Will You Be Right on Time?

A standard analog clock has an hour hand, a minute hand, and $60$ minute markers, $12$ of which are also hour markers. At a certain time, both hands point directly at minute markers (either of which may also be an hour marker), and the hour hand is $13$ markers ahead of the minute hand, measured clockwise. At what time does this happen?

The Fiddler, Zach Wissner-Gross, April 24, 2026(original post)

Solution

Number the markers $0$ to $59$. The minute hand points at a marker only at a whole minute $m$, sitting at marker $m$. At $h$ o’clock and $m$ minutes the hour hand has moved $5h + m/12$ markers from the top, a whole marker only when $12 \mid m$, i.e. $m \in \{0,12,24,36,48\}$. The clockwise lead is then $$\Bigl(5h + \tfrac{m}{12}\Bigr) - m \equiv 13 \pmod{60}.$$ Trying the five allowed minute values, only $m = 24$ gives an integer hour: $5h + 2 - 24 \equiv 13$, so $h = 7$. At $7\!:\!24$ the minute hand is at marker $24$, the hour hand at $37$, a lead of exactly $13$: $$\boxed{7\!:\!24}.$$

The computation

Test the marker condition directly: for each hour and each minute at which the hour hand also lands on a marker ($12\mid m$), check whether the hour hand leads the minute hand by exactly $13$ markers.

sol = [(h, m) for h in range(12) for m in (0, 12, 24, 36, 48)
       if ((5*h + m//12) - m) % 60 == 13]
print(sol)   # [(7, 24)]

Extra Credit

Is there a time when the hour, minute, and second hands, each sweeping continuously, together form two right angles, with one of the hands in the middle: that hand making $90^\circ$ with each of the other two, which then lie $180^\circ$ apart?

Solution

There is no such exact time. Writing the three hand angles as functions of the time and searching the full twelve-hour cycle, the closest the three hands come is at about $$\boxed{5\!:\!10\!:\!55.92},$$ where the minute hand sits between the hour and second hands: those two are $180.06^\circ$ apart, and the minute hand is $89.88^\circ$ and $90.06^\circ$ from them. The residual is small but nonzero, and refining the search does not drive it to zero, so the configuration is never achieved exactly.

The computation

Sweep the twelve-hour cycle in fine steps. At each instant, for each choice of middle hand, measure how far the configuration is from “outer hands $180^\circ$ apart, middle hand $90^\circ$ from each.” The smallest residual is small but stays nonzero.

import numpy as np
t = np.arange(0, 43200, 0.005)                 # seconds over 12 hours
ang = [(6*t) % 360, (0.1*t) % 360, (t/120) % 360]   # second, minute, hour
def gap(a, b):
    d = np.abs((a - b) % 360); return np.minimum(d, 360 - d)
best = np.full_like(t, 1e18)
for mid in range(3):
    o = [i for i in range(3) if i != mid]
    err = ((gap(ang[o[0]], ang[o[1]]) - 180)**2
           + (gap(ang[mid], ang[o[0]]) - 90)**2
           + (gap(ang[mid], ang[o[1]]) - 90)**2)
    best = np.minimum(best, err)
i = int(best.argmin()); s = t[i]
print(f"{int(s//3600)%12 or 12}:{int(s%3600//60):02d}:{s%60:06.3f}",
      f"min sq.error = {best.min():.4f}")   # 5:10:55.920  ~0.024 (not zero)