Sunday Times Teaser 3174 – Pensioners’ Outing

by BRG on July 22, 2023

by Edmund Marshall

Published Sunday July 23 2023 (link)

A group of pensioners took a trip in a minibus, which had 11 seats for passengers, three on the back row, and two on each of the four other rows. The ages in years of the passengers were all different double-digit integers greater than 64, and for each pair of those ages there was no common factor other than 1. All seats were occupied, and, on any particular row of seats, the digits in the ages of passengers were all different. The sum of the ages of passengers on the back row was the largest possible with these conditions.

What was the age in years of the most elderly passenger sitting on the back row?

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BRG permalink


from itertools import combinations
from math import gcd

# ages with different digits 64 < age < 100
ages = list(n for n in range(65, 100) if n % 11)

# group <n> ages from <ages> such that they are coprime
# and have no repeated digits among them
def group_ages(n, ages):
  gs = []
  for cn in combinations(ages, n):
    if all(gcd(a, b) == 1 for a, b in combinations(cn, 2)):
      dgts = ''.join(str(n) for n in cn)
      if len(dgts) == len(set(dgts)):
        gs.append(cn)
  return gs

# add <n> pairs from <prs>, accumulating ages 
# in <ags> and rows in <rws> (front to back)  
def add_pairs(n, prs, ags, rws):
  if n == 0:
    yield rws
  else:
    for i, p in enumerate(prs):
      if not ags.intersection(p):
        if all(gcd(m, n) == 1 for m in p for n in ags):
          yield from add_pairs(n - 1, prs[i + 1:], ags | set(p), [p] + rws)

# consider possible age triples for the back
# row (in decreasing order of their sum)
for g3 in sorted(group_ages(3, ages), key=sum, reverse=True):
  # and possible pairs for the four front rows   
  for prs in add_pairs(4, group_ages(2, ages), set(g3), [g3]):
    print(f"{prs[-1][2]} {prs}")
    exit()

from itertools import combinations

from math import gcd

# ages with different digits 64 < age < 100

ages = list(n for n in range(65, 100) if n % 11)

# group <n> ages from <ages> such that they are coprime

# and have no repeated digits among them

def group_ages(n, ages):

gs = []

for cn in combinations(ages, n):

if all(gcd(a, b) == 1 for a, b in combinations(cn, 2)):

dgts = ''.join(str(n) for n in cn)

if len(dgts) == len(set(dgts)):

gs.append(cn)

return gs

# add <n> pairs from <prs>, accumulating ages

# in <ags> and rows in <rws> (front to back)

def add_pairs(n, prs, ags, rws):

if n == 0:

yield rws

else:

for i, p in enumerate(prs):

if not ags.intersection(p):

if all(gcd(m, n) == 1 for m in p for n in ags):

yield from add_pairs(n - 1, prs[i + 1:], ags | set(p), [p] + rws)

# consider possible age triples for the back

# row (in decreasing order of their sum)

for g3 in sorted(group_ages(3, ages), key=sum, reverse=True):

# and possible pairs for the four front rows

for prs in add_pairs(4, group_ages(2, ages), set(g3), [g3]):

print(f"{prs[-1][2]} {prs}")

exit()

John Z. permalink

A bit slow…


from math import gcd

ages = tuple(sorted(set(range(65, 99)) - {66, 77, 88}))

# find tuples of 11 ages with no common factor other than 1
def oap(aa, sol = ()):
  if len(sol) == 11:
    yield sol
  else:
    # consider ages remaining
    for i, a in enumerate(aa):
      if not sol or all(gcd(a, b) == 1 for b in sol):
        yield from oap(aa[i + 1:], sol + (a,))

# arrange bus into 5 rows of 2 and one of 3 with no common digits on a row
def bus(aa, rr =()):
  if len(rr) == 10:
    # does last row have no common digits?
    if len({*divmod(rr[8], 10), *divmod(rr[9], 10), *divmod(aa[0], 10)}) == 6:
      # the busload organized by 3 rows of 4 and 1 of 3
      yield rr + aa
      
  else:
    # rows 1 - 4:
    for i, a in enumerate(aa):
      for b in aa[i + 1:]:
        # 2 person rows without common digits
        if len({*divmod(a, 10), *divmod(b, 10)}) == 4:
          yield from bus(tuple(set(aa) - {a, b}), rr + (a, b))
            

omax = 0

# generate gcd = 1 busloads
for a in oap(ages):
  # find seating arrangements meeting no common digit on a row requirement
  for b in bus(a):
    nmax = sum(b[-3:])
    if nmax > omax:
      omax = nmax
      amax = max(b[-3:])
      print(b, amax, omax)

print('maximum age of back of the bus:', amax)

from math import gcd

ages = tuple(sorted(set(range(65, 99)) - {66, 77, 88}))

# find tuples of 11 ages with no common factor other than 1

def oap(aa, sol = ()):

if len(sol) == 11:

yield sol

else:

# consider ages remaining

for i, a in enumerate(aa):

if not sol or all(gcd(a, b) == 1 for b in sol):

yield from oap(aa[i + 1:], sol + (a,))

# arrange bus into 5 rows of 2 and one of 3 with no common digits on a row

def bus(aa, rr =()):

if len(rr) == 10:

# does last row have no common digits?

if len({*divmod(rr[8], 10), *divmod(rr[9], 10), *divmod(aa[0], 10)}) == 6:

# the busload organized by 3 rows of 4 and 1 of 3

yield rr + aa

else:

# rows 1 - 4:

for i, a in enumerate(aa):

for b in aa[i + 1:]:

# 2 person rows without common digits

if len({*divmod(a, 10), *divmod(b, 10)}) == 4:

yield from bus(tuple(set(aa) - {a, b}), rr + (a, b))

omax = 0

# generate gcd = 1 busloads

for a in oap(ages):

# find seating arrangements meeting no common digit on a row requirement

for b in bus(a):

nmax = sum(b[-3:])

if nmax > omax:

omax = nmax

amax = max(b[-3:])

print(b, amax, omax)

print('maximum age of back of the bus:', amax)

John Z permalink

“Code first, think later” rides again!

Reply

Sunday Times Teaser 3174 – Pensioners’ Outing

by Edmund Marshall

Published Sunday July 23 2023 (link)

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