# Solution

(related to Problem: The Wapshaw's Wharf Mystery)

There are eleven different times in twelve hours when the hour and minute hands of a clock are exactly one above the other. If we divide $12$ hours by $11$ we get $1$ hr. $5$ min. $27\frac{3}{11}$ sec., and this is the time after twelve o'clock when they are first together, and also the time that elapses between one occasion of the hands being together and the next. They are together for the second time at $2$ hr. $10$ min. $54\frac{6}{11}$ sec. (twice the above time); next at $3$ hr. $16$ min. $21\frac{9}{11}$ sec.; next at $4$ hr. $21$ min. $49\frac{1}{11}$ sec. This last is the only occasion on which the two hands are together with the second hand "just past the forty-ninth second." This, then, is the time at which the watch must have stopped. Guy Boothby, in the opening sentence of his Across the World for a Wife, says, "It was a cold, dreary winter's afternoon, and by the time the hands of the clock on my mantelpiece joined forces and stood at twenty minutes past four, my chambers were well-nigh as dark as midnight." It is evident that the author here made a slip, for, as we have seen above, he is $1$ min. $49\frac{1}{11}$ sec. out in his reckoning.

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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