[Number Theory]
Problem: Compute the least positive integer $n$ such that the set of angles

$$
\left\{123^{\circ}, 246^{\circ}, \ldots, n \cdot 123^{\circ}\right\}
$$

contains at least one angle in each of the four quadrants.

Solution: The first angle is $123^{\circ}$, which is in Quadrant II, the second $\left(246^{\circ}\right)$ is in Quadrant III, and the third is in Quadrant I, because $3 \cdot 123^{\circ}=369^{\circ} \equiv 9^{\circ} \bmod 360^{\circ}$. The missing quadrant is IV, which is $270^{\circ}-246^{\circ}=24^{\circ}$ away from the second angle in the sequence. Because $3 \cdot 123^{\circ} \equiv 9^{\circ} \bmod 360^{\circ}$, the terminal ray of the $(n+3)^{\mathrm{rd}}$ angle is rotated $9^{\circ}$ counterclockwise from the $n^{\text {th }}$ angle. Thus three full cycles are needed to reach Quadrant IV starting from the second angle: the fifth angle is $255^{\circ}$, the eighth angle is $264^{\circ}$, and the eleventh angle is $273^{\circ}$. So $n=11$.

Final answer: 11