Clock Puzzles

Aclock face is a small machine for turning time into angles, and most of Dudeney’s clock puzzles ask us to run the machine backwards. The whole theory fits on a postcard. The rim has sixty minute divisions. The minute hand goes once round, sixty divisions, every hour, and the hour hand covers five divisions in the same hour, so the minute hand gains fifty-five divisions an hour on the hour hand, which is $\tfrac{11}{12}$ of a turn. The hands move smoothly and continuously, never in jumps, and the second hand of a watch goes round once a minute.

Throughout, a time is measured in minutes (or seconds) after twelve o’clock, and a hand’s position in divisions clockwise from XII. At $t$ minutes after twelve the minute hand stands at $t$ divisions and the hour hand at $t/12$, both counted modulo 60. Almost every answer comes from setting two of these positions equal, or a fixed distance apart, and solving one linear equation. Fractions of a minute are exact here: $5\tfrac{5}{11}$ minutes means just that, and the answers keep them. The last three puzzles of Dudeney’s section are about calendars and clocks that keep bad time rather than about hands.

No. 57. What Was the Time?

No. 58. A Time Puzzle

No. 59. A Puzzling Watch

No. 60. The Wapshaw’s Wharf Mystery

No. 61. Changing Places

No. 62. The Club Clock

No. 63. The Stop-Watch

No. 64. The Three Clocks

No. 65. The Railway Station Clock

No. 66. The Village Simpleton

What Was the Time?

The crux is that the professor’s recipe falls as the day goes on while the clock rises, so they can agree only once in each half-day. Let it be $t$ hours after noon, with $t$ below 12 so that the clock reads $t$. A quarter of the time since noon is $t/4$, and noon to-morrow is $24-t$ hours away, so $$\frac{t}{4}+\frac{24-t}{2}=t, \qquad\text{that is}\qquad 12=\frac{5t}{4},\qquad t=9\tfrac{3}{5}.$$ It was 9.36 in the evening. As a check, a quarter of 9 h 36 min is 2 h 24 min, half of the remaining 14 h 24 min is 7 h 12 min, and the two add up to 9 h 36 min.

Could the professor have been speaking after midnight? Then the clock reads $t-12$, and noon to-morrow, strictly the noon of the day after he speaks, is $48-t$ hours off. The equation becomes $t/4+(48-t)/2=t-12$, so $t=28\tfrac45$, which is not a time before the next noon. There is no morning answer. A looser speaker who meant only “the next noon” would get $t/4+(24-t)/2=t-12$, so $t=19\tfrac15$, which is 7.12 in the morning: a quarter of 19 h 12 min is 4 h 48 min, half of the 4 h 48 min still to noon is 2 h 24 min, and together they make 7 h 12 min. But a man speaking at breakfast calls that noon “to-day”, so Dudeney’s single answer is the right one for his words.

Answer 9.36 p.m.

A Time Puzzle

This is one linear equation once both times are measured from the same zero. Let it be $x$ minutes before six. Then it is now $180-x$ minutes past three, and fifty minutes ago it was $130-x$ minutes past three. The puzzle says $130-x=4x$, so $x=26$. It is now 5.34; fifty minutes ago it was 4.44, which is 104 minutes past three, four times 26.

Notice that “minutes past three” has to be allowed to run beyond sixty. If it were held below sixty, then $130-x<60$ would force $x>70$ and $4x<60$ would force $x<15$, so there would be no answer at all.

Answer 26 minutes

A Puzzling Watch

The crux is that the meetings of the hands are counted by the watch’s own gears. The minute hand gains fifty-five divisions an hour on the hour hand and must gain sixty to lap it, so on any watch’s own dial the hands meet every $\tfrac{60}{55}=\tfrac{12}{11}$ of an hour, which is $65\tfrac{5}{11}$ minutes as that watch shows them. This holds however fast or slow the watch runs.

So if the sixty-five minutes were read off the same watch, the friend’s observation is impossible. The only sensible reading is that he timed the meetings by a correct clock. Then the watch shows $65\tfrac{5}{11}$ minutes while $65$ true minutes pass: it gains, by $\tfrac{5}{11}$ of a minute in 65 minutes, or $$\frac{5}{11}\cdot\frac{60}{65}=\frac{60}{143}\ \text{minute}$$ in each true hour, about 25 seconds. Measured per hour of the watch’s own face the gain is $\tfrac{5}{11}\cdot 60\big/\tfrac{720}{11}=\tfrac{5}{12}$ of a minute. Dudeney gives the first figure and makes the same point about the two readings.

Answer it gains $\tfrac{60}{143}$ min per true hour

The Wapshaw’s Wharf Mystery

The hands of an accurate watch are together every $\tfrac{12}{11}$ of an hour, that is $\tfrac{720}{11}=65\tfrac{5}{11}$ minutes, so in twelve hours they meet eleven times, at $\tfrac{720k}{11}$ minutes after twelve for $k=0,1,\dots,10$. The crux is that the second hand tells these eleven meetings apart. The fraction of a minute at the $k$th meeting is the fractional part of $\tfrac{5k}{11}$, so the second hand stands at $$60\cdot\Bigl\{\frac{5k}{11}\Bigr\}=\frac{300k \bmod 660}{11}\ \text{seconds}.$$ Because 5 and 11 have no common factor, the eleven values of $5k \bmod 11$ are all different, and the second-hand readings are $$\begin{gather*} 0,\ 27\tfrac{3}{11},\ 54\tfrac{6}{11},\ 21\tfrac{9}{11},\ 49\tfrac{1}{11},\ 16\tfrac{4}{11},\\ 43\tfrac{7}{11},\ 10\tfrac{10}{11},\ 38\tfrac{2}{11},\ 5\tfrac{5}{11},\ 32\tfrac{8}{11} \end{gather*}$$ for $k=0,\dots,10$. Only $k=4$ lands just past the forty-ninth second. The watch stopped at $4\cdot\tfrac{720}{11}=261\tfrac{9}{11}$ minutes after twelve, which is 4 h 21 min $49\tfrac{1}{11}$ s. The dial cannot say morning or afternoon; a night-time robbery points to the morning. Dudeney agrees.

He adds a literary footnote. Guy Boothby opens Across the World for a Wife with the hands of a mantelpiece clock “joined” at twenty past four. By the list above they cannot be together then: the nearest meeting is 1 min $49\tfrac{1}{11}$ s later.

Answer 4 h 21 min $49\tfrac{1}{11}$ s

Changing Places

Name each time by its hour and minutes. At $a$ hours $m$ minutes the minute hand is at $m$ divisions and the hour hand at $5a+\tfrac{m}{12}$. At $b$ hours $n$ minutes they are at $n$ and $5b+\tfrac{n}{12}$. The hands change places when $$n=5a+\frac{m}{12},\qquad m=5b+\frac{n}{12}.$$ The crux is that these two equations have exactly one solution for each pair of hours. Substituting the first into the second gives $m=5b+\tfrac{5a}{12}+\tfrac{m}{144}$, so $$m=\frac{720b+60a}{143},\qquad n=\frac{720a+60b}{143}.$$ These are real minutes provided $m<60$, that is $12b+a<143$, which holds for all hours from 0 (twelve o’clock) to 11 except $a=b=11$. When $a=b$ the two times are the same moment and the hands are simply together, which is no change of places. So each unordered pair of different hours gives exactly one pair of times, and a twelve-hour round has $\binom{12}{2}=66$ of them. With both times between three o’clock and midnight, both hours come from $3,4,\dots,11$, and there are $\binom{9}{2}=36$ pairs. The earliest is $a=3$, $b=4$: 3 h $21\tfrac{57}{143}$ min with 4 h $16\tfrac{112}{143}$ min.

For the second question, the minute hand at the time in hour $a$ stands at $\tfrac{720b+60a}{143}$, and its distance from IX, which is 45 divisions, is $$\frac{|720b+60a-6435|}{143}=\frac{60\,\lvert 12b+a-107\tfrac14\rvert}{143}.$$ Since $12b+a$ is a whole number, the best it can do is 107, a quarter away, and with $a$ and $b$ between 3 and 11 the only way to make 107 is $b=8$, $a=11$. That is the time 11 h $44\tfrac{128}{143}$ min, partnered with 8 h $58\tfrac{106}{143}$ min, and its minute hand is $\tfrac{60}{143}\cdot\tfrac14=\tfrac{15}{143}$ of a division short of IX. Dudeney has 36 and this time. If one asked only that at least one time of each pair fall after three o’clock, the count would be 63 instead.

Dudeney closes with a question from a Civil Service column: how soon after twelve does a clock whose two hands are the same length become ambiguous? The answer is the first changing-places time, $a=0$, $b=1$, at $5\tfrac{5}{143}$ minutes past twelve, because its partner is a moment that reads just as sensibly with the hands swapped.

Answer 36 pairs; 11 h $44\tfrac{128}{143}$ min

The Club Clock

Work in seconds after twelve and in divisions of the dial. At $t$ seconds the second hand is at $t$, the minute hand at $t/60$ and the hour hand at $t/720$, all modulo 60. The crux is that “midway” is a single equation: the second hand lies on a line bisecting the two hands when twice its position equals the sum of theirs, modulo 60, $$2t-\frac{t}{60}-\frac{t}{720}=\frac{1427\,t}{720}\equiv 0 \pmod{60},$$ so $t=43200k/1427$ for a whole number $k$. That equation cannot tell the two halves of the bisecting line apart. One half runs through the angle between the hands, which is where “midway between” them is; the other points the opposite way. Each step from $k$ to $k+1$ adds a little over thirty seconds, which swings the second hand half-way round while the other two barely move, so the solutions almost always alternate between the two halves.

The picture fixes $k=1397$: $t=42291\tfrac{1143}{1427}$ s, or 11 h 44 min $51\tfrac{1143}{1427}$ s, the only solution between 11.44.30 and 11.45. Next comes $k=1398$, at 11 h 45 min $22\tfrac{106}{1427}$ s, when the minute hand is near 45.37 and the hour hand near 58.78, but the second hand is at 22.1, pointing directly away from the gap between them. Then $k=1399$ gives 11 h 45 min $52\tfrac{496}{1427}$ s: second hand at 52.35, minute hand at 45.87, hour hand at 58.82, and $52.35$ is the midpoint. Dudeney gives exactly this and warns against the 22-second time. Over twelve hours there are 713 genuine “midway” moments.

Answer 11 h 45 min $52\tfrac{496}{1427}$ s

The Stop-Watch

First pin down the picture.

The minute hand gains $\tfrac{11}{12}$ of a division a minute, so the minute hand is 20 divisions, a third of the rim, clockwise of the hour hand at $t=\tfrac{12}{11}\cdot 20=21\tfrac{9}{11}$ minutes after twelve and every $\tfrac{720}{11}$ minutes after that. The one a little after 9.05 is 9 h $5\tfrac{5}{11}$ min. Then the hour hand is at $45\tfrac{5}{11}$, the minute hand at $5\tfrac{5}{11}$ and the second hand at $27\tfrac{3}{11}$. The gaps are 20 divisions from hour hand to minute hand, $21\tfrac{9}{11}$ from minute hand to second hand, and $18\tfrac{2}{11}$ from second hand back to hour hand.

The crux is Dudeney’s own: hold the watch up to a mirror. Reflecting the dial in the line from XII to VI sends each position $x$ to $60-x$, and since all three hands are proportional to the time, it sends the picture at time $t$ to the picture at time $720-t$. Distances are unchanged by a reflection. So $720-545\tfrac{5}{11}=174\tfrac{6}{11}$ minutes, which is 2 h $54\tfrac{6}{11}$ min with the second hand at $32\tfrac{8}{11}$, shows the same three gaps.

Is there any other time? The hour and minute hands must be a third apart, which happens at only 22 moments in twelve hours, and checking the second hand at each leaves just the two above. So the next occasion after 9.05 is 2 h $54\tfrac{6}{11}$ min, as Dudeney says.

One caution. In the mirror the hands run the other way round the dial. If “the same distances” also means the same clockwise order of hour, minute and second hands, the mirror time does not count, and the answer is the same reading twelve hours on.

Answer 2 h $54\tfrac{6}{11}$ min, second hand at $32\tfrac{8}{11}$ s

The Three Clocks

A twelve-hour clock shows twelve whenever its reading is a whole number of half-days. After $T$ hours of true time, clock A reads $T$, clock B reads $T(1+\tfrac{1}{1440})$ and clock C reads $T(1-\tfrac{1}{1440})$. A needs $T$ to be a multiple of 12. Given that, B and C need their gain and loss, $\tfrac{T}{1440}$ hours, to be a multiple of 12 too, so $T$ is a multiple of $12\times1440$ hours, which is 720 days. The crux is only this: B must gain a full twelve hours, a minute a day, and C lose them. The first meeting is at noon, 720 days on.

The rest is calendar. From noon on 1 April 1898 to 1 April 1899 is 365 days, and to 1 April 1900 is another 365, because 1900, though divisible by four, is a century year not divisible by 400 and so is not a leap year. Seven hundred and twenty days is ten days short of that, which lands on 22 March 1900, a Thursday. Dudeney set the puzzle in 1898 precisely to see who knew about 1900, and his answer agrees. Anyone who counted 29 days in February 1900 gets 21 March.

Answer noon, Thursday 22 March 1900

The Railway Station Clock

The hands point in opposite directions when the minute hand is thirty divisions ahead of the hour hand, which happens at $t=\tfrac{12}{11}(30+60j)$ minutes after twelve, $j=0,1,\dots,10$. The hour hand then makes an angle of $t/2$ degrees with XII, and the straight line of the two hands makes the same angle, reduced below $180^\circ$. The crux: these lines come only in steps of $\tfrac{180}{11}\approx16.36$ degrees, so there are just eleven directions to compare with the wall.

The wall is 861 inches by 124, so each diagonal makes an angle $\theta$ with the horizontal where $\tan\theta=\tfrac{124}{861}$, giving $\theta\approx8.195^\circ$. In the language of the clock, the line of the hands must lie at $90^\circ-\theta$ or $90^\circ+\theta$ from XII. The candidates closest are $j=2$, when the line is at $\tfrac{900}{11}\approx81.82^\circ$, and $j=8$, when it is at $\tfrac{1080}{11}\approx98.18^\circ$. Both miss the diagonal by $0.0135^\circ$; every other opposition misses by more than $16^\circ$.

So there are two answers, one for each diagonal. At $j=2$ the time is 2 h $43\tfrac{7}{11}$ min: the hour hand is just above III and the minute hand just below IX, and the hands lie along the diagonal rising from bottom left to top right. At $j=8$ the time is 9 h $16\tfrac{4}{11}$ min: the hour hand is just above IX and the minute hand just below III, along the other diagonal. Dudeney gives only the first.

Neither is exact, which Dudeney does not mention. For the hands at 2.43$\tfrac{7}{11}$ to be truly parallel to a diagonal the wall would need $\tan\theta=\tan\tfrac{90^\circ}{11}\approx 0.14378$, while $\tfrac{124}{861}\approx0.14402$. A wall 10 ft 4 in. high would have to be about 862.4 inches long, 71 ft 10.4 in., to fit. Dudeney plainly chose his measurements to fit this angle to within an inch and a half, and at the scale of a station clock the miss is invisible, but in the strict sense of “exact” the puzzle has no solution.

Answer 2 h $43\tfrac{7}{11}$ min or 9 h $16\tfrac{4}{11}$ min (both approximate)

The Village Simpleton

The crux is to turn each clause into a shift of days. If to-day is $T$, then “when the day after to-morrow is yesterday” is three days on, $T+3$, and “when the day before yesterday was to-morrow” is three days back, $T-3$. Number the days from Sunday as 0, working modulo 7. Two days are equally far from Sunday, going round the week whichever way is shorter, when their numbers are equal or negatives of each other. Equal would need $T+3\equiv T-3$, that is $6\equiv0$, which is false. Negatives need $T+3\equiv-(T-3)$, so $2T\equiv0$, and since 2 has an inverse modulo 7 this forces $T\equiv0$. It was Sunday.

Check: three days on is Wednesday, three days back was Thursday, and each has two days between it and Sunday. If “as far from Sunday” meant counting only forwards, or only backwards, the two days would have to coincide, which they never do, so the countryman’s riddle has exactly one answer. Dudeney has Sunday too.

Answer Sunday