5. Tim has a working analog 12 -hour clock with two hands that run continuously (instead of, say, jumping on the minute). He also has a clock that runs really slow-at half the correct rate, to be exact. At noon one day, both clocks happen to show the exact time. At any given instant, the hands on each clock form an angle between $0^{\circ}$ and $180^{\circ}$ inclusive. At how many times during that day are the angles on the two clocks equal?
Answer: 33
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 $30 x^{\circ}$ and $360 x^{\circ}$, and those of the slow clock are at $15 x^{\circ}$ and $180 x^{\circ}$. The two angles are thus $330 x^{\circ}$ and $165 x^{\circ}$, of course after removing multiples of $360^{\circ}$ and possibly flipping sign; we are looking for solutions to
$$
330 x^{\circ} \equiv 165 x^{\circ} \quad\left(\bmod 360^{\circ}\right) \text { or } 330 x^{\circ} \equiv-165 x^{\circ} \quad\left(\bmod 360^{\circ}\right) .
$$

In other words,
$$
360 \mid 165 x \text { or } 360 \mid 495 x \text {. }
$$

Or, better yet,
$$
\frac{165}{360} x=\frac{11}{24} x \text { and } / \text { or } \frac{495}{360} x=\frac{11}{8} x
$$
must be an integer. Now $x$ is any real number in the range $[-12,12)$, so $11 x / 8$ ranges in $[-16.5,16.5)$, an interval that contains 33 integers. For any value of $x$ such that $11 x / 24$ is an integer, of course $11 x / 8=3 \times(11 x / 24)$ is also an integer, so the answer is just 33 .