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Jim Randell 9:43 am on 10 February 2026 Permalink Reply
Tags: by: Victor Bryant ( 142 )   

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  Jim Randell 9:44 am on 10 February 2026 Permalink | Reply

  You need to take into account the date the puzzle was originally published when solving this puzzle.

  This Python program runs in 72ms. (Internal runtime is 1.12ms).

  from datetime import (date, timedelta)
  from enigma import (powers, inf, first, le, repeat, inc, peek, irange, is_npalindrome, printf)
  
  # date of puzzle (all dates must be earlier than this)
  pub = date(2007, 6, 10)
  
  # consider numbers up to 200_000
  (sqs, cbs, p4s) = (first(powers(1, inf, k), count=le(200000)) for k in [2, 3, 4])
  
  # is a date a weekend (= Saturday (6) or Sunday (7))
  is_weekend = lambda d: d.isoweekday() in {6, 7}
  
  # consider powers of 4
  for p4 in p4s:
  
    # look for a higher cube, potentially within the same month
    for cb in reversed(cbs):
      if not (cb > p4): break
      if cb - p4 > 30: continue
  
      # consider possible dates for the cube in the current year (before publication)
      for dcb in repeat(inc(timedelta(days=-1)), pub):
        if dcb.year < pub.year: break
        # must be on a weekend
        if not is_weekend(dcb): continue
        # calculate date for the power of 4
        dp4 = dcb - timedelta(days=cb - p4)
        # must be in the same month
        if not (dp4.month == dcb.month): continue
  
        # look for earlier perfect squares
        for sq in sqs:
          if not (sq < p4): break
          # calculate date for the square
          dsq = dp4 - timedelta(days=p4 - sq)
          y = dp4.year - dsq.year
          if not (0 < y < 2): continue
          # must be on a weekend
          if not is_weekend(dsq): continue
  
          # when is the next palindromic date (after cb)
          pl = peek(n for n in irange(cb + 1, inf) if is_npalindrome(n))
          dpl = dp4 + timedelta(days=pl - p4)
  
          # calculate founding date
          df = dp4 - timedelta(days=p4 - 1)
  
          # output solution
          printf("p4 = D{p4} = {dp4}; cb = D{cb} = {dcb}; sq = D{sq} = {dsq}; pl = D{pl} = {dpl}; D1 = {df}")
  

  Solution: The next palindromic number will occur on 20th June 2007.

  The square number is day 50176 (= 224^2) on 2006-01-07 (Saturday).

  The perfect cube is day 50653 (= 37^3) on 2007-04-29 (Sunday).

  The power of 4 is day 50625 (= 15^4) on 2007-04-01 (Sunday).

  And the next palindromic number after 50653 is 50705 on 2007-06-20 (Wednesday), i.e. 10 days after the puzzle was published.

  If the ledger has been in continual operation from the day of founding, then day 1 was 1868-08-23 (Sunday).

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