clock related challenge A person leaves his house between 4.00 and 5.00 pm. He carefully notes the position of the minute hand and hour hand when he leaves the house. He returns back between 7.00 and 8.00 pm.He notices that the hour hand and the minute hand have exactly interchanged their positions. What time did he leave his house?
 A: $\def\deg{{^\circ}}$ The key thing with these clock puzzles is that the hands are not independent; when the minute hand goes forward by some angle $x$, the hour hand also goes forward by the amount $x\over 12$.  The hour hand turns $360\deg$ in 12 hours, so $30\deg$ in one hour, and that is $\frac12\deg$ per minute. The minute hand turns $6\deg$ in one minute.
If the time is h:mm, where $0\le mm \lt 60$ and $0\le h \lt 12$, then the minute hand is at position $6\deg\cdot mm$ degrees and the hour hand is at position $h\cdot30\deg+mm\cdot{1\over 2}\deg$ degrees—the $h\cdot30\deg$ term is how far it has turned to get the the beginning of hour $h$, and then the $mm\cdot{1\over 2}\deg$ is how many degrees it has turned in the $mm$ minutes since the beginning of the hour. 
Now we need two times, 4:mm and 7:nn, where the hand positions are reversed. The first has the minute hand at $6\deg mm$ and the hour hand at $120\deg + mm\cdot{1\over 2}\deg$. The second has the minute hand  at $6\deg nn$ and the hour hand at $210\deg + nn\cdot{1\over 2}\deg$. So we need:
$$
\begin{eqnarray}
6\deg mm & = & 210\deg + nn\cdot{1\over 2}\deg \\
6\deg nn & = & 120\deg + mm\cdot{1\over 2}\deg 
\end{eqnarray}
$$
Dividing through by $6\deg$ gives:
$$
\begin{eqnarray}
mm & = & 35  + {nn\over 12} \\
nn & = & 20 + {mm\over 12} 
\end{eqnarray}
$$
By substitution we get $mm = 36+\frac{132}{143}$ minutes, and $nn = 23+\frac{11}{143}$ minutes. The two times are therefore approximately 4:36:55.359  and 7:23:04.615.
A: There are 7 pairs of different positions of the clock hands, given for n=0,...,6
by 12*n/13 and 12-12*n/13, in HOURS, not minutes.  The first pair is 0 0
The previous example corresponds to n=5.
Proof:
Observe that the minute hand moves 12 times faster than the hour hand.  The total path length traversed by the minute hand is 12*t.
It is also equal to n total rotations MINUS the path length of the hour hand, which is t.
Therefore 12*t = n -t, which simplifies to t = n/13.
This is the fractional length of the circular arc of the clock face.  The corresponding hour value is twelve times this value, 12*n/13.
The symmetrical hour is 12-12*n/13, symmetrical with respect to the 12<->6 vertical axis of the clock face. 
