S2T2

Jim Randell 9:38 am on 3 October 2025 Permalink Reply
Tags: by: Graham Smithers ( 82 )   

• 

  Jim Randell 9:39 am on 3 October 2025 Permalink | Reply

  If we suppose the square number is ABCDE, and the cube is FGHIJ then the combined number is one of:

  AFBGCHDIEJ
  FAGBHCIDJE

  And the corresponding prime positions (2nd, 3rd, 5th, 7th) form a prime.

  So the prime is:

  AFBGCHDIEJ → FBCD
  FAGBHCIDJE → AGHI

  Here is a straightforward solution using the [[ SubstitutedExpression() ]] solver from the enigma.py library.

  It runs in 114ms. (Internal runtime is 43ms).

  from enigma import (SubstitutedExpression, is_prime, printf)
  
  # use the SubstitutedExpression solver to find the square and the cube
  p = SubstitutedExpression(
    ["is_square(ABCDE)", "is_cube(FGHIJ)"],
    answer="(ABCDE, FGHIJ, AFBGCHDIEJ, FBCD, FAGBHCIDJE, AGHI)",
    # [optional] squares cannot end in 2, 3, 7, 8
    d2i={ 0: 'AF', 2: 'E', 3: 'E', 7: 'E', 8: 'E' }
  )
  
  # look for viable solutions
  for (sq, cb, n1, p1, n2, p2) in p.answers(verbose=0):
    (f1, f2) = (is_prime(p1), is_prime(p2))
    if f1 or f2:
      printf("square = {sq}, cube = {cb}")
      if f1: printf("-> n = {n1}, prime = {p1}")
      if f2: printf("-> n = {n2}, prime = {p2}")
      printf()
  

  Solution: The 10-digit number is: 5204137869.

  So the 4-digit prime is 2017, the odd digits form 50176 (= 224^2), and the even digits form 24389 (= 29^3).

  ---

  For a faster runtime we can collect the squares and cubes up front.

  The following Python program has an internal runtime of 479µs.

  from enigma import (
    defaultdict, irange, sq, cb, diff, cproduct,
    interleave, restrict, is_prime, join, printf
  )
  
  # index numbers (with distinct digits) by digit content
  def numbers(seq):
    d = defaultdict(list)
    for n in seq:
      s = str(n)
      # check digits are distinct
      if len(s) != len(set(s)): continue
      d[join(sorted(s))].append(s)
    return d
  
  # find candidate squares and cubes
  sqs = numbers(sq(x) for x in irange(100, 316))
  cbs = numbers(cb(x) for x in irange(22, 46))
  
  # consider possible cubes
  for (kc, vs) in cbs.items():
    # and look for corresponding squares
    ks = join(diff("0123456789", kc))
    for (square, cube) in cproduct([sqs[ks], vs]):
      # look for possible interleavings
      for (a, b) in [(square, cube), (cube, square)]:
        n = join(interleave(a, b))
        # extract the digits in (1-indexed) prime positions
        p = int(restrict(n, [1, 2, 4, 6]))
        if is_prime(p):
          printf("square = {square}, cube = {cube} -> n = {n}, p = {p}")
  

  LikeLike