Teaser 2688: Mother’s Day
From The Sunday Times, 30th March 2014 [link]
Today we are having a family get-together to celebrate Mother’s Day. My maternal grandmother, my mother and I have each written down our date of birth in the form “ddmmyy”. This gives us three six-figure numbers and, surprisingly, both of the ladies’ numbers are multiples of mine. Furthermore, all of the digits from 0 to 9 occur somewhere in the three six-figure numbers.
What is my mother’s six-figure date of birth?
[teaser2688]
S2T2 
Jim Randell 9:05 am on 22 November 2022 Permalink |
At first I found many possible solutions to this puzzle.
But if we require that the 6-digit numbers formed from the date cannot have a leading zero, then this narrows down the solution space considerably (and essentially restricts the three dates to the form 10xxxx, 20yyyy, 30zzzz, and the multiples being 1, 2, 3).
This Python program runs in 129ms.
Run: [ @replit ]
from datetime import (date, timedelta) from enigma import ( fdiv, inc, repeat, nconcat, nsplit, catch, subsets, append, tuples, union, join, printf ) # extract date as a 6-digit number def number(d): if d.day < 10: return None return nconcat(d.day, d.month, d.year % 100, base=100) # check viable generation gap, a -> b def gap(a, b): y = fdiv((b - a).days, 365.2422) return not (y < 16 or y > 50) # consider dates for the setters birthdate for d in repeat(inc(timedelta(days=-1)), date(2014, 3, 30)): if d.year < 1922: break # construct the number "ddmmyy" n = number(d) if n is None: continue # look for proper multiples that also give a valid date mn = n ds = list() while True: mn += n if mn > 311299: break (dd, mm, yy) = nsplit(mn, base=100) for y in (1900 + yy, 1800 + yy): if y < 1892: continue md = catch(date, y, mm, dd) if md is None or number(d) is None: continue ds.append(md) # look for a set of 3 plausible ages for ss in subsets(sorted(ds), size=2): ss = append(ss, d) if not all(gap(a, b) for (a, b) in tuples(ss, 2)): continue # check all digits are used if len(union(nsplit(number(d)) for d in ss)) < 10: continue # output dates printf("{ss}", ss=join(ss, sep=" -> "))The program works backwards from 2014 to 1922 looking for sets of 3 dates that make a plausible set of birthdates for the three generations.
It finds 2 possible situations, as below (ages shown are on the date the puzzle was published (2014-03-30)):
The second of these is only just plausible, so presumably the first provides the required answer. (I would have preferred the puzzle eliminated one of these solutions by an explicit condition).
Solution: The mother’s date of birth is: 200546 (i.e. 20th May 1946).
To see solutions where the 6-digit number formed from the date is permitted to have a leading zero, the check at line 9 can be removed. In this case the program finds 115 solutions.
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