Divisibility rules/Rule for 2 and powers of 2 proof

A number $N$ is divisible by $2^n$ if the last ${n}$ digits of the number are divisible by $2^n$ .

Proof

An understanding of basic modular arithmetic is necessary for this proof.

Let $N = a_ka_{k-1}\cdots a_1a_0$ be the base-ten expression for $N$ , where the $a_i$ are digits.

Thus

$N = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0.$

Taking $N$ mod $2^n$ gives

$N$	$= 10^k a_k + 10^{k-1} a_{k-1} + \cdots + 10 a_1 + a_0$
	$\displaystyle \equiv 10^{n-1} a_{n-1} + 10^{n-2} a_{n-2} + \cdots + a_1 + a_0$
	$\equiv a_{n-1}a_{n-2}\cdots a_1a_0 \pmod{2^n}$

Thus, if the last $n$ digits of $N$ are divisible by $2^n$ then $N$ is divisible by $2^n$ .