Sunday Times Teaser 3293 – Three Sisters

by BRG on November 2, 2025

by Victor Bryant

Published Sunday November 02 2025

I have three granddaughters Jay, Kay and Elle. I set Jay and Kay a test and asked each of them to produce a list of positive numbers that used just nine digits with no digit occurring more than once. I wanted the majority of the numbers in each list to be odd and the majority of the numbers in each list to be perfect squares.

Jay’s list (which contained more numbers than Kay’s) added up to give the year of Elle’s birth, whereas Kay’s list added up to give the year in which Elle will be 25.

In one of the lists the highest number was a perfect square.

What (in increasing order) were the numbers in that list?

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John Z. permalink


from itertools import combinations, permutations

dig = set('0123456789')

# perfect squares without repeated digits
psq = {i * i for i in range(1, int(2050 ** 0.5))
            if len(set(str(i * i))) == len(str(i * i))}

# si: index into num, resn: the result numbers, resd: the digits in the result
def solver(si, resn = [], resd = set()):
  # does number of digits used = 9
  if len(resd) == 9:
    # Elle's dob is less than 25 years ago; loose (50%) odd number criterion
    if sum(resn) > 2000 and sum(q % 2 for q in resn) >= len(resn) // 2: 
      yield resn
  # verify that number of perfect squares will be majority of numbers in list
  elif nsq > (len(resn) + 1) // 2:
    for i, n in enumerate(num[si:]):
      # sum must be less than max year and no repeat of digits
      if 2050 > sum(resn) + n[0] and not resd.intersection(n[1]):
        yield from solver(si + i + 1, resn + [n[0]], resd | n[1])
        
sols = []
# get 2, 3, 4, or 5 perfect squares to be used in lists
for nsq in range(2, 6):
  for sc in combinations(psq, nsq):
    cs = ''.join(str(s) for s in sc)
    css = set(cs)

    # check for sums of psq ok and no duplicate characters 
    if sum(sc) > 2025 or len(cs) != len(css):
      continue
    # other numbers made from digits not in perfect squares
    num = tuple((nn, set(n)) for j in range(1, 5)
          for n in permutations(dig - css, j) if n[0] != '0' and
          2000 > (nn := int(''.join(n))) and nn not in psq)
    
    # find lists of Jay's and Kay's numbers
    for sol in solver(0, list(sc), css):
      sols.append(tuple(sorted(sol)))  
   
Jayset, Kayset = set(), set()
# consider pairs of the lists of numbers found
for jk in combinations(sols, 2):
  # Jay’s list (which contained more numbers than Kay’s)
  Kay, Jay = sorted(jk, key = len)
  sk, sj = sum(Kay), sum(Jay)
  # tightened odd number criterion for Kay's list (50% or more for Jay)
  ok = sum(q % 2 for q in Kay) > len(Kay) // 2
  # square is last in list
  sqjl, sqkl = Jay[-1] in psq, Kay[-1] in psq

  # apply the Teaser criteria 
  if sk - sj == 25 and len(Jay) > len(Kay) and sqjl !=  sqkl and ok:
    Jayset |= {Jay}; Kayset |= {Kay}
    print('Jay:', Jay, sj, '  Kay:', Kay, sk)    
    print("\nAnswer", sqjl * ("Jay's list: " + str(Jayset.pop())),
                      sqkl * ("Kay's list: " + str(Kayset.pop())))

from itertools import combinations, permutations

dig = set('0123456789')

# perfect squares without repeated digits

psq = {i * i for i in range(1, int(2050 ** 0.5))

if len(set(str(i * i))) == len(str(i * i))}

# si: index into num, resn: the result numbers, resd: the digits in the result

def solver(si, resn = [], resd = set()):

# does number of digits used = 9

if len(resd) == 9:

# Elle's dob is less than 25 years ago; loose (50%) odd number criterion

if sum(resn) > 2000 and sum(q % 2 for q in resn) >= len(resn) // 2:

yield resn

# verify that number of perfect squares will be majority of numbers in list

elif nsq > (len(resn) + 1) // 2:

for i, n in enumerate(num[si:]):

# sum must be less than max year and no repeat of digits

if 2050 > sum(resn) + n[0] and not resd.intersection(n[1]):

yield from solver(si + i + 1, resn + [n[0]], resd | n[1])

sols = []

# get 2, 3, 4, or 5 perfect squares to be used in lists

for nsq in range(2, 6):

for sc in combinations(psq, nsq):

cs = ''.join(str(s) for s in sc)

css = set(cs)

# check for sums of psq ok and no duplicate characters

if sum(sc) > 2025 or len(cs) != len(css):

continue

# other numbers made from digits not in perfect squares

num = tuple((nn, set(n)) for j in range(1, 5)

for n in permutations(dig - css, j) if n[0] != '0' and

2000 > (nn := int(''.join(n))) and nn not in psq)

# find lists of Jay's and Kay's numbers

for sol in solver(0, list(sc), css):

sols.append(tuple(sorted(sol)))

Jayset, Kayset = set(), set()

# consider pairs of the lists of numbers found

for jk in combinations(sols, 2):

# Jay’s list (which contained more numbers than Kay’s)

Kay, Jay = sorted(jk, key = len)

sk, sj = sum(Kay), sum(Jay)

# tightened odd number criterion for Kay's list (50% or more for Jay)

ok = sum(q % 2 for q in Kay) > len(Kay) // 2

# square is last in list

sqjl, sqkl = Jay[-1] in psq, Kay[-1] in psq

# apply the Teaser criteria

if sk - sj == 25 and len(Jay) > len(Kay) and sqjl != sqkl and ok:

Jayset |= {Jay}; Kayset |= {Kay}

print('Jay:', Jay, sj, ' Kay:', Kay, sk)

print("\nAnswer", sqjl * ("Jay's list: " + str(Jayset.pop())),

sqkl * ("Kay's list: " + str(Kayset.pop())))