7, etc., the decimal representation of the fraction will start with a few non-recurring decimals before the recurring part.
Thus, 1/4 = 0.25 (4 is just 2x2, and since the denominator can be factorized using only 2 and 5, the decimal representation does not include any recurring part). On the other hand, 1/9 = 0.111111. . . because 9 is 3x3. Thus the denominator has prime factors other than 2 and 5, thus making its decimal representation a recurring one. Furthermore, 1/6 = 0.1666666. . . In this case, 6 = 2x3. Thus one of the factors is a 2 or 5 while the other factor is not. That is why in this case, the decimal representation contains recurring digits, but the recurring digits start after at least one non-recurring digit.
In this lesson, we are going to start this series of lessons on recurring decimals by concentrating on a simple way to calculate the reciprocals of prime numbers like 3, 7, 11, etc. Thus, we will be finding the decimal representation of 1/3, 1/7, etc. The usual way in which this is done is by long division. We perform a long division of 1 by the given denominator as many times as is required to actually get the decimal representation.
In the case of a number like 1/3, this is quite simple. The recurring decimal is just one digit long, and an astute person can recognize that the digit has started repeating right away. He/she can then stop the division after just a few steps, saving a lot of effort. But what about a number like 1/7 or 1/17? These are the cases in which our shortcut can make a big difference.
Before we explain the shortcut, we need to understand another property of recurring decimals. And that concerns the last digit of the recurring part of the decimal representation of the reciprocal of numbers that end in 1, 3, 7 or 9. Note that these numbers can not have 2 or 5 as their prime factors, so the reciprocals of these numbers consist of purely recurring digits. The product of the last digit of the recurring decimal representation of a number and the last digit of the denominator (in this case, 1, 3, 7, or 9) is always 9.
Thus, if a number ends in 1, the decimal representation of its reciprocal will have repeating decimals that end in 9 (1/11 is 0.0909. . ., for instance). If a number ends in 3, the decimal representation of its reciprocal will have repeating decimals that end in 3 (1/3 = 0.33333. . ., for instance). If a number ends in 7, the decimal representation of its reciprocal will have repeating decimals that end in 7 (1/7 = 0.142857142857. . ., for instance). And finally, if a number ends in 9, its denominator will have a decimal representation that has a repeating decimals ending in 1 (1/9 is 0.111111. . ., as an example).
The last digit of the recurring decimal representation of the fraction is referred to as the sesanyanka of the fraction (sesanyanka literally means last digit).
The shortcut that we are about to explore in this lesson is based on two vedic sutras. The first one is Anurupyena, which means "proportionally". We have seen applications of this sutra in other areas such as multiplication. Finding the recurring decimal representation of fractions is another application of this sutra. The other sutra we will use says Sesanyankena Caramena. Literally, this means "the remainders by the last digit". Our knowledge of the last digit of the recurring decimal representation of a reciprocal (the sesanyanka) is going to come in handy in the application of this sutra, which is why we spent some time going over them!
The shortcut works as follows:
- Pretend that you are going to use long division to find the reciprocal of the given number ending in 1, 3, 7, or 9. As such, you would put a decimal point on the quotient line, then add a zero to the dividend (1 in this case) and then attempt to perform a division. This will result in a quotient and a remainder (in the case of divisors less than 10, the quotient will be non-zero, for divisors greater than 10, the quotient will be 0)
- We are not interested in the quotient though. We are interested in the remainder. More precisely, we are interested in the ratio of the remainder to 1. Call this ratio our multiplier. Since we are taking ratios with respect to 1, the first remainder is always our ratio in this technique
- Write down the first remainder
- Now, multiply the first remainder by the multiplier, and cast out the divisor (keep subtracting the divisor from it until the number left is less than the divisor). This becomes the next remainder. Write it next to the first remainder, separated from it by a comma.
- Now multiply the next remainder by the multiplier, and again cast out the divisor. Once again write this remainder next to the previous ones as before
- Repeat step 5 until at some point, the remainder becomes 1. Add the one to the list of remainders and stop repeating step 5
- Now, multiply the last digit of each remainder by the last digit of the recurring decimal appropriate for the denominator, the sesanyanka (remember that the sesanyanka of a divisor that ends in 1 is 9, that for a divisor that ends in 3 is 3, that for a divisor that ends in 7 is 7 and that for a divisor that ends in 9 is 1), and put the last digits of the products down in order. This step is what is meant by the vedic sutra, Sesanyankena Caramena, the remainders by the last digit
- At the end of the process, you have a set of digits that represents the recurring decimal representation of the given reciprocal!
- We pretend to perform long division. We add a zero to the end of 1 to get 10. This gives us a quotient of 1 and a remainder of 3.
- We don't care about the quotient. But our remainder being 3 tells us that our multiplier is 3
- We write down 3
- Multiplying 3 by our multiplier, 3, gives us 9. Casting out our divisor, 7, gives us 2. This becomes our next remainder. Write it next to the 3, giving us 3,2
- Multiplying 2 by our multiplier 3, we get 6. Casting out 7 from 6 leaves us with 6. This gives us 3,2,6
- Multiplying 6 by 3 and casting out 7's gives us 4. Our next remainder, 5, comes from multiplying 4 by 3 and casting out 7's. When we multiply 5 by 3 and cast out 7's, our remainder becomes 1. This is our signal to stop. Our list of remainders now reads 3, 2, 6, 4, 5, 1
- The last digit of the product of 3 and 7 (remember that the sesanyanka of 7 is 7 itself) is 1. The last digit of the product of 2 and 7 is 4, and so on.
- We ultimately get 142587, and we say that 0.142857. . . is the recurring decimal form of 1/7. We can verify that 7x0.142857142857142857 . . . is very nearly 1 (as we increase the number of times we include 142857 in the product, the number of 9's in the product increases, and when the number of 142857's is infinite, you get 1 as the product).
Let us now apply it to something a little more challenging, such as 1/13. Instead of writing out all the steps, I am just going to show you how we get the multiplier and the set of remainders. Note that 13 is more than 10, so our first remainder becomes 10, making our multiplier 10. Using this multiplier on successive remainders and casting out 13's gives us the following series of remainders:
10, 9, 12, 3, 4, 1
We now multiply each of these remainders by the sesanyanka of 13, which is 3. We get 076923 as our series of digits, telling us that 1/13 is 0.076923 . . . One can verify that this is indeed true.
Similarly, let us do 1/31. Once again, since 10 is less than 31, our first remainder and our multiplier become 10. This then leads to the following series of remainders:
10, 7, 8, 18, 25, 2, 20, 14, 16, 5, 19, 4, 9, 28, 1
The sesanyanka of 31 is 9, so we now have to multiply each of these remainders by the 9 and put down just the last digits. We get 032258064516129, telling us that 1/31 is 0.032258064516129 . . .
Now, there is another time-saving shortcut in the last step that you might have spotted if you were astute: Since the answer depends only on the last digit of the product of the remainders and the sesanyanka, we don't actually have to multiply the entire remainder by the sesanyanka to get the series of digits in the answer. Simply multiplying the last digit of each remainder by the sesanyanka, and then taking the last digit of that product will suffice for this exercise, thus obviating the need to perform multiplications such as 28x9, 19x9, etc. We only need to perform 8x9 or 9x9 instead to get the digits in the final answer!
We will now calculate 1/49 using this method. The first remainder is 10 since 10 is smaller than 49. Thus our multiplier is also 10. This gives us the following series of remainders:
10, 2, 20, 4, 40, 8, 31, 16, 13, 32, 26, 15, 3, 30, 6, 11, 12, 22, 24, 44, 48, 39, 47, 29, 45, 9, 41, 18, 33, 36, 17, 23, 34, 46, 19, 43, 38, 37, 27, 25, 5, 1
Yes, it is a long series, but if you were to try to find 1/49 using long division, you would probably run out of room on the paper, whereas with this method, the entire series can be worked out mentally with no risk of running out of paper or sanity! The sesanyanka of 49 is 1, so the actual decimal representation of 1/49 is obtained by simply multiplying the last digit of each remainder by 1 and writing them down in order. Technically, we multiply by 1, but we all know that what this involves is simply taking the last digits of the remainders and putting them down in order! Thus 1/49 = 0.020408163265306122448979591836734693877551. . .
It is difficult to verify the last few digits of the answer since very few calculators have enough precision to handle and/or display this many digits, but if you are brave enough, you can do the long division to verify the answer!
Given that we started our vedic mathematics lessons by finding reciprocals of numbers that end in 9, we have come full circle, and found the reciprocal of a number ending in 9 once again. We used a different technique this time, but what we did in the very first lesson is a direct extension of this method, that does not involve finding the remainders and then deriving the digits of the decimal representation from them. We are going to continue in the next lessons by expanding on this methodology to extend it beyond merely reciprocals. Along the way, we will take a look once more at reciprocals of numbers that end in 9.
I know that some of the material covered in this lesson is going to appear confusing. I have tried making the explanations as detailed as possible, but detail and simplicity lie in the eyes of the beholder. With practice comes an ease that makes everything look obvious, making explanations appear unnecessary, but for a beginner, it is not obvious, and a more detailed explanation would have been more welcome. I understand this, and am welcome to any suggestions for making the material more elaborate with more detailed explanations. Just let me know through the comments if something is not clear and needs more elaboration, or better examples. In the meantime, good luck, and happy computing!