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 .
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 .
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 .