## Vedic Mathematics: Squares of numbers ending in 5

Consider $25^2$ Here the number is 25. We have to find out the square of the number. For the number 25, the last digit is 5 and the ‘previous’ digit is 2. Hence, ‘one more than the previous one’, i.e. 2+1=3. The formula, in this context, gives the procedure to multiply the previous digit 2 by one more than it self, that is, by 3.

It becomes the L.H.S (left hand side) of the result.

i.e. 2 X 3 = 6.

The R.H.S (right hand side) of the result is $5^2$, that is, 25.

Thus,

$25^2 = 2*(2+1)$ and 25 = 625

In the same way,

$35^2 = 3*(3+1)$ and 25 = 3*4 and 25 = 1225

$65^2 = 6*(6+1)$ and 25 = 6*7 and 25 = 4225

$105^2 = 10*(10+1)$ and 25 = 10*11 and 25 = 11025

$135^2 = 13*(13+1)$ and 25 = 13*14 and 25 = 18225