S2T2

Jim Randell 9:58 pm on 9 January 2026 Permalink Reply
Tags: by: Bernardo Recaman ( 6 )   

• 

  Jim Randell 10:03 pm on 9 January 2026 Permalink | Reply

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

  It runs in 226ms. (Internal runtime of the generated code is 126ms).

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  "pow(F + L + O + W + E + R, P) = FLOWER"
  
  "FACTOR % A = 0"
  "FACTOR % R = 0"
  "FACTOR % T = 0"
  
  --answer="(FLOWER, ART)"
  

  Solution: FLOWER ART = 390625 751.

  And:

  FLOWER = 390625 = 25^4 = (3 + 9 + 0 + 6 + 2 + 5) ^ 4.
  FACTOR = 378105 is divisible by 7, 5, 1.

  LikeLike

• 

  Ruud 6:11 am on 10 January 2026 Permalink | Reply

  import istr
  
  for f, l, o, w, e, r, p in istr.permutations(range(10), 7):
      if sum(istr("=flower")) ** p == istr("=flower"):
          for a, c, t in istr.permutations(range(10), 3):
              if istr("=flowerpact").all_distinct() and all(istr("=factor").is_divisible_by(x) for x in (a, r, t)):
                  print(istr("=flower"), istr("=art"))
  

  LikeLike

  • 

    ruudvanderham 11:32 am on 10 January 2026 Permalink | Reply

    This version is *much* faster. Note that it required the latest version of istr (1.1.16):

    import istr
    
    for p in range(2, 10):
        for flower in istr.power_ofs(p, 100000, 1000000):
            if sum(flower) ** p == flower:
                istr.decompose(flower, "flower")
                for a, c, t in istr.permutations(set(istr.range(10))-set(flower)-{p}, 3):
                    if istr("=flowerpact").all_distinct() and all(istr("=factor").is_divisible_by(x) for x in (a, r, t)):
                        print(istr("=flower"), istr("=art"))
    

    LikeLike

• 

  GeoffR 2:48 pm on 10 January 2026 Permalink | Reply

  Note: P = 5 is max value for int_pow() can accomodate without an error, but OK for a solution.

  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 1..9:F; var 1..9:A; var 1..9:R; var 2..5:P;  
  var 0..9:L; var 0..9:O; var 0..9:W; var 0..9:E;
  var 1..9:T; var 0..9:C;
  
  constraint all_different ([F, A, R, P, L, O, W, E, T, C]);
  
  var 100000..999999:FLOWER == 100000*F + 10000*L + 1000*O
  + 100*W + 10*E + R;
  
  var 100000..999999:FACTOR = 100000*F + 10000*A + 1000*C
  + 100*T + 10*O + R;
  
  var 100..999:ART = 100*A + 10*R + T;
  
  var 2..54:SUM = F + L + O + W + E + R;
  
  constraint int_pow(SUM, P, FLOWER);
  
  constraint FACTOR mod A == 0 /\ FACTOR mod R == 0 /\ FACTOR mod T == 0;
  
  solve satisfy;
  output["FLOWER = " ++ show(FLOWER) ++ ", ART = " ++ show(ART)];
  
  % FLOWER = 390625, ART = 751
  % ----------
  % ==========
  
  
  

  LikeLike

• 

  Brian Gladman 7:40 pm on 10 January 2026 Permalink | Reply

  from itertools import permutations
  
  # convert digits to numbers and vice versa
  dgts_to_nbr = lambda dgts: int(''.join(map(str, dgts)))
  nbr_to_dgts = lambda nbr: tuple(int(d) for d in str(nbr))
  
  # the digit sum of FLOWER is between the minimum 
  # and maximum sums of six digits from 0..9
  for sm in range(15, 40):
    # the possible powers to give sm ** P six digits
    for P in (4, 5):
      if 100_000 <= (FLOWER := sm ** P) < 1_000_000:
        t = (F, L, O, W, E, R) = nbr_to_dgts(FLOWER)
        # check that three digits remain unallocated
        if len(r3 := set(range(10)).difference(t + (P,))) == 3:
          # consider the order of the remaining digits for A, C and T
          for A, C, T in permutations(r3):
            FACTOR = dgts_to_nbr((F, A, C, T, O, R))
            # FACTOR must be divisible by A, R and T
            if FACTOR % (A * R * T) == 0:
              ART = dgts_to_nbr((A, R, T))
              print(f"FLOWER ART = {FLOWER} {ART} [{FACTOR = }, {P = }]")
  

  LikeLike