What will be the time when the hour, minute, and second hands will all coincide in between 6 O'Clock and 7 O'Clock?

How should I approach this type of question? Can anyone explain it?

  • $\begingroup$ 6:33:33? Maybe? $\endgroup$
    – Mithical
    Commented Jul 13, 2016 at 17:06

4 Answers 4



You really just need a couple equations

h = hi + min/60

h/12 = min/60 = sec/60

hi is the integer hour before the time you want to find (hi=6) h, min, sec are the actual hour, minute and second they intersect.

Substitute min/60 for h/12 and the equation becomes

h = hi + h/12 
(11/12)*h = hi 
h = 12/11*hi 
h = 12/11*6
**h = 72/11**

min = h/12*60
min = 6/11*60
sec = min = 6/11*60

h = 6.5454
min = 32.7272
sec = 32.7272

For an actual time drop the decimal points of the hour and minute and your left with

**final time is 6:32:32.727**


As per @Poolsharker answer these equations are incorrect another formula that should have been considered is

min = mini + sec/60
59/60*min = mini

From the previous final time mini = 32

min = 60/59*32
min = 32.5424 

which does not match the previous min time so they will not all 3 coincide


When we are a fraction $f$ of the way through a 12-hour "day" -- so $f=1$ means 12 hours have passed since noon or midnight -- the positions of the hands as fractions of a whole turn are: $f$ (for the hour hand), $12f$ (for the minute hand), and $720f$ (for the second hand), where integer differences are not visible. So for all three to coincide we need the differences $11f$ and $719f$ to be integers. Since 11 and 719 are coprime, this requires $f$ to be an integer, which means that the only 3-way coincidences happen with all hands pointing to the 12.

In particular, there is none between 6 and 7 o'clock.

(The bit about 11 and 719 being coprime may not be perfectly clear. Suppose $11f=m$ and $719f=n$ where $m,n$ are integers. Then $11n=719m$, so $11|719m$; since 11 is prime this requires either $11|719$ (no!) or $11|m$. So write $m=11k$; then $11f=m=11k$ so $f=k$ and $f$ is an integer.)

  • 1
    $\begingroup$ I think this is the most elegant solution out of the ones posted so far. $\endgroup$ Commented Jul 13, 2016 at 19:53

From my calculations...

all three hands will not intersect 'exactly' between 6 O'clock and 7 O'clock

The way I approached it isn't elegant but I think my calculations are correct:

Clock face makes up 360 degrees. When the minute hand moves 360 degrees, the hour hand has moved 30 degrees (1/12). So I did my calculations by moving the minute hand to where the hour hand is, and then adjusting the hour hand because of the minute hand movement.

So at 6:00, hour hand (hh) is at 180 degrees, minute hand (mh) is at 0.
Move mh 180 degrees, move hh 180 * (1/12) = 15 degrees.
Move mh 15 degrees, move hh 15 * (1/12) = 1.25 degrees.
Move mh 1.25 degrees, move hh 1.25 (1/12) = 0.104 degrees. and so on...

The calculations start repeating at the hour hand having moved 16.3636. So if we are 16.3636 degrees between the 6 and the 7 and 30 degrees = 5 minutes, then 16.3636 / (30/5) = 2.7272 minutes. Now we convert 0.7272 into seconds (0.7272 * 60) and we get 43.632. So I believe that the hour hand and minute hand we be nearly exact at 6:32:43.6, however that does leave the second hand outside. The best approximation would be 6:32:32.72 as mentioned by gtwebb.

Side note

While checking my calculations, I found this site calculating all the times the hour hand and minute hand intersect, good future reference http://www.kodyaz.com/articles/how-many-times-a-day-clocks-hands-overlap.aspx


There's no three pointers coincidence between 6 o'clock and 7 o'clock.

We can solve this problem using equations of motion.

The equation of motion in terms of angles of a pointer is:

$$\theta=\theta_i+\dot{\theta} t$$

where $\theta_i$ is the initial angle (at $t=0$), $\dot{\theta}$ the angular frequency and, $t$ the time.

We have three bodies: hours ($h$), minutes ($m$) and seconds ($s$) pointers.

Consider that the initial condition is at 6 o'clock ($t=0$). For a polar referential system where $\theta=0$ in 12 o'clock, with a direct clock direction for this angle, we know that at the initial condiditon, we have:

$$ \theta_h^i=\pi \text{ (rad)}, \hspace{5pt} \theta_m^i=0 \text{ (rad)}, \hspace{5pt} \theta_s^i=0 \text{ (rad)}$$

In relation to the angular frequencies, we have:

$$ \dot{\theta_h}=\frac{2\pi}{12\times3600} \text{(rad/s)}, \hspace{5pt} \dot{\theta_m}=\frac{2\pi}{3600} \text{ (rad/s)}, \hspace{5pt} \dot{\theta_s}=\frac{2\pi}{60} \text{ (rad/s)}$$

The condition for three pointers coincidence is:

$$\theta_h(t)=\theta_h(t)-2\pi n=\theta_h(t)- 2\pi m$$

where $n$ e $m$ are integers (the subtraction by $2\pi n$ and $2\pi m$ cancels the cyclicality of the pointers). We need that $n\in[0,1]$, $m\in[0,60]$ and $0\leq t\leq3600$ s for the coincidence be between 6 and 7 o'clock.

So, solving the system of equations of $\theta_h(t)=\theta_h(t)-2\pi n$ and $\theta_h(t)-2\pi n=\theta_h(t)- 2\pi m$ in order to $t$ am $m$ we get:

$$t=\frac{21600}{11}\left(1+2n\right) \text{ (s)},\hspace{15pt} m=\frac{719n+354}{11}$$

I tried to find the first value of $n$ that satisfies the conditions of $n$ and $m$ being positive integers, by substituition of $n=1,2,3...$ on the $m(n)$ expression and verified if $m$ was an integer. I got $n=5$ which means that $t=21600 s=6h$. Because $t=0$, at 6 o'clock, the coincidence happens at 12 o'clock, which means that it's impossible that it happens between 6 and 7 o'clock (remember that $t(n)$ is a crescent function with $n$, and $n=5$ is the lowest value for coincidence).


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