Introduction to Vedic Mathematics Series

One key skill that is tested in a B School entrance test is speed at which you can do the problems. Most papers, given enough time, can be solved in totality by most candidates and it is the speed that acts as the filter between the grain and the chaff. Hence being faster at number crunching will give you an added advantage by giving extra time for tackling the ‘other’ aspects of different problems. In this regard it is advisable to learn some simple tricks that can speed up your number crunching. Vedic mathematics is one such method. In this series of articles we shall abstract Vedic mathematics to a set of tricks and propose to introduce these tricks in a manner which will have the most impact from a B School entrance test perspective. As a disclaimer we state the obvious - that these tricks have not been invented by us and are grateful to various sources of knowledge and we are acting just as a compiler and maybe putting it in context.

To the uninitiated, Vedic Mathematics is a mathematical elaboration of sixteen simple mathematical formulae (and 13 sub-formulae) from the Vedas as brought out by Sri Bharati Krishna Tirthaji. We will now introduce these one by one focussing on the practical side of the same as we learn them.

Ekadhikena Purvena

This is a Sutra or formula which means: “By one more than the previous one”. This is one of the simplest formulas and since it’s easily applicable, it can help us hit the ground running. This has application in multiple numerical operations and we shall see them one by one

Squaring of Numbers ending in 5:

One of the simplest applications of this formula is in squaring of numbers that end in 5.
Consider the number, say, 75.
Now 75 = 7 | 5 i.e. last digit is 5 and previous digit is 7.
Now ‘one more than the previous one’ is 7+1, which is ‘8’.
In this particular application the procedure to remember is to “multiply the previous digit by the number given by the formula, which is one more than itself.”
Hence 7x8 = 56. This forms the LHS of the answer. The RHS of the answer is 5² which is 25
Hence 75² = 7x(7+1) | 5² = 56 | 25 = 5625!

Similarly
95²= 9 X (9+1) | 25 = 9*10/ 25 = 9025
115²= 11 X 12 | 25 = 132| 25 = 13225
205² = 20*21 | 25 = 420 | 25 = 42025
2005² (!!!) = 200*201 | 25 = 40200 |25 = 4020025

(I know you are wondering how to do 200*201? Well it’s not that hard but it can get harder and we shall soon cover them)
Try out the same for some numbers before proceeding to the next part of the article. It is advisable to I initially do it on paper and then try doing it in the mind. If you see that you are going wrong without writing it down, you can try to see , what are the minimal things that you need to write to get this done on paper since the tests, you would have limited space to write down.
On a side note this technique can be extended to find squares of multiples of numbers ending in 6 and 4 by combining another technique.

For example you need say, 74² and 76². Start with 75² which we have found out to be 5625.
For 74² , add 75 and 74 = 149. Now subtract 149 from 5625 which is (5625-150+1) = 5475+1 = 5476. Presto that’s 74².  As quick check of the subtraction to an extent 4² is 16 hence 74² has to end in 6 and this has ended.

For 75² - add 75 and 76 which is 151 add this to 5625 which is 5776. Which is nothing but 76².

This is a technique that can be used for finding preceding and succeeding squares of any number whose square you already know but the technique illustrates, even in this early stage, how different concepts can be combined to give fast results.
As far as objective questions are concerned the takeaway is a bit more. With choices available you won’t have to do most of the subtraction / addition / multiplication etc. For example

Imagine the question “What is the square of 46²?”

a) 2116 b) 2006 c) 2014 d) 2016

Thought process

i.    c eliminated since end digit is 4
ii.    45² = 4x5 | 25 = 2025
iii.    d less than 2025 eliminated
iv.    45+46 = around 90 = 100-10 => around 2115 => ‘a’ is the answer. Without the addition step.

Important Note

As we go through the different techniques we may find different ways of doing the same thing. The trick is to understand all the techniques and select and apply the one that is most appropriate to the situation given the circumstance. For example adding 100 and 101 and then adding 201 to another result might be easy but adding 34 and 35 and then adding 69 to another result might not be. The trick is to practise enough to understand what to apply.