Sunday Times Teaser 3314 – Number Unobtainable!

by BRG on 29 March 2026

by Stephen Hogg

Published Sunday March 29 2026 (link)

Dyall recalls six-digit phone numbers as a sequence of two three-figure values. His wife, Belle, recalls them as a sequence of three two-figure values. Their courting-days’ phone numbers had no zeroes. For each of their numbers, Dyall recalled that the two three-figure values in order made a descending set of primes, while Belle recalled that the three two-figure values in order made a descending set of primes. Back then their town’s residential numbers never began with “7”, “8” or “9”. Dyall told Tim these things, but not the numbers. He then stated that choosing a digit at random to be told its value would give a one in three chance of Tim being able to work out Dyall’s old phone number with certainty. In Belle’s case the chance was greater still.

Give Belle’s old phone number.

From → Uncategorized

4 Comments Leave one →

BRG permalink


from itertools import combinations
from collections import Counter
  
prms = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31}
prms |= {2} | {n for n in range(37, 1000, 2) if all(n % p for p in prms)}

# for collecting phone numbers as digits
pnds = []

# choose three two digit primes less than 70 
# in descending order for the phone numbers
ps = [n for n in range(11, 70, 2) if n in prms][::-1]
for p3 in combinations(ps, 3):
  # collect the six digits of the phone number
  dgts = [d for ds in (divmod(n, 10) for n in p3) for d in ds]
  # convert into two three digit primes ...
  r, s = 10 * p3[0] + p3[1] // 10, 100 * (p3[1] % 10) + p3[2]
  # ... in descending order
  if r > s and {r, s}.issubset(prms):
    pnds.append(dgts)

# collect digits in all phone numbers indexed on their positions
pos = list(Counter(dp) for dp in zip(*pnds))

# for each digit position in each phone number find the number 
# of digits that identify the phone number
for pn in pnds:
  n = sum(pos[i][d] == 1 for i, d in enumerate(pn))
  if n == 2:
    d_pn = ''.join(map(str, pn))
  elif n > 2:
    b_pn = ''.join(map(str, pn))
else: 
  print(f"Belle: {b_pn} (Dyall: {d_pn}).")

from itertools import combinations

from collections import Counter

prms = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31}

prms |= {2} | {n for n in range(37, 1000, 2) if all(n % p for p in prms)}

# for collecting phone numbers as digits

pnds = []

# choose three two digit primes less than 70

# in descending order for the phone numbers

ps = [n for n in range(11, 70, 2) if n in prms][::-1]

for p3 in combinations(ps, 3):

# collect the six digits of the phone number

dgts = [d for ds in (divmod(n, 10) for n in p3) for d in ds]

# convert into two three digit primes ...

r, s = 10 * p3[0] + p3[1] // 10, 100 * (p3[1] % 10) + p3[2]

# ... in descending order

if r > s and {r, s}.issubset(prms):

pnds.append(dgts)

# collect digits in all phone numbers indexed on their positions

pos = list(Counter(dp) for dp in zip(*pnds))

# for each digit position in each phone number find the number

# of digits that identify the phone number

for pn in pnds:

n = sum(pos[i][d] == 1 for i, d in enumerate(pn))

if n == 2:

d_pn = ''.join(map(str, pn))

elif n > 2:

b_pn = ''.join(map(str, pn))

else:

print(f"Belle: {b_pn} (Dyall: {d_pn}).")

John Z. permalink


from itertools import combinations
#2 digit primes up to 70
p = (3, 5, 7)
p2d = tuple(x for x in range(69, 10, -2) if all(x % q for q in(p)))

p3d = tuple(x for x in range(111, 1000, 2) if all(x % q for q in (p + p2d)))

# candidate phone numbers
pns = []

# consider Belle's 2 digit combinations
for b1, b2, b3 in combinations(p2d, 3):
  # Dyall's 3 digit numbers:
  b2t, b2u = divmod(b2, 10)
  d1 = b1 * 10 + b2t
  d2 = b2u * 100 + b3
  if d1 > d2 and d1 in p3d and d2 in p3d:
    print(b1,b2,b3, '------', d1, d2)
    pns += [divmod(b1, 10) + (b2t, b2u) + divmod(b3, 10)]

print()
# the number of candidate phone numbers
npn = len(pns)

# rotate phone numbers to prepare count of digit repeats in columns
snp = tuple(zip(*pns))

# count digit repeats in columns then rotate back to line up with phone numbers
dc = tuple(zip(*tuple(tuple(s.count(dgt) for dgt in s) for s in snp)))

# find digits of a phone number which are unique in 1/3 or more columns 
for i, dci in enumerate(dc):
  if dci.count(1) * 3 > npn:
    print('Belle:', ''.join(map(str,pns[i])))
  elif dci.count(1) * 3 == npn:
    print('Dyall:', ''.join(map(str,pns[i])))

from itertools import combinations

#2 digit primes up to 70

p = (3, 5, 7)

p2d = tuple(x for x in range(69, 10, -2) if all(x % q for q in(p)))

p3d = tuple(x for x in range(111, 1000, 2) if all(x % q for q in (p + p2d)))

# candidate phone numbers

pns = []

# consider Belle's 2 digit combinations

for b1, b2, b3 in combinations(p2d, 3):

# Dyall's 3 digit numbers:

b2t, b2u = divmod(b2, 10)

d1 = b1 * 10 + b2t

d2 = b2u * 100 + b3

if d1 > d2 and d1 in p3d and d2 in p3d:

print(b1,b2,b3, '------', d1, d2)

pns += [divmod(b1, 10) + (b2t, b2u) + divmod(b3, 10)]

print()

# the number of candidate phone numbers

npn = len(pns)

# rotate phone numbers to prepare count of digit repeats in columns

snp = tuple(zip(*pns))

# count digit repeats in columns then rotate back to line up with phone numbers

dc = tuple(zip(*tuple(tuple(s.count(dgt) for dgt in s) for s in snp)))

# find digits of a phone number which are unique in 1/3 or more columns

for i, dci in enumerate(dc):

if dci.count(1) * 3 > npn:

print('Belle:', ''.join(map(str,pns[i])))

elif dci.count(1) * 3 == npn:

print('Dyall:', ''.join(map(str,pns[i])))

BRG permalink

Hi John,

Your line 30 is ingenious but I am not sure that is the best approach for readability (although your comment gives a good clue!).

I think I would have parsed it in this way: to make the python code easier to follow


# count digit repeats in columns
dc = (tuple(s.count(dgt) for dgt in s) for s in zip(*pns))

# find digits of a phone number which are unique in 1/3 or more columns 
for i, dci in enumerate(zip(*dc)):

# count digit repeats in columns

dc = (tuple(s.count(dgt) for dgt in s) for s in zip(*pns))

# find digits of a phone number which are unique in 1/3 or more columns

for i, dci in enumerate(zip(*dc)):

John Z. permalink

Hi Brian,

Lines 26 – 30 could also be replaced by the following which avoids zipping and zipping back by using a more ‘manual’ approach to counting repeat digits:


# count digit repeats in columns for each phone number
dc = tuple(tuple(sum(pns[r][c] == pns[r2][c] for r2 in range(npn))
             for c in range(6)) for r in range(npn))

# count digit repeats in columns for each phone number

dc = tuple(tuple(sum(pns[r][c] == pns[r2][c] for r2 in range(npn))

for c in range(6)) for r in range(npn))

Sunday Times Teaser 3314 – Number Unobtainable!

by Stephen Hogg

Published Sunday March 29 2026 (link)

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta