Sunday Times Teaser 3306 – Pin-up

by BRG on 1 February 2026

by Victor Bryant

Published Sunday February 01 2026 (link)

Over the years I have used seven different PINs. The first consisted of a four-digit number (the first digit not being zero), and thereafter each new PIN was obtained by swapping over two of the digits of the previous PIN; each time this increased the PIN.

The first PIN (not surprisingly!) was divisible by 1, the second was divisible by 2, the third by 3, and so on, with the seventh divisible by 7 (in fact the seventh PIN was the only one divisible by 7).

What was the third PIN?

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BRG permalink


# work backwards from the last (highest) PIN
# cnt = PIN number; p = PIN; ps = PIN sequence
def pins(cnt, p, ps=()):
  ps = (p,) + ps
  if cnt == 1:
      yield ps
  else:
    # create the next PIN by exchanging two digits
    for i in (0, 1, 2):
      for j in range(i + 1, 4):
        s = list(str(p))
        s[i], s[j] = s[j], s[i]
        t = int(''.join(s))
        # all PINs have four digits and are divisible by the
        # PIN number; only the initial PIN here (that is the
        # last PIN) is divisible by 7
        if 1000 <= t < p and t % (cnt - 1) == 0 and t % 7:
          yield from pins(cnt - 1, t, ps)

# the last PIN is divisible by 7 and 3 (because the sum of
# its digits is the same as that of the PIN divisible by 3)
for p in range(9996, 999, -21):
  for sp in pins(7, p):
    print(f"3rd PIN = {sp[2]} {sp}")

# work backwards from the last (highest) PIN

# cnt = PIN number; p = PIN; ps = PIN sequence

def pins(cnt, p, ps=()):

ps = (p,) + ps

if cnt == 1:

yield ps

else:

# create the next PIN by exchanging two digits

for i in (0, 1, 2):

for j in range(i + 1, 4):

s = list(str(p))

s[i], s[j] = s[j], s[i]

t = int(''.join(s))

# all PINs have four digits and are divisible by the

# PIN number; only the initial PIN here (that is the

# last PIN) is divisible by 7

if 1000 <= t < p and t % (cnt - 1) == 0 and t % 7:

yield from pins(cnt - 1, t, ps)

# the last PIN is divisible by 7 and 3 (because the sum of

# its digits is the same as that of the PIN divisible by 3)

for p in range(9996, 999, -21):

for sp in pins(7, p):

print(f"3rd PIN = {sp[2]} {sp}")

John Z. permalink

Brilliant solution, Brian! I’m a bit uncomfortable with changing the value of a parameter within a recursive function, though. I would have preferred something like this:


# work backwards from the last (highest) PIN
# cnt = PIN number; p = PIN; ps = PIN sequence
def pins(cnt, ps):
  if cnt == 1:
      yield ps
  else:
    # create the next PIN by exchanging two digits
    for i in (0, 1, 2):
      for j in range(i + 1, 4):
        s = list(str(ps[0]))
        s[i], s[j] = s[j], s[i]
        t = int(''.join(s))
        # all PINs have four digits and are divisible by the
        # PIN number; only the initial PIN here (that is the
        # last PIN) is divisible by 7
        if 1000 <= t < ps[0] and t % (cnt - 1) == 0 and t % 7:
          yield from pins(cnt - 1, (t,) + ps)

# the last PIN is divisible by 7 and 3 (because the sum of
# its digits is the same as that of the PIN divisible by 3)
for p in range(9996, 999, -21):
  for sp in pins(7, (p,)):
    print(f"3rd PIN = {sp[2]} {sp}")

# work backwards from the last (highest) PIN

# cnt = PIN number; p = PIN; ps = PIN sequence

def pins(cnt, ps):

if cnt == 1:

yield ps

else:

# create the next PIN by exchanging two digits

for i in (0, 1, 2):

for j in range(i + 1, 4):

s = list(str(ps[0]))

s[i], s[j] = s[j], s[i]

t = int(''.join(s))

# all PINs have four digits and are divisible by the

# PIN number; only the initial PIN here (that is the

# last PIN) is divisible by 7

if 1000 <= t < ps[0] and t % (cnt - 1) == 0 and t % 7:

yield from pins(cnt - 1, (t,) + ps)

# the last PIN is divisible by 7 and 3 (because the sum of

# its digits is the same as that of the PIN divisible by 3)

for p in range(9996, 999, -21):

for sp in pins(7, (p,)):

print(f"3rd PIN = {sp[2]} {sp}")

BRG permalink

Hi John,

I am guessing its my line 5 you feel uncomfortable with. I understand it
looks a bit out of place but my reason for this is that I prefer my ps
variable to be a pure output only variable and my p to be the input variable.
I can load ps from p in the call to pins but that seems a bit awkward so I
decided to do it as a part of the recursion. From the respective of the
recursion on line 19 it just means that the update to ps is delayed to
the start of the next recursion.

But its always good to develop and maintain a personal style!

Reply

Sunday Times Teaser 3306 – Pin-up

by Victor Bryant

Published Sunday February 01 2026 (link)

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