Vedic Maths (Square)

[Damomeera]

If we want to square a number ending with '5' we have to simply put the last two digits of the answer as 25 and the remaining digits are calculated as

a*(a+1)

here a is number in front of digit 5 in the number for which we've to find square.....

For example, square of (35) is 1225
Here we see last two digits are 25
and the remaining digits are calculated as 3*4

Similarly for square of 95 is 9025

Thanks

 

[scramble]

what about numbers not ending in 5?

 

[_Eduardo_]

what about numbers not ending in 5?

No way, only 5.

Just expand the binomial (10a+5)^2 = 100a^2 + 2*10a*5 + 25 = 100a(a+1) + 25

 

[scramble]

No way, only 5.

Just expand the binomial (10a+5)^2 = 100a^2 + 2*10a*5 + 25 = 100a(a+1) + 25

exactly!!! so what other method is there in Vedic maths?

 

[Damomeera]

For numbers having only 9
the calculation of its square is simple.
for example for 999 the square can be calculated as 999-1 followed by 000+1
the answer is 998001
in the same way for 99 the square is 9801
for 9999 is 99980001
and so on
I hope u understand the logic

 

[albbg]

You could be interested in this site:

https://www.vedicmaths.org/Bookstores/Free Books.asp

 

[scramble]

So i am guessing that for every number there must be a logical step!

 

[AarenAce]

I can't understand your question clearly