Thursday, 25 August 2011

Amazing way to calculate squares within seconds

"Ekadhikena Purveṇa" as we discussed for division is the Sanskrit term for "One more than the previous one" and very important Vedic Mathematics formula. We already know from the previous post that this formula can be used for both multiplication and division, here let us have a look at how it helps for multiplication:
This formula can be applied to multiplication of numbers that satisfy both the following conditions

  1. the same first digit
  2. the sum of their last unit digits is 10.
Even though it might so appear that this formula is useful in limited cased but an interesting category of numbers that satisfy both above requirements are all the numbers ending with 5. So when we need to derive squares for any number ending with 5, this formula is applicable straightaway. In this post, I am going to talk about limited application of this formula for deriving squares of numbers with last digit as 5.


Let us break the method into two parts:
  1. Deriving last two digits of answer
  2. Deriving all previous digits of answer
1. Deriving last two digits of answer

No application of mind, last two digits are 25, always.

2. Deriving all previous digits of answer

All the digits in the number before the last five would constitute base for us and let us call them B1.

Another base is B1 +1 (Because the formula is about one more than the previous one…)

Digits previous of 25 in the final answer would be simply [B1 x (B1+1)]


Let us understand this with an example of squaring 25

Last two digit of the answer are 25

For other digits, we need B1 and it goes as follows:

B1 = 2 (All the digits other than the last 5 in the number we want to derive square for)

B1 + 1 = 2 + 1 = 3

Therefore other digits of the answer preceding 25 are = 2 x 3 = 6

Hence the final answer would be 625.

So now squaring is not a headache anymore having understood vedic mathematics, right?


Learning just the shortcut method isn’t enough, so let us understand the logic how this operates. For explaining the logic we need to use the basic concepts of factorization:

We all know that (a+b)2 = a2 + 2ab + b2 

Now how does this square of numbers ending with 5 work in this case?

Let us continue with the above example:


(20 + 5)2

202 + (2x20x5) + 52 (Factorization using the above rule)

202 + (20x10) + 52 (Simplified with multiplication of 5x2 for middle term)

20 (20 + 10) + 52 (taking 20 common from first two terms)

20 (30) + 52

2 x 10 x 3 x 10 + 52

(2x3) x (10x10) + 52

(2x3) x (100) + 52

Now this expression looks like this [B1 x (B1+1)] x 100 + 52

Because in this expression two things are constant i.e. whatever is the B’s multiplication that shall at the end be multiplied by 100 and 25 shall be added so it is safe to make a rule that last two digits of the answer are going to be 25 and right preceding these digits we need to insert the multiplication of B1 & B1+1

Now you must feel awakened with the logical explanation of Vedic mathematics.

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