You probably could have just asked me to arrive at the right time ...

A standard analog clock includes an hour hand, a minute hand, and 60 minute markers, 12 of which are also hour markers.

At a certain time, both the hour hand and minute hand are both pointing directly at minute markers (either of which could also be an hour marker). The hour hand is 13 markers ahead (i.e., clockwise) of the minute hand. At what time does this occur?

Let's set up the problem. We can represent each hand by its phase on the clockface, that is the proportion of the entire rotation that the hand has already completed. Let $h(t)$ and $m(t)$ be the phase of the hour and minute hands, respectively, at time $t,$ measured in hours. Since the minute hand makes one full rotation every hour, we see that the minute hand's phase at time $t$ is always the fractional part of $t$, that is $m(t) = t - \lfloor t \rfloor.$ Similarly, since in one hour the hour hand only travels 1/12th of a rotation that we have $h(t) = t/12 - \lfloor t / 12 \rfloor.$

The difference in the phases can be transformed into the angle between the hour and minute hands, so we could look at $$\delta(t) = h(t) - m(t) = \left\lfloor t \right\rfloor - \left\lfloor \frac{t}{12} \right\rfloor - \frac{11t}{12}.$$ However, since the angle between two hands is always less than or equal to $180^\circ,$ or phase of $1/2,$ we instead should look at $$\Delta(t) = \delta(t) - \lfloor \delta(t) + 1/2 \rfloor = \begin{cases} -\frac{11}{12}t, & 0 \leq t \lt \frac{6}{11} \\ 1 - \frac{11}{12}t, & \frac{6}{11} \leq t \lt \frac{18}{11}; \\ \dots, & \\ k - \frac{11}{12}t, & \frac{6(2k-1)}{11} \leq t \lt \frac{6(2k+1)}{11}; \\ \dots, & \\ 11-\frac{11}{12}t, & \frac{138}{11} \leq t \lt 12. \end{cases}$$ We can see graphically that $\Delta$ is cyclical and that there are 11 full cycles within the 12 hour cycle of the clock face, that is, a period of $\tau = 12/11.$

Let's first try to find the solution to $$\Delta(t_0) = 1 - \frac{11t_0}{12} = \frac{13}{60},$$ with $6/11 \leq t_0 \lt 18/11.$ We see that we have $$t_0 = \frac{47}{55}.$$ Since $\Delta$ is periodic, we see that all of the solutions of the $\Delta = 13/60$ are of the form $$t_k = t_0 + k \tau = \frac{47}{55} + \frac{12k}{11} = \frac{47 + 60k}{55},$$ for $k = 0, 1, \dots, 10,$ as summarized in the table below.

$k$	$t_k$	hh:mm:ss
0	47/55	12:51:36
1	107/55	01:56:44
2	167/55	03:02:11
3	227/55	04:07:38
4	287/55	05:13:05
5	347/55	06:18:33
6	407/55	07:24:00
7	467/55	08:29:27
8	527/55	09:34:55
9	587/55	10:40:21
10	647/55	11:45:49

So we see that the appropriate time where the hour and minute hands are pointing at minute markers that are 13 markers apart is at 7:24.