S2T2

Jim Randell 8:21 am on 15 July 2025 Permalink Reply
Tags: by: Victor Bryant ( 140 )   

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  Jim Randell 8:21 am on 15 July 2025 Permalink | Reply

  Each digit is the last digit of a 3-digit prime, so the only possible digits are 1, 3, 9, 7.

  This Python program looks for solutions using increasing numbers of digits, and stops after the smallest possible sequences are found.

  It runs in 60ms. (Internal runtime is 841µs).

  Run: [ @codepad ]

  from enigma import (irange, inf, primes, multiset, subsets, tuples, rev, nconcat, wrap, uniq, printf)
  
  # up to 3-digit primes
  primes.expand(999)
  
  # possible digits
  digits = [1, 3, 7, 9]
  
  # generate numbers from <k> adjacent digits of <ns> (in both directions)
  @wrap(uniq)  # skip duplicate numbers
  def adj(ns, k=3):
    for ds in tuples(ns, k, circular=1):
      yield nconcat(ds)
      yield nconcat(rev(ds))
  
  # look for solutions using <n> digits
  def solve(n):
    # choose the n digits
    for ds in subsets(digits, size=n, select='R'):
      # the sum of the digits is prime
      if not (sum(ds) in primes): continue
      # each digit occurs a prime number of times
      m = multiset.from_seq(ds)
      if not all(v in primes for v in m.values()): continue
  
      # start with the smallest number
      n0 = m.min()
      # and choose an arrangement of the rest
      for ns in subsets(m.difference([n0]), size=len, select='mP', fn=list):
        if rev(ns) < ns: continue  # skip duplicate solutions
        ns.insert(0, n0)
        # check each 3 adjacent digits form a prime (in both directions)
        if not all(x in primes for x in adj(ns)): continue
        yield ns
  
  # find the smallest possible sequences
  for n in irange(3, inf):
    s = 0
    for ns in solve(n):
      printf("[n={n}] {ns} -> {t}", t=sum(ns))
      s += 1
    if s: break
  

  Solution: The total of the digits is 29.

  The digits in the circle are: (1, 9, 1, 9, 9). Consisting of 2 occurrences of the digit 1, and 3 occurrences of the digit 9.

  The 3-digit numbers formed are: 191, 199, 919, 991. Which are all prime.

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• 

  Ruud 8:03 am on 16 July 2025 Permalink | Reply

  import istr
  import collections
  import itertools
  
  primes = [i for i in istr.range(2, 1000) if i.is_prime()]
  candidates = {str(prime) for prime in primes if len(prime) == 3 and all(c in "1379" for c in prime)}
  for n in itertools.count(2):
      for x in itertools.product("1379", repeat=int(n)):
          s = "".join(x)
          if all(count in primes for count in collections.Counter(s).values()):
              s2 = str(s + s[0:2])
              if (
                  all(s2[i : i + 3] in candidates for i in range(len(s)))
                  and all(s2[i : i + 3][::-1] in candidates for i in range(len(s)))
                  and (sumx:=sum(map(int, x))) in primes
              ):
                  print(s,sumx)
                  assert False
  

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