»  Solutions to puzzles in my VDARE.com monthly Diary

  December 2020


In the Math Corner of my December Diary I offered the following brainteaser, borrowed from the December 29th "Mind-Bender for the Quarantined" from Dr. Peter Winkler at the National Museum of Mathematics. Feeling somewhat guilty about taking two brainteasers in succession from Dr. Winkler, I prefaced by writing:

And yes, I've poached it from their subscription list. By way of penance, I urge anyone who likes this kind of thing to sign up with them at mindbenders.momath.org. It's free!

OK, the puzzle: The Poorly-Designed Clock.

The hour and minute hands of a certain clock are indistinguishable. How many moments are there in a day when it is not possible to tell from this clock what time it is?


•  Preliminaries

In what follows I shall refer to the position of hands on the clock face in degrees clockwise (!) round from the 12 o'clock mark at the top. So if a hand is at the 3 o'clock mark, it's at 90°; at the 6 o'clock mark, it's at 180°; at the 9 o'clock mark, it's at 270°; and so on. If it's at the 12 o'clock mark I may call that 0° or I may call it 360°, whichever I think sheds more light.

Now let's follow the hands through a 12-hour cycle from midnight to noon. (I shall assume we have access to daylight so that we know which is which.) Where are the hands at M minutes after H o'clock?

Well, the minute hand is at 6M degrees — that's easy.

The hour hand is trickier. At precisely H o'clock it's at 30H degrees. It then progresses through the next 60 minutes to 30H + 30 degrees, so its speed is half a degree per minute. At M minutes after H o'clock, therefore, it's at 30H + M/2 degrees.

A good warm-up exercise is to figure when the hour and minute hands are at exactly the same place. Answer: When 60M = 30H + M/2. That simplifies to 11M = 60H. So:

Note there are just eleven of those instants; and that, turning back now to the original problem, we know the exact time at each one of them even if the two hands are identical.

OK, to the original problem of that poorly-designed clock.

•  Reader solution

Something here is conceptually difficult for a lot of readers. Surely (said many) the only time you can tell what time it is on this clock are at those eleven moments when the two hands precisely coincide. At any other time you can't tell!

That's wrong, as some doodling will show you. Suppose, for example, that one of the hands is at the 12-minute mark on the clock face (i.e. at 72°, a bit past the 2 o'clock mark) while the other is at the 36-minute mark (at 216°, a bit past the 7 o'clock mark). That is unambiguously 7:12 and the first hand mentioned is unambiguously the minute hand. If it were the hour hand it would be showing 2/5ths of the way — which is to say, 24 minutes — from 2 o'clock to 3 o'clock. Then the other hand, the minute hand, would be at 144°, the 24-minute mark, not 216°, the 36-minute mark, see?

You almost always can tell what time it is, even without the help of the hands being different lengths.

Well, I diddled with this in an inconclusive way, i.e. kept coming up with different answers. Readers did much better. The most ingenious reader solution was graphical. Here's the graph. (Better resolution here.)


You have to not mind that the horizontal and vertical scales are different. Sorry about that.

Both scales go from 0 to 360. A point on this graph with co-ordinates (h, m) represents a configuration where the hour hand is at h degrees, the minute hand at m degrees.

Let's take a trip across this graph with the hour hand, starting at midnight. That's (0,0). By 1 o'clock we've reached (30,360), which is the same as (30,0). We have traversed that nearly-vertical red line over at the left.

From 1 o'clock to 2 o'clock we traverse the second, blue, nearly-vertical line. (This is one of those graphs that makes more sense if you imagine it rolled into a cylinder, the top and bottom edges taped together.) And so on, for twelve nearly-vertical lines in a 12-hour spell.

Now let's take a trip with the minute hand. From midnight to 1 o'clock we traverse the lowest, purple, nearly-horizontal line, from (0,0) to (360,30).

From 1 o'clock to 2 o'clock we traverse the second, black, nearly-horizontal line. (So that cylinder should be bent round to join the left edge to the right edge — a torus!) And so on, for twelve nearly-horizontal lines in a 12-hour spell.

The trick now it to see that an ambiguity occurs where these lines cross. Look at those intersection points I've marked with wee circles, for example. They are symmetrical about the y = x diagonal (thick orange line). The lower one has co-ordinates (158.6, 103.2), to one place of decimals. The other is of course (103.2, 158.6). Translating from degrees to the marked minutes on the clock face: One hand is at 17.2 minutes, a bit past the 3 o'clock mark, the other is at 26.4 minutes, a tad past the 5 o'clock mark. The time could be 17.2 minutes past 5 or 26.4 minutes past 3, you can't tell, to any number of decimal places.

So the question now resolves to: How many of those intersections are there? Answer: 144. Twelve of them, however — two actually the same, (0,0) and (360,360) — are on that diagonal; and those points are the eleven occasions when the two hands precisely coincide, when we do know what time it is.

So in this 12-hour period we've mapped, there are 132 moments when we can't tell what time it is. In a day, therefore, there are 264. (Again, I'm assuming we know whether it's a.m. or p.m., I guess by looking out the window at the quality of the light.)

Answer: 264.

•  However, Dr. Winkler gets the same result a different way.


Note that we can tell what time it is when the hour and minute hands coincide, even though we can't tell which hand is which; this happens 22 times a day, since the minute hand goes around 24 times while the hour hand goes around twice, in the same direction.

This reasoning turns out to be good practice for the proof. Imagine that we add to our clock a third "fast" hand, which starts at 12 at midnight and runs exactly 12 times as fast as the minute hand.

Now we claim that whenever the hour hand and the fast hand coincide, the hour and minute hands are in an ambiguous position. Why? Because later, when the minute hand has traveled 12 times as far, it will be where the fast hand (and thus also the hour hand) is now, while the hour hand is where the minute hand is now.

Conversely, by the same reasoning, all ambiguous positions occur when the hour hand and fast hand coincide.

So, we need only compute the number of times a day this coincidence occurs. The fast hand goes around 12 x 12 x 2 = 288 times a day, while the hour goes around just twice, so this happens 286 times.

Of these 286 moments, 22 are times when the hour hand and minute hand (thus all three hands) are coincident, leaving 264 ambiguous moments.

(Dr. Winkler further adds: This delightful problem was posed by Andy Latto, a Boston-area software engineer, at the Gathering for Gardner IV, one of a series of conferences held in Atlanta in honor of Martin Gardner. It can be solved algebraically or geometrically, with sufficient care and patience, but the above proof, supplied to Andy by Michael Larsen, a mathematics professor at Indiana University, is the slickest argument I've seen. The idea of a third hand, instead of a second clock, came from David Gale.)