Sunday Times Brain Teaser 737

by G.H. Dickson

Our old watchmaker works weekdays 9.30 am to 5 pm as regular as, well, clockwork. I recently asked him to adjust two 8-day striking clocks – the sort which, when fully-wound, will run for nearly eight days. They were keeping different times and each was wrong by an exact number of minutes per day, both less than an hour.

He immediately wound the clocks fully and set them to the right time, which was an exact number of minutes after the hour. He then put them on a shelf for later observation.

The next Monday, when he went to take the clocks down to adjust them, he found that both were simultaneously starting to strike 8 o’clock, which was some hours plus an exact number of minutes past the correct time.

What day and exact time was it when he originally set them?

One Comment Leave one →

brian gladman permalink

Using a routine from my number theory library number_theory.py


from number_theory import divisors

days = ( 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 
         'Saturday', 'Sunday' )

# Let the time elapsed from when the clocks are synchronised be t 
# minutes and let clock one lose k1 minutes per day and clock two
# gain k2 minutes per day  (to give a 12 hour difference in eight
# days requires one clock to run slow and the other to run fast).
# If t1 and t2 are the times on clocks one and two, we then have:
#
#   t1 = (1 - k1 / 1440).t  ;  t2 = (1 + k2 / 1440).t
#
# But these clock times differ by a multiple (m) of 12 hours so:
#
#   (k1 + k2).t / 1440 = 720.m  ==>  t = 720.1440.m / (k1 + k2)
#
# Since t is less than 8 days:
#
#   720.1440.m /(k1 + k2) < 8.1440  ==>  k1 + k2 > 90.m
# 
# Since k1 and k2 are both less than 60 this shows that m = 1 and
# that k1 + k2 divides 720.1440 with 90 < k1 + k2 < 120.

for k1_plus_k2 in divisors(720 * 1440):
  if 90 < k1_plus_k2 < 120:
    t = 720 * 1440 // k1_plus_k2
    for k2 in range(k1_plus_k2 - 59, 60):
      # The difference between t2 and t is k2 * t / 1440, which must
      # be a whole number of minutes
      m, r = divmod(k2 * t, 1440)
      if r == 0:
        
        # Since t2 is at 8:00 and is later than the correct time (on
        # or after the opening time) t2 must be 8:00 PM. This is 'm'
        # minutes after the correct time, which is hence 20 * 60 - m
        # minutes on the 2nd Monday. The clocks were set 't minutes
        # before this, which is at (7 * 1440 + 20 * 60 - m - t) when
        # measured from the start of the previous week.
        day, m = divmod(7 * 1440 + 20 * 60 - (m + t), 1440)
        
        # The time when set must be between 9:30 AM and 5:00 PM
        if 9 * 60 + 30 <= m < 17 * 60:
          print('{} at {}:{} AM.'.format(days[day], *divmod(m, 60)))

from number_theory import divisors

days = ( 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday',

'Saturday', 'Sunday' )

# Let the time elapsed from when the clocks are synchronised be t

# minutes and let clock one lose k1 minutes per day and clock two

# gain k2 minutes per day (to give a 12 hour difference in eight

# days requires one clock to run slow and the other to run fast).

# If t1 and t2 are the times on clocks one and two, we then have:

# t1 = (1 - k1 / 1440).t ; t2 = (1 + k2 / 1440).t

# But these clock times differ by a multiple (m) of 12 hours so:

# (k1 + k2).t / 1440 = 720.m ==> t = 720.1440.m / (k1 + k2)

# Since t is less than 8 days:

# 720.1440.m /(k1 + k2) < 8.1440 ==> k1 + k2 > 90.m

# Since k1 and k2 are both less than 60 this shows that m = 1 and

# that k1 + k2 divides 720.1440 with 90 < k1 + k2 < 120.

for k1_plus_k2 in divisors(720 * 1440):

if 90 < k1_plus_k2 < 120:

t = 720 * 1440 // k1_plus_k2

for k2 in range(k1_plus_k2 - 59, 60):

# The difference between t2 and t is k2 * t / 1440, which must

# be a whole number of minutes

m, r = divmod(k2 * t, 1440)

if r == 0:

# Since t2 is at 8:00 and is later than the correct time (on

# or after the opening time) t2 must be 8:00 PM. This is 'm'

# minutes after the correct time, which is hence 20 * 60 - m

# minutes on the 2nd Monday. The clocks were set 't minutes

# before this, which is at (7 * 1440 + 20 * 60 - m - t) when

# measured from the start of the previous week.

day, m = divmod(7 * 1440 + 20 * 60 - (m + t), 1440)

# The time when set must be between 9:30 AM and 5:00 PM

if 9 * 60 + 30 <= m < 17 * 60:

print('{} at {}:{} AM.'.format(days[day], *divmod(m, 60)))

Sunday Times Brain Teaser 737

by G.H. Dickson

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