Consider a standard twelve-hour clock whose hour and minute hands move continuously. Let $m$ be an integer, with $1 \leq m \leq 720$. At precisely $m$ minutes after 12:00, the angle made by the hour hand and minute hand is exactly $1^{\circ}$. Determine all possible values of $m$.
The minute hand makes a full revolution of $360^{\circ}$ every 60 minutes, so after $m$ minutes it has swept through $\frac{360}{60} m=6 m$ degrees. The hour hand makes a full revolution every 12 hours (720 minutes), so after $m$ minutes it has swept through $\frac{360}{720} m=m / 2$ degrees. Since both hands started in the same position at 12:00, the angle between the two hands will be $1^{\circ}$ if $6 m-m / 2= \pm 1+360 k$ for some integer $k$. Solving this equation we get

$$
m=\frac{720 k \pm 2}{11}=65 k+\frac{5 k \pm 2}{11}
$$

Since $1 \leq m \leq 720$, we have $1 \leq k \leq 11$. Since $m$ is an integer, $5 k \pm 2$ must be divisible by 11, say $5 k \pm 2=11 q$. Then

$$
5 k=11 q \pm 2 \Rightarrow k=2 q+\frac{q \pm 2}{5}
$$

If is now clear that only $q=2$ and $q=3$ satisfy all the conditions. Thus $k=4$ or $k=7$ and substituting these values into the expression for $m$ we find that the only possible values of $m$ are 262 and 458.