Show that all numbers $a$ are congruent modulo 8 to its units digit in base 1000

So as the title says, I want to show that all numbers $a$ are congruent modulo 8 to its units digit in base 1000. I believeI'm more confused as to what 'its units digit in base 1000' mean more than anything, so clarification would be helpful.

A hint on the problem would be welcomed as well, though I don't think that should be too difficult as I have a proposition that says something similar to what I'm trying to prove, I just don't quite understand the wording in that case either.

For reference, the proposition I believe I should use is as follows: "$7$ (respectively $11, 13$) divides $a$ iff $7$ (respectively $11, 13$) divides the alternating sum of the "digits" of $a$ in base 1000."

Answer

" I believeI'm more confused as to what 'its units digit in base 1000' mean more than anything, so clarification would be helpful. "

A non-negative integer (actually any real, but lets assume a non-negative integer) can be written in base, $1000$ (or any base $b$) by expressing it as:

$a = \sum_{i=0}^n a_i * 1000^i; 0 \le a_i < 1000; a_i \in \mathbb Z$

for appropriate values of and appropriately many, $a_i$s.

The units digit is $a_0$.

So the question is asking you to prove:

$a \equiv a_0 \mod 8$.

Once the question is clarified, I hope and trust the actual reason is fairly straightforward.

==== old response (when I misunderstood the exact question but had the general idea-- but still technically an incorrect answer) =======

The question is asking: Let $a = \sum_{i=0}^n a_i *1000^i$. Let $b=\sum_{i=0}^m b_i*1000^i $ with $a_0 = b_0$.

Prove that $a \equiv b \mod 8$.

That's the question. Can you prove it?