Sunday Times Teaser 2837 – Back and Forth

by Danny Roth

Published: 5 February 2017 (link)

George and Martha have a digital 24-hour clock that always displays the time as four digits (eg, 0217). During one lazy weekend George noted two particular times when the four-digit display was palindromic. He calculated the number of minutes from the first of these times to the second and he discovered that the answer was a three-figure palindrome. When he reported all the details to Martha, she commented that the sum of the eleven digits in the three palindromes was also palindromic.

What were the two palindromic times?

2 Comments Leave one →

Brian Gladman permalink


# requires Python 3.6+

from itertools import permutations

# list four digit palindromic times
ls = []
for h in range(24):
  a, b = divmod(h, 10)
  m = 10 * b + a
  if m < 60:
    # save the hour, minute and digit sum for each time
    ls += [(h, m, 2 * (a + b))]

# the two times are either on the same day or a day apart
for day in (0, 24):
  # consider the possible differences between two times
  for (h1, m1, ds1), (h2, m2, ds2) in permutations(ls, 2):
    
    # ... which must be a three digit palindrome    
    d = (60 * (h2 + day) + m2) - (60 * h1 + m1)
    if 100 < d < 1000 and d // 100 == d % 10:
      # the sum of the eleven digits is a (two digit) palindrome
      if not (sum(int(x) for x in str(d)) + ds1 + ds2) % 11: 
        d1, d2 = f'{h1:02d}:{m1:02d}', f'{h2:02d}:{m2:02d}'
        if day:
          print(f'{d1} (Saturday) and {d2} (Sunday).')
        else:
          print(f'{d1} and {d2}')

# requires Python 3.6+

from itertools import permutations

# list four digit palindromic times

ls = []

for h in range(24):

a, b = divmod(h, 10)

m = 10 * b + a

if m < 60:

# save the hour, minute and digit sum for each time

ls += [(h, m, 2 * (a + b))]

# the two times are either on the same day or a day apart

for day in (0, 24):

# consider the possible differences between two times

for (h1, m1, ds1), (h2, m2, ds2) in permutations(ls, 2):

# ... which must be a three digit palindrome

d = (60 * (h2 + day) + m2) - (60 * h1 + m1)

if 100 < d < 1000 and d // 100 == d % 10:

# the sum of the eleven digits is a (two digit) palindrome

if not (sum(int(x) for x in str(d)) + ds1 + ds2) % 11:

d1, d2 = f'{h1:02d}:{m1:02d}', f'{h2:02d}:{m2:02d}'

if day:

print(f'{d1} (Saturday) and {d2} (Sunday).')

else:

print(f'{d1} and {d2}')

Jim Randell permalink

This program considers when there is a corresponding minute for each hour to make a list of palindromic times, and then considers pairs of times from this list that satisfy the remaining conditions.


from itertools import product

# record palindromic times
ts = list()
for h in range(0, 24):
  (a, b) = divmod(h, 10)
  m = 10 * b + a
  if m < 60:
    # record the displayed time (as a string), and the time (in minutes)
    ts.append((f"{h:02d}{m:02d}", h * 60 + m))

# now consider two palindromic times (the second time could be the following day)
for ((s1, t1), (s2, t2), t3) in product(ts, ts, (0, 1440)):

  # calculate the difference in minutes
  t = t2 + t3 - t1

  # it should be a 3-digit palindrome
  if not(99 < t < 1000): continue
  (ab, c) = divmod(t, 10)
  (a, b) = divmod(ab, 10)
  if a != c: continue

  # calculate the digits sum
  ds = sum(int(d) for d in s1 + s2 + str(t))

  # it should also be a palindrome
  if ds > 9 and ds % 11 > 0: continue

  # output a solution
  x=(" (next day)" if t3 else "")
  print(f"{s1} -> {s2}{x} = {t} minutes, digit sum = {ds}")

from itertools import product

# record palindromic times

ts = list()

for h in range(0, 24):

(a, b) = divmod(h, 10)

m = 10 * b + a

if m < 60:

# record the displayed time (as a string), and the time (in minutes)

ts.append((f"{h:02d}{m:02d}", h * 60 + m))

# now consider two palindromic times (the second time could be the following day)

for ((s1, t1), (s2, t2), t3) in product(ts, ts, (0, 1440)):

# calculate the difference in minutes

t = t2 + t3 - t1

# it should be a 3-digit palindrome

if not(99 < t < 1000): continue

(ab, c) = divmod(t, 10)

(a, b) = divmod(ab, 10)

if a != c: continue

# calculate the digits sum

ds = sum(int(d) for d in s1 + s2 + str(t))

# it should also be a palindrome

if ds > 9 and ds % 11 > 0: continue

# output a solution

x=(" (next day)" if t3 else "")

print(f"{s1} -> {s2}{x} = {t} minutes, digit sum = {ds}")

Sunday Times Teaser 2837 – Back and Forth

by Danny Roth

Published: 5 February 2017 (link)

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