In this article we will learn about yet another simple but powerful Sutra (formula) of Vedic Mathematics called
Ekadhikena Purvena in Sanskrit means “One more than the previous“.
So in mathematical calculations involving this formula, we generally consider the given number and another number whose value is one greater than the given number.
Ekadhikena Purvena in Multiplication
Ekadhikena Purvena sutra can be used to quickly multiply any two numbers of two digits (say AB and AC) whose
- First digits are the same and (A=A)
- The second digits add up to 10 (B+C=10)
For example to multiply 47 x 43, here is how we go about it.
- Take the first digit which is common in both numbers, in the above case 4
- Apply Ekadhikena to it – i.e. increment it by one. So it becomes 5.
- Now write down the product of 5×4 which is 20.
- Next write down the product of the other two digits which is 7×3=21
- The answer is 2021
How simple isnt it? You dont even require a pencil and paper, nor do you require to do any carrying over of remainders.
Let us consider another example. To multiply 51×59. The answer is
- 5 x (5+1) = 5×6 =30
- For the second part we need to round it to two digits in case it is just a single digit, hence it becomes 09
- So the answer is 3009
Ekadhikena Purvena in Finding Squares
The above method can also easily be implemented to find the squares of numbers ending with five, because the sum of their last two digits is 5+5 which is 10, which hence meets the requirement of the formula.
So to find 75×75, all we need to do is 7×8,5×5 which is 5625. Instant calculation, isn’t it?
Ekadhikena Purvena in Converting Fractions to Decimals
How easy it is to calculate the recurring decimal of fractions like 1/19 whose denominators end with 9? Using Ekadhikena Purvena this calculation becomes as simple as… well, seeing is believing here
- Take 1 which is the “Purvena” (The digit previous to 9).
- Increase it by one which makes it 2
- Now start with 1 and go on multiplying it by 2 as follows:
- 21 which is [2x1,1]
- 421 which is [2x2,21]
- 8421 which is [2x4,421]
- 68421 which is [2x8,8421]. We now have one carrying since 2×8=16
- 368421 which is [(2x6)+1,68421]. We now have one carrying since (2×6)+1=13
- 7368421 which is [(2x3)+1,368421]
- 47368421 which is [(7x2),7368421]. We now have one carrying since (2×7)=14
- 947368421 which is [(4x2)+1,47368421]
Ok we should now stop here. Why?
Well, we have figured out half of the digits in the answer. And finding the remaining half is even more simpler.
But before that, how did we find that we were done with half of the answer?
The rule is, take the denomination and remove 1 from it. So 19-1=18. Now this gives us the total number of digits in the complete answer for 1/19. And since we have now found 9 digits we are half way through the answer.
For the remaining half simply find the inverse of each digit in the so far discovered half. Inverse of a digit is that digit which makes the total of the two 9. So inverse of 1 is 8, 2 is 7 and so on till 9 is 0.
The inverse of 947368421 hence is 052631578. Ok now that we have both the halves of the answer. The complete answer is just stick the two together after a decimal, which is 0.052631578947368421
Simple isn’t it? If you dont find this simple, then try calculating 1/19 using the traditional method taught in the schools
Even the normal calculators will not give you such an accurate decimal representation of these fractions, and most importantly with little practice you can simply go on writing down the digits in the answer from right to left without the need of any additional worksheet.
Now one might wonder the need to remember so many different formula for different calculations. Just imagine what if you were taught these kind of calculations right from your school days. Well, expertise is after all all about getting used what we regulary do.
Its not as if in the schools we were taught one or two methods of doing calculations. We as students, even otherwise had to learn numerous methods of computations. So if learning these vedic mathematical lessons is going to help us do lightning fast calculations without using calculators, then why not?