# Divisibility rules/Rule for 2 and powers of 2 proof

A number is divisible by if the last digits of the number are divisible by .

## Contents

## Proof

### Basic Idea

Let the number be where k and p are integers and . Since is , is a multiple of , meaning is also a multiple of . As long as p is a multiple of , then is a multiple of . Since has trailing 0's, is the last digits of the number .

### Concise

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

Let the base-ten representation of be where the are digits for each and the underline is simply to note that this is a base-10 expression rather than a product. If has no more than digits, then the last digits of make up itself, so the test is trivially true. If has more than digits, we note that:

Taking this we have

because for , . Thus, is divisible by if and only if

is. But this says exactly what we claimed: the last digits of are divisible by if and only if is divisible by .