Number of chords to be drawn to return the starting point Given a pair of concentric circles, chords AB,BC,CD,...   of the outer circle are drawn such that they all touch the inner circle. If ∠ABC = 75°;how many chords can be drawn before returning to the starting point?

Thought to start by joining points of contact with the centre,but couldn't got any further idea. Any help is appreciated.
Question Source: IOQM Olympiad
 A: Method using modular arithmetic
Notice that $\triangle AOB$ is isosceles and therefore $\angle AOB = 105^\circ$.
Each successive chord adds $105^\circ$ to this angle so we want to know when this would be a multiple of $360^\circ$. In other words, we want
$$
105n \equiv 0 \pmod {360}
$$
This implies
$$
7n \equiv 0 \pmod { 24}
$$
Since $\gcd{(7, 24)} = 1$, the minimum $n$ that satisfies this equation is $\boxed{n = 24}$
Method using Complex Numbers
Without loss of generality, assume that the outer circle has a radius of $1$. We can rotate the initial diagram such that the point $A$ corresponds to the point $(1, 0)$ on the coordinate plane.
Every time we draw a chord, the angle increases by $105^\circ$, so essentially we are rotating the complex number by $105^\circ$ each time.
Let $z = e^{i \cdot 0}$. The next chord point (in our case $B$) will occour at $z = e^{i \cdot \frac{7\pi}{12}}$ and $C$ will occour at $z = e^{i \cdot \frac{7\pi}{6}}$ and so on.
We want to know, when $z = e^{i \cdot \frac{7\pi}{12} \cdot n}$ will return to the point $A$ at $(1, 0)$. Clearly, this will happen when:
$$
\frac{7\pi}{12}\cdot n = 2\pi k
$$
In other words, for some integer $n$ we want:
$$
7n = 24k
$$
The above simply corresponds to the modular equation:
$$
7n \equiv 0 \pmod {24}
$$
Since $\gcd(7,24)=1$, the minimum n that satisfies this equation is $\boxed{n=24}$.
Verification
If we use complex numbers, we actually know the coordinates of each point. For example,
$$
e^{i \cdot \frac{7\pi}{12}} \implies (\cos(105^\circ), \ \sin(105^\circ))
$$
Using this we can be sure that the only solution occurs when $n = 24$.
