Divisibility rules/Rule for 5 and powers of 5 proof
A number is divisible by
if the last
digits are divisible by that power of 5.
Proof
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
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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
.