Remainder of all 3's

Let's call $x = \frac{n}{383}$ , then $n = 383x \equiv 333 \pmod{1000} \Rightarrow$ $47\cdot 383x = 18001x \equiv x \equiv 47\cdot 333 \equiv 651 \pmod{1000}$ Note, Bonus : Does x really exist?

After trying out the first few (4, 5, ... - digit "all 3") values, we will find, that n could possibly have many more digits.

Then we write n in the following form:

$k = \frac {n}{383} \iff n = 383k$

383k = 100000m + 33333 , $k, \ m \in \mathbb {Z}$

What we are looking for, is the number formed of the last 3 digits of k.

Now, what we basically have is a simple multiplication with missing digits:

$383 × ... \square \square \square = ... 33 333$

We can also turn it into a division, which we will be doing it in an unusual way, "backwards" (starting from the last digit of the number, we divide; if the divisor is known to be a factor, we can do so):

$... 33 333 ÷ 383 = ... \square \square \square$

Let's solve this:

$... 33 333 ÷ 383 = ... \boxed {6} \boxed {5} \boxed {1}$

-383 × 1

...32950

-3830 × 5

...13800

We got these $\overline {abc}$ digits by considering the possible digits, when multiplying a number by 3:

3c = ...3, the only integer solution between for 0 ≤ c ≤ 9 (decimal digit) is c = 1

3b = ...5 (5 is the place value of the Tens in ...2950), the only integer solution between for 0 ≤ b ≤ 9 (decimal digit) is b = 5

3a = ...8 (8 is the place value of the Hundreds in ...13800), the only integer solution between for 0 ≤ a ≤ 9 (decimal digit) is a = 6

Therefore, the remainder when dividing k by 1000 (the last 3 digits of k) is: $\boxed {651}$

Okay well what's missing from other solutions is the fact that $x$ really exists. Well, to do the bonus question from @Guillermo Templado note that

$10^{p-1} \equiv 1 (\text{mod p)}$ so $10^{382} \equiv 1 (\text{mod 383})$

i.e $383|9999....9999$ where there are 382 9s.

Now $383|3333....3333$ (382 threes) because if $p|ab$ then either $p|a$ or $p|b$ where $p$ is prime, then sub in $p=383, a=3, b=3333...3333$ and we're done.