Sunday Times Teaser 3269 – Combination Lock

by BRG on May 18, 2025

by Andrew Skidmore

Published Sunday May 18 2025 (link)

Phil buys a four-figure combination lock where each dial can be set from 0 to 9, and needs to decide what number to set to unlock it. He opts for a number with all different digits and a leading zero so he only has to remember a three-figure number. When the chosen number is set for the lock to open, nine other four-figure numbers are visible. For instance, if he chooses 0123, then 1234, 2345, 3456, 4567, 5678, 6789, 7890, 8901 and 9012 are all visible. As a retired Maths teacher with a natural interest in numbers he examines the other nine numbers that are displayed when the lock is set to open. None of these numbers are prime but two of them are perfect squares.

What three-figure number does Phil need to remember?

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BRG permalink


from itertools import permutations

# create sets of four digit primes and squares
pr = {3, 5, 7}
pr |= {n for n in range(11, 100, 2) if all(n % p for p in pr)}
pr = {n for n in range(1001, 10000, 2) if all(n % p for p in pr)}
# create the set of four digit squares with four different digits
sq = {sq for n in range(32, 100) if len(set(str((sq := n * n)))) == 4}

# consider all three digit numbers without zeros
for p3 in permutations(range(1, 10), 3):
  n4 = (0,) + p3
  # make the nine number sequence
  n9 = [int(''.join(str((d + i) % 10) for d in n4)) for i in range (1, 10)]
  # consider only sequences with no primes ...
  if not any(n in pr for n in n9):
    # ... and exactly two squares
    if sum(bool(n in sq) for n in n9) == 2:
      print(f"Phil's number is {int( ''.join(str(n) for n in p3))}.")

from itertools import permutations

# create sets of four digit primes and squares

pr = {3, 5, 7}

pr |= {n for n in range(11, 100, 2) if all(n % p for p in pr)}

pr = {n for n in range(1001, 10000, 2) if all(n % p for p in pr)}

# create the set of four digit squares with four different digits

sq = {sq for n in range(32, 100) if len(set(str((sq := n * n)))) == 4}

# consider all three digit numbers without zeros

for p3 in permutations(range(1, 10), 3):

n4 = (0,) + p3

# make the nine number sequence

n9 = [int(''.join(str((d + i) % 10) for d in n4)) for i in range (1, 10)]

# consider only sequences with no primes ...

if not any(n in pr for n in n9):

# ... and exactly two squares

if sum(bool(n in sq) for n in n9) == 2:

print(f"Phil's number is {int( ''.join(str(n) for n in p3))}.")

This is a bit faster:


# create the set of four digit primes
pr = {3, 5, 7}
pr |= {n for n in range(11, 100, 2) if all(n % p for p in pr)}
pr4d = {str(n) for n in range(1001, 10000, 2) if all(n % p for p in pr)}
# create the set of four digit squares with four different digits
sq4d = {sq for n in range(32, 100) if len(set((sq := str(n * n)))) == 4}

# consider all possible squares
for s in sq4d:
  # make the nine number sequence from this square
  n9 = {''.join(str(((int(d) + i) % 10)) for d in s)
                                      for i in range (1, 10)}

  # consider any sequences with one more square ...
  if len(t := n9 & sq4d) == 1:
    # ... and no primes
    if s < t.pop() and not n9 & pr4d:
      r = {n[1:] for n in n9 if n[0] == '0'}.pop()
      print(f"Phil's number is {r} [{', '.join(sorted(n9 | {s}))}].")

# create the set of four digit primes

pr = {3, 5, 7}

pr |= {n for n in range(11, 100, 2) if all(n % p for p in pr)}

pr4d = {str(n) for n in range(1001, 10000, 2) if all(n % p for p in pr)}

# create the set of four digit squares with four different digits

sq4d = {sq for n in range(32, 100) if len(set((sq := str(n * n)))) == 4}

# consider all possible squares

for s in sq4d:

# make the nine number sequence from this square

n9 = {''.join(str(((int(d) + i) % 10)) for d in s)

for i in range (1, 10)}

# consider any sequences with one more square ...

if len(t := n9 & sq4d) == 1:

# ... and no primes

if s < t.pop() and not n9 & pr4d:

r = {n[1:] for n in n9 if n[0] == '0'}.pop()

print(f"Phil's number is {r} [{', '.join(sorted(n9 | {s}))}].")

John Z permalink


from itertools import product

# generate primes to 100, then to 10000 (might be easier to use isprime)
p100 = ((2, 3, 5, 7) +
        tuple((p for p in range(11, 100, 2) if all(p % x for x in (3, 5, 7)))))
p10000 = tuple(p for p in range(101, 10000, 2) if all(p % x for x in p100))

sqs = tuple(x * x for x in range(100)) # squares from 0 up to 10000

# consider all possible four digit numbers that start with an initial zero
for d in product((0,), range(1, 10), range(10), range(10)):

  if len(set(d)) != 4: # digits must be different
    continue
  
  # generate other visible numbers
  n = [sum(((x + k) % 10) * 10 ** (3 - i) for i, x in enumerate(d))
       for k in range(1, 10)]

  # are any of the visible numbers prime?
  if any(nn in(p10000) for nn in(n)):
    continue

  # are there two perfect squares in the visible numbers?
  if sum(nn in(sqs) for nn in(n)) == 2:
    print("Phil's number:", ''.join(str(dd) for dd in d[1:]))
    print('Other visible numbers:', str(n)[1:-1])

from itertools import product

# generate primes to 100, then to 10000 (might be easier to use isprime)

p100 = ((2, 3, 5, 7) +

tuple((p for p in range(11, 100, 2) if all(p % x for x in (3, 5, 7)))))

p10000 = tuple(p for p in range(101, 10000, 2) if all(p % x for x in p100))

sqs = tuple(x * x for x in range(100)) # squares from 0 up to 10000

# consider all possible four digit numbers that start with an initial zero

for d in product((0,), range(1, 10), range(10), range(10)):

if len(set(d)) != 4: # digits must be different

continue

# generate other visible numbers

n = [sum(((x + k) % 10) * 10 ** (3 - i) for i, x in enumerate(d))

for k in range(1, 10)]

# are any of the visible numbers prime?

if any(nn in(p10000) for nn in(n)):

continue

# are there two perfect squares in the visible numbers?

if sum(nn in(sqs) for nn in(n)) == 2:

print("Phil's number:", ''.join(str(dd) for dd in d[1:]))

print('Other visible numbers:', str(n)[1:-1])

Sunday Times Teaser 3269 – Combination Lock

by Andrew Skidmore

Published Sunday May 18 2025 (link)

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