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Jim Randell 2:19 pm on 26 June 2026 Permalink Reply
Tags: by: Danny Roth ( 91 )   

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  Jim Randell 2:20 pm on 26 June 2026 Permalink | Reply

  This Python program considers possible birth years between 1900 and 2000.

  It factorises the year into candidate values for the “birth product” and the “smaller number”. It then finds the day of the year that corresponds to the candidate “birth product” and verifies that the day × month product gives the same value. If this test succeeds then we have found a date with the required properties.

  It runs in 66ms. (Internal runtime is 941µs).

  from datetime import (date, timedelta)
  from enigma import (irange, divisors_pairs, printf)
  
  # consider years
  for y in irange(1900, 2000):
  
    # divide year in <smaller number> and <birth product>
    for (sn, bp) in divisors_pairs(y):
      if bp > 366 or sn == bp: continue
  
      # check day bp in year y
      d = date(y, 1, 1) + timedelta(days=bp - 1)
      # does it have the correct product?
      if not (d.month * d.day == bp and d.year == y): continue
  
      # output viable date
      printf("year {y} = {sn} * {bp}; day {bp} = {d}")
  

  Solution: The student’s DOB is: 30/03/1980. The professor’s DOB is: 30/05/1950.

  For the student we have:

  30 × 3 = 90
  and 30th March is the 90th day of 1980
  also:
  90 × 22 = 1980

  For the professor we have:

  30 × 5 = 150
  and 30th May is the 150th day of 1950
  also:
  150 × 13 = 1950

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  Ruud 6:39 pm on 26 June 2026 Permalink | Reply

  import datetime
  
  
  day = datetime.datetime(1900, 1, 1)
  
  while day < datetime.datetime(2000, 1, 1):
      if day.day * day.month == (daynr := (day - datetime.datetime(day.year, 1, 1)).days + 1):
          multiplier, rest = divmod(day.year, daynr)
          if rest == 0 and multiplier < daynr:
              print(f"{day.day}/{day.month}/{day.year}")
      day += datetime.timedelta(days=1)
  

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  Ruud 11:41 am on 27 June 2026 Permalink | Reply

  Different approach:

  from math import ceil
  
  for is_leapyear in (False, True):
      for daynr, (month, day) in enumerate(
          ((month, day) for month in range(1, 13) for day in range(1, [0, 31, 28 + is_leapyear, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31][month] + 1)), 1
      ):
          if month * day == daynr and ceil(1900 / daynr) < daynr:
              for year in range(ceil(1900 / daynr) * daynr, ceil(2000 / daynr) * daynr, daynr):
                  if (year % 4==0) == is_leapyear:
                      print(f"{day}/{month}/{year}")
  

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