Marking the prime points on a circle If you travel around a circle and mark all the points on the circle where the distance you travelled is a prime number, where you would go through many rotations*, do you end up marking the entire circumference of the circle as the number of points you pick goes to infinity? 
Or do you leave  "holes" in the circle by virtue of only marking the prime points of the circumference?
*I understand that you would have to define the circumference of your circle. Define it as a finite number, prime or not. How does the circumference you pick affect the answer/whether the circumference is prime or not?
Please let me know how I can explain this question further or whether this question does not make sense.
 A: Let's look at some examples, calling the starting point $0$
If you have a circle of circumference $2$, then you mark just two points. Point $0$ gets marked just once for the prime $2$, and all the odd primes fall at point $1$.
If the circumference is $6$ you miss points $0$ and $4$ and mark $2$ and $3$ just once, while $1$ and $5$ are each marked infinitely often.
This shows a link with Dirichelet's theorem on primes in arithmetic progression.
A different kind of question would be to take a circle of radius $1$ (circumference $2\pi$, so close to the $6$ of the previous example) and to ask whether the prime points form a dense set on the circumference of the circle.
A: EDIT: In view of the edit made to the question, it would appear that the following was not what OP had in mind. 
If you have a finite number, $n$, of points arrayed on a circle, and you go around picking every $m$th point, where $m$ is any number relatively prime to $n$ (doesn't matter whether $m$ is prime or not), you will pick up every one of the $n$ points, eventually. Otherwise, not. 
A: No, you'll never fill the circle. This is because:


*

*The set of prime is countable.

*But, the set of point on the circle is uncountable, i.e its cardinarlity is continuum.
It's like you can never fill the real line with rationals, because the set of rationals is countable, whereas the set of point in the real line is not.
You can have a quick look here.
