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).

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