## X*(X+1)*100+25 – a perfect square

By: Kannan Ramamoorthy On: Thu 24 October 2013
In: Mathematics
Tags: #Mathematics #Numbers #Speed Math

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)`.