# 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."

• If we represent a number $N$ as $1000^k a_k+1000^{k-1}a_{k-1}+\cdots+1000a_1+a_0$, where are the $a_i$ are between $0$ and $999$, then $a_0$ is the units "digit" of $N$ to the base $1000$. Now the rest should be one line, even a very short hint would be a solution. – André Nicolas Feb 16 '16 at 3:26
• In base $1000$, what are the factors of the base? Do you know about the difference between bases $2$ and $10$ and $16$? – abiessu Feb 16 '16 at 3:27
• Would it be to much of a hint if I pointed out that statement says nothing more or less than "1000 is divisible by 8"? – fleablood Feb 16 '16 at 4:50

" 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?

• Oh, and each $a_i$ and each $b_i$ is an integer between 0 and 999 inclusive. – fleablood Feb 16 '16 at 5:00
• Argh. No. That isn't the question after all. The question IS: prove $a = a_0 \mod 8$. Which is a very similar question with the same reason for an answer. But it is a different question. – fleablood Feb 16 '16 at 5:07