Sunday Times Teaser 3313 – Rod’s Rods

by BRG on 22 March 2026

by Victor Bryant

Published Sunday March 22 2026

Rodney has a set of small identical rods, which he uses to display numbers in “calculator” style. The “0” requires six rods, the “1” requires two rods, the “2”, “3” and “5” require five each, the “4” requires four, the “6” and “9” require six, the “7” requires three and the “8” requires seven. He recently showed me three numbers, each made by using the whole set. The lowest was a three-figure perfect square, the next was a palindromic prime, and the highest was a four-figure prime with its four different digits increasing from left to right.

What were those three numbers?

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BRG permalink


from number_theory import Primes

# The number of rods used for each digit
#       0  1  2  3  4  5  6  7  8  9
rods = [6, 2, 5, 5, 4, 5, 6, 3, 7, 6]

# the total number of rods used in a number
rod_cnt = lambda nbr: sum(rods[int(dgt)] for dgt in str(nbr))
  
# consider all three digit squares
for sr in range(10, 32):
  # find their rod counts
  sq_cnt = rod_cnt((sqr := sr ** 2))
  
  # consider three digit palindromic primes higher than
  # the square with the same rod count
  for p1 in Primes().range(max(100, sqr), 1000):
    if str(p1) == str(p1)[::-1] and sq_cnt == rod_cnt(p1):
      
      # consider four digit primes higher than the palindromic
      # prime whose digits increase from left to dight
      for p2 in Primes().range(max(1000, p1 + 1), 10000):
        a, b, c, d = str(p2)
        if a < b < c < d and sq_cnt == rod_cnt(p2):
          print(f"{sqr = }, {p1 = }, {p2 = }")

from number_theory import Primes

# The number of rods used for each digit

# 0 1 2 3 4 5 6 7 8 9

rods = [6, 2, 5, 5, 4, 5, 6, 3, 7, 6]

# the total number of rods used in a number

rod_cnt = lambda nbr: sum(rods[int(dgt)] for dgt in str(nbr))

# consider all three digit squares

for sr in range(10, 32):

# find their rod counts

sq_cnt = rod_cnt((sqr := sr ** 2))

# consider three digit palindromic primes higher than

# the square with the same rod count

for p1 in Primes().range(max(100, sqr), 1000):

if str(p1) == str(p1)[::-1] and sq_cnt == rod_cnt(p1):

# consider four digit primes higher than the palindromic

# prime whose digits increase from left to dight

for p2 in Primes().range(max(1000, p1 + 1), 10000):

a, b, c, d = str(p2)

if a < b < c < d and sq_cnt == rod_cnt(p2):

print(f"{sqr = }, {p1 = }, {p2 = }")

Frits permalink

is_prime() of enigma is a little bit faster in this program than the number_theory version.
Maybe this is happening as I also use enigma to measure the performance.


# from enigma import is_prime
from number_theory import is_prime
from itertools import product, combinations
from functools import reduce
 
d2n = lambda s: reduce(lambda x, y: 10 * x + y, s) 
 
# collect keys per value
revdict = lambda d: {v: [k1 for k1, v1 in d.items() if v1 == v] 
                     for v in set(d.values())}
 
# the total number of rods used in a number
rod_cnt = lambda n: sum(rods[int(dgt)] for dgt in str(n))
 
# number of segment per digit
#       0  1  2  3  4  5  6  7  8  9
rods = [6, 2, 5, 5, 4, 5, 6, 3, 7, 6]
 
# 4-digit primes with increasing digits
p4 = revdict({n: rod_cnt(n) for i4 in combinations(range(1, 10), 4)
                 if is_prime(n := d2n(i4))})
 
# 3-digit squares
sqs = {rc: [n2 for n in range(10, 32) if rod_cnt(n2 := n * n) == rc] 
            for rc in p4}
            
# palindromic primes (number xyyx is not prime, divisibile by 11)
pals = revdict({n: rod_cnt(n) for n in range(101, 1000, 2) if is_prime(n) and 
                   str(n) == str(n)[::-1] and rod_cnt(n) in sqs})
 
print("answer(s):")
for rc, ns in pals.items():
  for s, n, p in product(sqs[rc], ns, p4[rc]):
    if s >= n: continue
    print(f"{s}, {n} and {p}")

# from enigma import is_prime

from number_theory import is_prime

from itertools import product, combinations

from functools import reduce

d2n = lambda s: reduce(lambda x, y: 10 * x + y, s)

# collect keys per value

revdict = lambda d: {v: [k1 for k1, v1 in d.items() if v1 == v]

for v in set(d.values())}

# the total number of rods used in a number

rod_cnt = lambda n: sum(rods[int(dgt)] for dgt in str(n))

# number of segment per digit

# 0 1 2 3 4 5 6 7 8 9

rods = [6, 2, 5, 5, 4, 5, 6, 3, 7, 6]

# 4-digit primes with increasing digits

p4 = revdict({n: rod_cnt(n) for i4 in combinations(range(1, 10), 4)

if is_prime(n := d2n(i4))})

# 3-digit squares

sqs = {rc: [n2 for n in range(10, 32) if rod_cnt(n2 := n * n) == rc]

for rc in p4}

# palindromic primes (number xyyx is not prime, divisibile by 11)

pals = revdict({n: rod_cnt(n) for n in range(101, 1000, 2) if is_prime(n) and

str(n) == str(n)[::-1] and rod_cnt(n) in sqs})

print("answer(s):")

for rc, ns in pals.items():

for s, n, p in product(sqs[rc], ns, p4[rc]):

if s >= n: continue

print(f"{s}, {n} and {p}")

John Z. permalink


from itertools import combinations, product

# return dictionary of number of rods to value in iterator of integers
def rodcalc(x):
  # key: rod count; values: list of numbers with the given number of rods 
  x2r = dict()
  for i in x:
    r = sum(rods[int(a)] for a in str(i))
    x2r[r] = x2r.get(r, []) + [i]
  return x2r
    
rods = (6, 2, 5, 5, 4, 5, 6, 3, 7, 6)

#primes up to 100
p = (3, 5, 7)
p100 = (2,) + p + tuple(x for x in range(11, 101, 2) if all(x % q for q in(p)))

# three digit squares
sqrods = rodcalc((i * i for i in range(10, 32)))
  
# 3 digit palindromic primes (no 4 digit palindromic primes as XYYX % 11 = 0)
pprods = rodcalc((i for si in product('1379', '0123456789') if
      (i := int(''.join(si + si[0:1]))) and all(i % p for p in p100[:11])))

# 4 digit primes with increasing digits
p4drods = rodcalc((i for si in combinations('123456789', 4) if
                   (i := int(''.join(si))) and all(i % p for p in p100)))

# intersection of keys gives the number of rods in the set
for nrods in (sqrods.keys() & pprods.keys() & p4drods.keys()):
  for n1 in sqrods[nrods]:
    for n2 in pprods[nrods]:
      if n1 < n2:
        for n3 in p4drods[nrods]:
          print('Number of rods in the set:', nrods)
          print('The numbers are:', n1, n2, n3)

from itertools import combinations, product

# return dictionary of number of rods to value in iterator of integers

def rodcalc(x):

# key: rod count; values: list of numbers with the given number of rods

x2r = dict()

for i in x:

r = sum(rods[int(a)] for a in str(i))

x2r[r] = x2r.get(r, []) + [i]

return x2r

rods = (6, 2, 5, 5, 4, 5, 6, 3, 7, 6)

#primes up to 100

p = (3, 5, 7)

p100 = (2,) + p + tuple(x for x in range(11, 101, 2) if all(x % q for q in(p)))

# three digit squares

sqrods = rodcalc((i * i for i in range(10, 32)))

# 3 digit palindromic primes (no 4 digit palindromic primes as XYYX % 11 = 0)

pprods = rodcalc((i for si in product('1379', '0123456789') if

(i := int(''.join(si + si[0:1]))) and all(i % p for p in p100[:11])))

# 4 digit primes with increasing digits

p4drods = rodcalc((i for si in combinations('123456789', 4) if

(i := int(''.join(si))) and all(i % p for p in p100)))

# intersection of keys gives the number of rods in the set

for nrods in (sqrods.keys() & pprods.keys() & p4drods.keys()):

for n1 in sqrods[nrods]:

for n2 in pprods[nrods]:

if n1 < n2:

for n3 in p4drods[nrods]:

print('Number of rods in the set:', nrods)

print('The numbers are:', n1, n2, n3)

Sunday Times Teaser 3313 – Rod’s Rods

by Victor Bryant

Published Sunday March 22 2026

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