For any number

`2^n`

(or`5^n`

), 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.

For ex., say a number `120016`

, 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 `120016`

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 `abX1000`

, no doubt this is divisible by 8. Because `1000`

is a factor for it which is multiple of `8`

.

Now we just need to confirm if the number `cde`

is divisible by 8. The same applies for any number X , to confirm if it is divisible by `2^n`

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

The similar way we can prove for powers of `5`

!

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

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