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

Basic Idea

Let the number $N$ be $(10^n)k + p$ where k and p are integers and $p<(10^n)$ . Since $\frac{10^n}{2^n}$ is $5^n$ , $(10^n)$ is a multiple of $2^n$ , meaning $(10^n)k$ is also a multiple of $2^n$ . As long as p is a multiple of $2^n$ , then $N$ is a multiple of $2^n$ . Since $(10^n)k$ has $n$ trailing 0's, $p$ is the last $n$ digits of the number $n$ .

Concise

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

Let the base-ten representation of $N$ be $\underline{a_ka_{k-1}\cdots a_1a_0}$ where the $a_i$ are digits for each $i$ and the underline is simply to note that this is a base-10 expression rather than a product. If $N$ has no more than $n$ digits, then the last $n$ digits of $N$ make up $N$ itself, so the test is trivially true. If $N$ has more than $n$ digits, we note that: