3649, this was my default pin number that I got for my card.(I can reveal it, I have changed it now..;)). The interesting fact was that it is concatenation of 6² and 7². And one more fact is that any 2 consecutive number is a perfect square. 36,64 and 49…

# Tag: Numbers

## 2=>3=>8=>9

Just noted the above numbers 2,3,8,9 pressed in the Lift(Elevator) this morning..

Something Interesting with the levels.. 2^3 = 8 & 3^2=9..;)

## Our car number

We bought a new car which has its registration number as 5050.

This is more than just a fancy number, it is * sum of numbers from 1 to 100*.

(100X101)/2 = 5050..;)

## Number divisible by 2,4,8,16../5,25,125..

For any number * 2^n*(or

*), if you need to find out if it is a factor of number X, it is enough if you check the last n digits of the number X.*

**5^n**For ex., say a number 12**0016,** if I need to find if the number is divisible by 16(=2^4), I just need to check if the last 4 digits is divisible by 4. So here 12** 0016 **is divisible by 16, because the last 4 digit is divisible by 16.

Now, let’s not convinced just with some shortcut.

Let’s understand the concept by taking a 5 digit number represented by * abcde*, where a,b,c,d,e each represents some decimal from 0 to 9.

I need to find out if the number * abcde* is divisible by 8(=2^3).

Expressing the number * abcde* as

*ab000 +cde*So now, * ab000 *is nothing but

*, no doubt this is divisible by 8.*

**abX1000**Now we just need to confirm if the number * cde *is divisible by 8. That’s it for any number

*X*, to confirm if it is divisible by

*we just need to check if the last n digits is divisible by n.*

**2^n**The similar way we can prove for powers of 5..

## 2^n = (1+2+4+…2^(n-1))+1

This is one thing that I noticed. Then tried proving it, the below is the proof.

There is a proof that we can get it through binary(To be frank, I got this proof from my friend;), the way that I tried is simple, I just added the 1’s and kept taking out the 2’s).

Take the RHS and express it in binary,

The binary of

1 is 1,

2 is 10

4 is 100

And * 2^n-1 is 100..(n-1)0’s *

So when I express the sum 1+2+4+…2^(n-1) as binary,

I’ll get * 1111…(n-1 1’s) *

Now consider the RHS,

* 1111….(n-1 1’s)+1 *

So adding 1 at the last will keeps giving a 1 carried to its preceding position.

And at last you’ll get * 100..(n 0’s)*

Which when expressed in decimal is *2^n*

That’s it..:-)

## X*(X+1)*100+25 – a perfect square

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 will be

=4*(4+1)*100+25

=20*100+25

=2025

The process that we do above might seem tedious at first glance. But not. You just write X*(X+1) then 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).