# Divisibility 1

All natural numbers(composite or prime) have divisibility conditions. Let us derive some with some interesting ideas.

For 11

Div. Condition : Difference of the the sum of alternately placed digits is either zero or an integer multiple of 11.

Proof:

$10 \equiv\ -1 (mod 11)$

$\Leftrightarrow 10^n \equiv\ 1 (mod 11)$ if n is even.

$\Leftrightarrow 10^n \equiv\ -1 (mod 11)$ if n is odd.

If n is even ,

$\Leftrightarrow 11|10^n -1$

If n is odd ,

$\Leftrightarrow 11|10^n +1$

Take any random natural number X . Let its digits be $a_na_{n-1}a_{n-2}\hdots a_2a_1a_0$

$X$ can also be written as –

$X = [(10^n-1)a_n + (10^{n-1}+1)a_{n-1} + \hdots$

$+(10^2+1)a_2+(10^1+1)_1+0](=A)$

$+ a_n - a_{n-1} + a_{n-2} - a_{n-3} + \hdots + a_2 - a_1 + a_0$

but we know that $11 | A$

Hence if X is divisible by 11 , it should divide the difference of the sum of odd and even placed terms.Hence proved.

Case 2 : n is odd

Proceede similar to Case 1.

Take a random integer X with digits $a_na_{n-1}a_{n-2}a_{n-3}\hdots a_3a_2a_1a_0$

Case 1 : n – even

$X = 10^na_n + 10^{n-1}a_{n-1} + \hdots + 10^2a_2 +10^1a_1 + 10^0a_0$