Sunday Times Teaser 3283 – Die Hard?

by BRG on August 24, 2025

by Victor Bryant

Published Sunday August 24 2025 (link)

I have three identical standard dice (1 to 6 spots on each face, with 1 opposite 6, 2 opposite 5 and 3 opposite 4). I have placed them together in a row on a table and looked at the row from various sides. Regarding each face of a die as a digit (so that, for example, five spots would be regarded as a 5), I can read three 3-figure primes; one along the front of the row, one (the largest of the three) along the top of the row, and one along the back of the row viewed from behind. Furthermore, the total number of spots on the 11 visible faces is also a prime.

What, in increasing order, are the three 3-figure primes?

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BRG permalink


from itertools import product

#       A B C       back    )
#     ---------             )
#   D | E F G | H   top     )  viewed from the top
#     ---------             )
#       I J K       front   )

# primes up to 666
prms = {3, 5, 7, 11, 13, 17, 19, 23, 29}
prms |= {2} | {n for n in range(31, 666, 2) if all(n % p for p in prms)}

# given two dice faces anti-clockwise at a vertex, find the third
# face at this vertex (using a western die if 'western' is True)
def die_third_face(f, s, western=True):
  if s in {f, 7 - f}:
    raise ValueError
  oth = [i for i in range(1, 7) if i not in {f, s, 7 - f, 7 - s}]
  return oth[((f < s) == ((s - f) % 2)) == (western == (s + f > 7))]

# return the four die faces orthogonal to <x, 7 - x>
orth = lambda x: (v for v in range(1, 7) if v not in {x, 7 - x})

sols = set()
# consider the front face and back face spots
for a, b, c in product(range(1, 7), repeat=3):
  if ((bf := 100 * c + 10 * b + a) in prms
    and (ff := (777 - int(str(bf)[::-1]))) in prms):
    # consider the top face spots
    for e, f, g in product(orth(a), orth(b), orth(c)):
      # the top face prime is the largest of the three primes
      if (tf := 100 * e + 10 * f + g) in prms and tf > max(ff, bf):
          # left face, right face
          d, h = die_third_face(e, a), die_third_face(c, g)
          # the eleven visible faces have a prime number of spots
          if 21 + d + e + f + g + h in prms:
            # the top face has the largest prime
            sols.add(tuple(sorted((ff, bf, tf))))

print(f"Solution: {sols.pop()}")

from itertools import product

# A B C back )

# --------- )

# D | E F G | H top ) viewed from the top

# --------- )

# I J K front )

# primes up to 666

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

prms |= {2} | {n for n in range(31, 666, 2) if all(n % p for p in prms)}

# given two dice faces anti-clockwise at a vertex, find the third

# face at this vertex (using a western die if 'western' is True)

def die_third_face(f, s, western=True):

if s in {f, 7 - f}:

raise ValueError

oth = [i for i in range(1, 7) if i not in {f, s, 7 - f, 7 - s}]

return oth[((f < s) == ((s - f) % 2)) == (western == (s + f > 7))]

# return the four die faces orthogonal to <x, 7 - x>

orth = lambda x: (v for v in range(1, 7) if v not in {x, 7 - x})

sols = set()

# consider the front face and back face spots

for a, b, c in product(range(1, 7), repeat=3):

if ((bf := 100 * c + 10 * b + a) in prms

and (ff := (777 - int(str(bf)[::-1]))) in prms):

# consider the top face spots

for e, f, g in product(orth(a), orth(b), orth(c)):

# the top face prime is the largest of the three primes

if (tf := 100 * e + 10 * f + g) in prms and tf > max(ff, bf):

# left face, right face

d, h = die_third_face(e, a), die_third_face(c, g)

# the eleven visible faces have a prime number of spots

if 21 + d + e + f + g + h in prms:

# the top face has the largest prime

sols.add(tuple(sorted((ff, bf, tf))))

print(f"Solution: {sols.pop()}")

GeoffR permalink


#       f e d 
#     +-------+
#   m | r s t | n
#     +-------+
#       a b c

from itertools import permutations
from number_theory import is_prime

western_triples = {
  (1,2,3), (1,3,5), (1,5,4), (1,4,2),   # around face 1
  (6,2,4), (6,4,5), (6,5,3), (6,3,2),   # around face 6
  (2,1,4), (2,4,6), (2,6,3), (2,3,1),   # around face 2
  (5,1,3), (5,3,6), (5,6,4), (5,4,1),   # around face 5
  (3,1,2), (3,2,6), (3,6,5), (3,5,1),   # around face 3
  (4,1,5), (4,5,6), (4,6,2), (4,2,1),   # around face 4
}

prms = tuple(n for n in range(101, 667, 2) if is_prime(n))

# We only need to permute abc and rst here
# and then set def values from abc
for abc, rst in permutations(prms, 2):
  s_abc, s_rst = (str(x) for x in (abc, rst))
  if set(s_abc + s_rst) & set('0789'):
    continue
  
  a, b, c = (int(x) for x in s_abc)
  r, s, t = (int(x) for x in s_rst)
  d, e, f = 7 - c, 7 - b, 7 - a
  if not is_prime((_def := 100 * d + 10 * e + f)):
    continue
  
  # Order three 3-digit numbera 
  if rst > max(abc, _def):
    # find LH face on 1st die
    for m in range(1, 7):
      if m in (a, r, f, 7 - a, 7 - r, 7 - f):
        continue
      if (m, r, f) in western_triples:
              
      # find RH end of 3rd die
        for n in range(1, 7): 
          if n in (c, t, d, 7 - c, 7 - t, 7 - d):
            continue
          if (n, t, c) in western_triples:
            if is_prime(a + b + c + d + e + f + r + s + t + m + n):
              print(m, abc, _def, rst, n)
              print("Tuples:", (m, r, f), (n, t, c))
              print()

# f e d

# +-------+

# m | r s t | n

# +-------+

# a b c

from itertools import permutations

from number_theory import is_prime

western_triples = {

(1,2,3), (1,3,5), (1,5,4), (1,4,2), # around face 1

(6,2,4), (6,4,5), (6,5,3), (6,3,2), # around face 6

(2,1,4), (2,4,6), (2,6,3), (2,3,1), # around face 2

(5,1,3), (5,3,6), (5,6,4), (5,4,1), # around face 5

(3,1,2), (3,2,6), (3,6,5), (3,5,1), # around face 3

(4,1,5), (4,5,6), (4,6,2), (4,2,1), # around face 4

}

prms = tuple(n for n in range(101, 667, 2) if is_prime(n))

# We only need to permute abc and rst here

# and then set def values from abc

for abc, rst in permutations(prms, 2):

s_abc, s_rst = (str(x) for x in (abc, rst))

if set(s_abc + s_rst) & set('0789'):

continue

a, b, c = (int(x) for x in s_abc)

r, s, t = (int(x) for x in s_rst)

d, e, f = 7 - c, 7 - b, 7 - a

if not is_prime((_def := 100 * d + 10 * e + f)):

continue

# Order three 3-digit numbera

if rst > max(abc, _def):

# find LH face on 1st die

for m in range(1, 7):

if m in (a, r, f, 7 - a, 7 - r, 7 - f):

continue

if (m, r, f) in western_triples:

# find RH end of 3rd die

for n in range(1, 7):

if n in (c, t, d, 7 - c, 7 - t, 7 - d):

continue

if (n, t, c) in western_triples:

if is_prime(a + b + c + d + e + f + r + s + t + m + n):

print(m, abc, _def, rst, n)

print("Tuples:", (m, r, f), (n, t, c))

print()

BRG permalink

Here are some changes to simplify and speed up the code. I have also changed the notation for the die faces to match those I used in my first comment so that the two solutions are easier to compare.


#       A B C       back    )
#     ---------             )
#   D | E F G | H   top     )  viewed from the top
#     ---------             )
#       I J K       front   )

from itertools import combinations
from number_theory import is_prime

western_triples = {
  (1,2,3), (1,3,5), (1,5,4), (1,4,2),   # around face 1
  (6,2,4), (6,4,5), (6,5,3), (6,3,2),   # around face 6
  (2,1,4), (2,4,6), (2,6,3), (2,3,1),   # around face 2
  (5,1,3), (5,3,6), (5,6,4), (5,4,1),   # around face 5
  (3,1,2), (3,2,6), (3,6,5), (3,5,1),   # around face 3
  (4,1,5), (4,5,6), (4,6,2), (4,2,1),   # around face 4
}
# map die faces <x> and <y> to the third face <z> anti-clockwise
xy2z = {(x, y): z for x, y, z in western_triples}

prms = tuple(n for n in range(101, 667, 2)
             if is_prime(n) and not set(str(n)).intersection('0789'))

# permute primes for CBA and EFG and then set IJK values from ABC
for cba, efg in combinations(prms, 2):
  s_cba, s_efg = (str(x) for x in (cba, efg))

  # extract the individual digits
  c, b, a = (int(x) for x in s_cba)
  e, f, g = (int(x) for x in s_efg)
  i, j, k = 7 - a, 7 - b, 7 - c
  ijk = 100 * i + 10 * j + k
  # the back face must be prime
  if not is_prime((ijk := 100 * i + 10 * j + k)):
    continue
  
  # the top face prime is the largest
  if efg > max(cba, ijk):
    # find and check the left and right end faces <d> and <h>
    if (d := xy2z.get((e, a), 0)) and (h := xy2z.get((g, k), 0)):
      # the total number of dice spots is prime
      if is_prime(a + b + c + i + j + k + e + f + g + d + h):
        abc = 100 * a + 10 * b + c
        print(f"{abc = } | {d = } {efg = } {h = } | {ijk = }")
        print(f"die triples used: ({e}, {a}) -> {d} ({g}, {k}) -> {h}\n")

# A B C back )

# --------- )

# D | E F G | H top ) viewed from the top

# --------- )

# I J K front )

from itertools import combinations

from number_theory import is_prime

western_triples = {

(1,2,3), (1,3,5), (1,5,4), (1,4,2), # around face 1

(6,2,4), (6,4,5), (6,5,3), (6,3,2), # around face 6

(2,1,4), (2,4,6), (2,6,3), (2,3,1), # around face 2

(5,1,3), (5,3,6), (5,6,4), (5,4,1), # around face 5

(3,1,2), (3,2,6), (3,6,5), (3,5,1), # around face 3

(4,1,5), (4,5,6), (4,6,2), (4,2,1), # around face 4

}

# map die faces <x> and <y> to the third face <z> anti-clockwise

xy2z = {(x, y): z for x, y, z in western_triples}

prms = tuple(n for n in range(101, 667, 2)

if is_prime(n) and not set(str(n)).intersection('0789'))

# permute primes for CBA and EFG and then set IJK values from ABC

for cba, efg in combinations(prms, 2):

s_cba, s_efg = (str(x) for x in (cba, efg))

# extract the individual digits

c, b, a = (int(x) for x in s_cba)

e, f, g = (int(x) for x in s_efg)

i, j, k = 7 - a, 7 - b, 7 - c

ijk = 100 * i + 10 * j + k

# the back face must be prime

if not is_prime((ijk := 100 * i + 10 * j + k)):

continue

# the top face prime is the largest

if efg > max(cba, ijk):

# find and check the left and right end faces <d> and <h>

if (d := xy2z.get((e, a), 0)) and (h := xy2z.get((g, k), 0)):

# the total number of dice spots is prime

if is_prime(a + b + c + i + j + k + e + f + g + d + h):

abc = 100 * a + 10 * b + c

print(f"{abc = } | {d = } {efg = } {h = } | {ijk = }")

print(f"die triples used: ({e}, {a}) -> {d} ({g}, {k}) -> {h}\n")

Frits permalink

@Brian,

Your diagram is confusing to me. From it I would assume IJK and CBA were
primes.

In line 26 you can use combinations iso permutations as “prms” is ordered.
In line 23 an isdisjoint() could be used as the intersection content is not
important.

With a little analysis we can derive the value of 5 variables and then can
calculate one side face.

Reply
- BRG permalink
  
  Thanks for the code review. I have added your improvements and
  corrected the definitions in the code.
  
  Reply

BRG permalink

Here is another approach.


#       A B C       back    )
#     ---------             )
#   D | E F G | H   top     )  viewed from the top
#     ---------             )
#       I J K       front   )

# primes up to 67
prms = {3, 5, 7, 11, 13, 17, 19, 23, 29}
prms |= {2} | {n for n in range(31, 67, 2) if all(n % p for p in prms)}
# primes that can be made with three dice
dprms = tuple(n for n in range(101, 667, 2)
          if all(n % p for p in prms) and not (set(str(n)).intersection('0789')))

# given two dice faces anti-clockwise at a vertex, find the third
# face at this vertex (using a western die if 'western' is True)
def die_third_face(f, s, western=True):
  if s in {f, 7 - f}:
    raise ValueError
  oth = [i for i in range(1, 7) if i not in {f, s, 7 - f, 7 - s}]
  return oth[((f < s) == ((s - f) % 2)) == (western == (s + f > 7))]

# consider primes for CBA (and IJK)
for cba in dprms:
  c, b, a = (int(x) for x in str(cba))
  i, j, k = 7 - a, 7 - b, 7 - c
  if (ijk := 100 * i + 10 * j + k) in dprms:

    # consider primes for EFG (the largest of the three primes)
    for efg in dprms[dprms.index(max(cba, ijk)):]:
      e, f, g = (int(x) for x in str(efg))
      # find valid left and right end faces D and H
      try:
        d, h = die_third_face(e, a), die_third_face(g, k)
      except ValueError:
        continue
      # the total number of dice spots is prime
      if 21 + d + e + f + g + h in prms:
        abc = 100 * a + 10 * b + c
        print(f"{abc = } | {d = } {efg = } {h = } | {ijk = } "
              f"-> {tuple(sorted((abc, efg, ijk)))}")

# A B C back )

# --------- )

# D | E F G | H top ) viewed from the top

# --------- )

# I J K front )

# primes up to 67

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

prms |= {2} | {n for n in range(31, 67, 2) if all(n % p for p in prms)}

# primes that can be made with three dice

dprms = tuple(n for n in range(101, 667, 2)

if all(n % p for p in prms) and not (set(str(n)).intersection('0789')))

# given two dice faces anti-clockwise at a vertex, find the third

# face at this vertex (using a western die if 'western' is True)

def die_third_face(f, s, western=True):

if s in {f, 7 - f}:

raise ValueError

oth = [i for i in range(1, 7) if i not in {f, s, 7 - f, 7 - s}]

return oth[((f < s) == ((s - f) % 2)) == (western == (s + f > 7))]

# consider primes for CBA (and IJK)

for cba in dprms:

c, b, a = (int(x) for x in str(cba))

i, j, k = 7 - a, 7 - b, 7 - c

if (ijk := 100 * i + 10 * j + k) in dprms:

# consider primes for EFG (the largest of the three primes)

for efg in dprms[dprms.index(max(cba, ijk)):]:

e, f, g = (int(x) for x in str(efg))

# find valid left and right end faces D and H

try:

d, h = die_third_face(e, a), die_third_face(g, k)

except ValueError:

continue

# the total number of dice spots is prime

if 21 + d + e + f + g + h in prms:

abc = 100 * a + 10 * b + c

print(f"{abc = } | {d = } {efg = } {h = } | {ijk = } "

f"-> {tuple(sorted((abc, efg, ijk)))}")

Sunday Times Teaser 3283 – Die Hard?

by Victor Bryant

Published Sunday August 24 2025 (link)

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