Brain-Teaser 2: Stop watch
From The Sunday Times, 5th March 1961 [link]
Setting one’s watch can be a tricky business, especially if it has a sweep second hand; for, unlike the hour and minute hands, the second hand is independent of the winder. The other day, when trying to set my watch by the midday time signal, I managed to get the hour hand and minute hand accurately aligned at 12 o’clock just as the pips signalled noon, but the second hand escaped me — in fact, on the pip of twelve it was just passing the 5-second mark. I can’t say that it was exactly on the 5-second mark, but it was within a second or two of it one way or the other. I didn’t bother to adjust it further in case I should finish up with the other hands wrong as well.
That night I forgot to wind my watch and, the next morning, found that (not unnaturally) it had stopped. I noticed that the hour hand, the minute hand and the second hand were all exactly aligned one above another.
(1) At exactly what time had my watch stopped?
(2) Where exactly was the second hand pointing (to the very fraction of a second) on the pip of twelve noon the previous day?
[teaser2]
S2T2 
Jim Randell 8:45 am on 9 July 2020 Permalink |
See also: Enigma 1761, Enigma 383, Enigma 404, Enigma 409.
In each 12 hour period there are 11 times when the hour and minute hand coincide.
This Python program looks at each of these eleven times, and works out far the second hand is ahead of its expected position if it is also coincident with the hour and minute hands. And if the second hand is between 3 and 7 seconds ahead of where it should be, then we have a solution to the puzzle.
Run: [ @repl.it ]
from fractions import Fraction as F from enigma import irange, sprintf as f, printf # format fractional seconds def fmt(s): (n, r) = divmod(s, 1) return f("{n:02d} + {r} s") # number of seconds in a period of the clock (12 hours) T = 12 * 60 * 60 # find times at which the hour and minute hands coincide for i in irange(0, 10): t = F(T * i, 11) (h, ms) = divmod(t, 60 * 60) (m, s) = divmod(ms, 60) # the second hand should be at s, but is actually at ms/60 # the difference should be between 3 and 7 seconds d = (F(ms, 60) - s) % 60 if not(d < 3 or d > 7): printf("i={i} -> t={t} = {h:02d}:{m:02d}:{s}; diff = {d}", s=fmt(s), d=fmt(d))Solution: (1) The watch stopped at 8:43:38 + 2/11. (2) At exactly 12 noon the second hand was showing 5 + 5/11 seconds.
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