Harvard-MIT Mathematics Tournament

Solution:

A tricky thing about this problem may be that the angles on the two clocks might be reversed and would still count as being the same (for example, both angles could be $90^{\circ}$, but the hour hand may be ahead of the minute hand on one clock and behind on the other).

Let $x$, $-12 \leq x < 12$, denote the number of hours since noon. If we take $0^{\circ}$ to mean upwards to the "XII" and count angles clockwise, then the hour and minute hands of the correct clock are at $30x^{\circ}$ and $360x^{\circ}$, and those of the slow clock are at $15x^{\circ}$ and $180x^{\circ}$. The two angles are thus $330x^{\circ}$ and $165x^{\circ}$, of course after removing multiples of $360^{\circ}$ and possibly flipping sign; we are looking for solutions to
$$330x^{\circ} \equiv 165x^{\circ} \pmod{360^{\circ}} \text{ or } 330x^{\circ} \equiv -165x^{\circ} \pmod{360^{\circ}}$$
In other words,
$$360 \mid 165x \text{ or } 360 \mid 495x.$$
Or, better yet,
$$\frac{165}{360}x = \frac{11}{24}x \text{ and/or } \frac{495}{360}x = \frac{11}{8}x$$
must be an integer. Now $x$ is any real number in the range $[-12, 12)$, so $11x/8$ ranges in $[-16.5, 16.5)$, an interval that contains 33 integers. For any value of $x$ such that $11x/24$ is an integer, of course $11x/8 = 3 \times (11x/24)$ is also an integer, so the answer is just 33.