When the Clock's Hands Line Up

Written by Josh on November 30, 2014

A friend of mine challenged me a while ago to find all the times where the hands of the clock are 180° from each other. The easy example of this is 6:00, but there are a total of 11 times in 12 hours where this occurs.

This may seem like a simple problem at first, but it is anything but. Below, I will prove the times where the clock’s hour and minute hands align.


First, let’s define some variables.

\begin{align} H &= \text{hour (1-12)} \ M &= 60(H-int(H)) = \text{minute (1-60)} \ D_h &= 30H = \text{hour in degrees} \ D_m &= 6M = \text{minute in degrees} \end{align}

We can then reduce $D_m$ to terms of $H$.

\begin{align} D_m &= 6(60(H-int(H)) \ &= 360(H-int(H)) \end{align}

$H-int(H)$ uses the $int$ function (which is comparable to the floor function in math.h or FLOOR in Excel) to get the decimal part of a number. $int(x)$ returns the largest integer that is not larger than x. Essentially, it chops off the decimal part of a (positive) number. For example, int(\pi) = 3, int(0.999) = 0, and int(1) = 1. In this context, $H-int(H)$ removes the whole number part of $H$, leaving only the decimal part. (The rules are slightly different for negative numbers, but that doesn’t matter in this case.)

\begin{align} decimal(x) &= x - int(x) \ decimal(37.5) &= 37.5 - int(37.5) \ &= 37.5 - 37 = 0.5 \ decimal(6) &= 6 - int(6) \ &= 6 - 6 = 0 \end{align}

This is useful because the minute hand of a clock does not depend on the hour, only the portion past the hour. If we define the hour as a decimal number, such as 3.5 for 3:30, the minute hand is positioned at 0.5 * 360 = 180\degrees.

At this point, the angles of both hands have been defined; all we need to do is solve the equation for D_h - D_m = 180.

\begin{align} D_h - D_m &= 180 \ %30H - 360(H-int(H)) &= 180 \% \frac{30H - 360(H-int(H))}{30} &= \frac{180}{30} \ H - 12(H-int(H)) &= 6 \ H - 12H + 12int(H) &= 6 \ 12int(H) - 11H &= 6 \ 12int(H) &= 11H + 6 \ 11H + 6 &= 12int(H) \end{align}

We know that at 6:00 the hour and minute hands line up, so let’s plug in 6.0 for $H$.

\begin{align} 11(6) + 6 &= 12int(6) \ 66 + 6 &= 12*6 \ 72 &= 72 \end{align}

That works out, so we know the equation is correct. Let’s try 7:00.

\begin{align} 11(7) + 6 &= 12int(7) \ 77 + 6 &= 12*7 \ 83 &\ne 84 \end{align}

7:00 doesn’t quite work. We need one more on the left than the right. In most cases, that would be impossible, but ignore that for a minute. The left side of this equation is 11H + 6. We want this to equal 84, so let’s solve for $H$.

\begin{align} 11H + 6 &= 84 \ -6 &\hphantom{=} -6 \ \frac{11H}{11} &= \frac{78}{11} \\ H &= \frac{78}{11} = 7{\frac1{11}} \end{align}

So when H = 7{\frac1{11}}, $11H + 6 = 84$. Now we need to put our new value of $H$ into $12int(H)$. This gives us 12int(7{\frac1{11}}). The $int$ function takes off the fractional part, leaving us with $12*7 = 84$, the same as before.

\begin{align} H &= 7{\frac1{11}} \ 11 * 7{\frac1{11}} + 6 &= 12int(7{\frac1{11}}) \ 78 + 6 &= 12 * 7 \ 84 &= 84 \end{align}

7{\frac1{11}} = 7.\overline{09} \approx 7 \text{ hours, } 5.5 \text{ minutes}, so at about 7:05 the hour and minute hands are 180° apart. There is 1{\frac1{11}} hours between these two times where the hands line up, so it is a reasonable assumption that every 1{\frac1{11}} hours the hands line up.

\begin{align} 7{\frac1{11}} + 1{\frac1{11}} &= 8{\frac2{11}} \\ + 1{\frac1{11}} &= 9{\frac{3}{11}} \\ + 1{\frac1{11}} &= 10{\frac{4}{11}} \\ + 1{\frac1{11}} &= 11{\frac{5}{11}} \\ + 1{\frac1{11}} &= 12{\frac{6}{11}} \\ + 1{\frac1{11}} &= 13{\frac{7}{11}} \\ + 1{\frac1{11}} &= 14{\frac{8}{11}} \\ + 1{\frac1{11}} &= 15{\frac{9}{11}} \\ + 1{\frac1{11}} &= 16{\frac{10}{11}} \\ + 1{\frac1{11}} &= 17{\frac{11}{11}} = 18 \end{align}

At this point, we’ve gone until we hit a whole number. Most of these times, though, are after 12 o’clock. In these cases, we can apply the modulus function, $mod$. This gives us the remainder after dividing. For example, $13 \bmod 12 = 1$. It essentially says that the highest number is 11, then you wrap around to zero. (Formally, this is known as modular arithmetic.) We stopped adding at the whole number 18. $18 \bmod 12 = 6$, so we looped back around to six o’clock. If we plug any of these numbers back into the original equation for $H$, we will prove that at these times the hands of the clock are aligned.

\begin{align} 11 * 8{\frac2{11}} + 6 &= 12int(8{\frac{2}{11}}) \ 90 + 6 &= 12 * 8 \ 96 &= 96 \end{align}

When the hour is 8{\frac2{11}}, the hands line up. Multiplying \frac2{11} by 60 gives us around 10.9 minutes. At around 8:11, the hands are aligned. I won’t plug in every number here, but they all work out except for 18. While 18 doesn’t work, it is equivalent to 6:00, which we know works. This shows the domain of the equation: [6, 18). (In inequality notation, $6 \le H \lt 18$.)

In conclusion the clock hands align starting at 6:00 and every 1{\frac1{11}} hours afterwards. This works out to the following times: 6:00, 7:05, 8:11, 9:16, 10:22, 11:27, 12:33, 1:38, 2:44, 3:49, and 4:55. Five o’clock is the only hour where the hands do not align. After 4:55, they next align at 6:00, restarting the cycle. The times listed are approximate; I rounded the decimal part of the minute up to the nearest whole number.


You can graph this problem using WolframAlpha or by downloading this Grapher file if you use a Mac. Grapher is an application included with Mac OS X that nobody seems to know about that can graph most equations. It does not show solutions, though. WolframAlpha confirms the solutions I came to, as does an actual clock. I created the equations above without assistance, but initially solved the equation using WolframAlpha. Today, while writing this post, I figured out how to solve it without a graph.

This page uses MathJax and LaTeX to display equations nicely.

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