TITLE: Minute hand and hour hand interchange
QUESTION [9 upvotes]: A person who left home between $4$ p.m. and $5$ p.m. returned between $5$ p.m.
and $6$ p.m. and found that the hands of his watch has exactly changed places. When
did he go out?
My attempt: The dial of a clock is divided into $60$ equal divisions. In one hour, the minute
hand makes one complete revolution, i.e., it moves through $60$ divisions and the hour
hand moves through $5$ divisions.
Suppose, when the man went out, the hour hand was $x$ divisions ahead of the point
labeled $12$ on the dial, where $20 < x < 25$ (as he went out between $4$ p.m. and $5$ p.m.).
Also suppose, when the man returned, the hour hand was $y$ divisions ahead of zero
mark and $25 < y < 30$.
Since the minute hand and hour hand exactly interchanged places during the interval
that the man was out, the minute hand was at y when he went out and at $x$ when he
returned.
Since the minute hand moves $12$ times as fast as the hour hand, we can say that the angle that the minute hand sweeps will be $12$ times that swept by the hour hand. But, I am not able to express this in terms of $x$ and $y$.
Any constructive hint is appreciated.

REPLY [3 votes]: HINT:
Let's consider a time as a number between $0$ and $12$.
At time $x=$ the position of the hour hand, the position of the minute hand equals $12 \{x\}$, where $\{x\}$ is the fractional part of $x$. If $x = a_0. a_1 a_2 \ldots_{(12)}$ written in base $12$, then $12\{x\}= a_1.a_2 a_3 \ldots_{(12)}$.
Let $x$, $y$ be the departure, and arrival times. Then we have the system
$$
y= 12\{x\}\\  
x = 12\{y\}
$$
and the conditions $4\le x\le 5$, $5\le y \le 6$. We conclude that in base $12$ we have
$$ x= 4.5454\ldots_{(12)} \\
y = 5.4545\ldots_{(12)}$$