```
= 4*(4+1)*100+25
= 20*100+25
= 2025
```

Once gone through a shortcut to find square of a number that ends with 5.

For any number that is of format `X5`

i.e, `10X+5`

. The square will be `X*(X+1)*100+25`

.

To say it with example, say a number `45`

,which is `4*10+5`

. Hence here the `X`

is `4`

. So the square can be calculated quickly as below,

```
= 4*(4+1)*100+25
= 20*100+25
= 2025
```

The process that we do above might seem tedious at first glance. But actually not! You just calculate `X*(X+1)`

then put 25 next to that. At the place of ones and tens you are getting `0`

anyhow and adding `25`

is same as appending `25`

to `X*(X+1)`

.

Just tried to write the proof for the above.

So, we wanted to square the number `10X+5`

. So see how the equation evolves to get the another form.

```
= 100x^2 + 100x + 25
= 100x(x+1) + 25
= x(x+1)*100 + 25
```

So whatever we see in the format `X(X+1)*100 + 25`

, is a perfect square with the square root as `10x+5`

.

Hence `9025`

is a perfect square with root as `95`

.

`13225`

is a perfect square with root as `115`

.

`38025`

is a perfect square with root as `195`

.

We just have to see if the number, after stripping off the `25`

, is of format `X*(X+1)`

.

*For any feedback or corrections, please write in to: *** kannangce@rediffmail.com **

comments powered by Disqus