Sunday Times Teaser 2671 – On the Face of It

by Robin Naylor

At the start of last year my wife and I were each given clocks which ran fast by a whole number of minutes per day with my wife’s running faster than my own.

When we first looked at the clock hands they showed the same time and it turned out that this would happen at least once more during the year.

It also turned out that my clock would be right at least once during every month in the year but this was not true for my wife’s clock.

How many minutes did each clock gain in a day?

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brian gladman permalink


from itertools import count
from collections import defaultdict

# days in the months of a leap year
month_days = (0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)

# find the days in a year when a clock running fast will be correct
# again given the first correct day and the minutes per day gained;
# then return how many times this occurs in each month in the year.
def days(first_day, gain):
  # a count of the clock rollovers in each month
  in_month = [0] * 12
  for mult in count(0):
    # the next rollover
    rollover = first_day + int(mult * (720 / gain))
    # while it is in the year
    if rollover >= 366:
      break
    # find which month it is in and add to this month's count
    for m in range(12):
      rollover -= month_days[m + 1]
      if rollover < 0:
        in_month[m] += 1
        break
  return in_month

# gains for start dates in January that have and don't have
# 'rollovers' in each month of the year respectively
me, wife = defaultdict(list), defaultdict(list)
for first_day in range(31):
  # for minutes gained per day (limits set by the need to
  # have at least one rollover in two months but not one
  # for certain in February whatever the starting date)
  for gain in range(12, 26):
    # find the counts of 'rollovers' in each month
    counts = days(first_day, gain)
    if 0 in counts:
      # my wife's clock doesn't have 'rollovers' in every month
      wife[first_day].append(gain)
    else:
      # but mine does
      me[first_day].append(gain)

fs = 'The clocks gain {:d} and {:d} minutes per day.'
# for a shared start date in January
for d in set(me.keys() & wife.keys()):
  # the gains for wife's and my own clock
  for wc_gain in wife[d]:
    for mc_gain in me[d]:
      # to have the two clocks agreeing again in a year
      # the wife's clock must gain two or more minutes 
      # per day more then mine
      if wc_gain >= mc_gain + 2:
        print(fs.format(mc_gain, wc_gain))

from itertools import count

from collections import defaultdict

# days in the months of a leap year

month_days = (0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)

# find the days in a year when a clock running fast will be correct

# again given the first correct day and the minutes per day gained;

# then return how many times this occurs in each month in the year.

def days(first_day, gain):

# a count of the clock rollovers in each month

in_month = [0] * 12

for mult in count(0):

# the next rollover

rollover = first_day + int(mult * (720 / gain))

# while it is in the year

if rollover >= 366:

break

# find which month it is in and add to this month's count

for m in range(12):

rollover -= month_days[m + 1]

if rollover < 0:

in_month[m] += 1

break

return in_month

# gains for start dates in January that have and don't have

# 'rollovers' in each month of the year respectively

me, wife = defaultdict(list), defaultdict(list)

for first_day in range(31):

# for minutes gained per day (limits set by the need to

# have at least one rollover in two months but not one

# for certain in February whatever the starting date)

for gain in range(12, 26):

# find the counts of 'rollovers' in each month

counts = days(first_day, gain)

if 0 in counts:

# my wife's clock doesn't have 'rollovers' in every month

wife[first_day].append(gain)

else:

# but mine does

me[first_day].append(gain)

fs = 'The clocks gain {:d} and {:d} minutes per day.'

# for a shared start date in January

for d in set(me.keys() & wife.keys()):

# the gains for wife's and my own clock

for wc_gain in wife[d]:

for mc_gain in me[d]:

# to have the two clocks agreeing again in a year

# the wife's clock must gain two or more minutes

# per day more then mine

if wc_gain >= mc_gain + 2:

print(fs.format(mc_gain, wc_gain))

Sunday Times Teaser 2671 – On the Face of It

by Robin Naylor

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