S2T2

Jim Randell 9:07 am on 23 March 2021 Permalink Reply
Tags: by: Susan Bricket ( 17 )   

• 

  Jim Randell 9:08 am on 23 March 2021 Permalink | Reply

  We can use the [[ SubstitutedExpression ]] solver from the enigma.py library to solve this puzzle directly.

  The following run file executes in 122ms.

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  "TRIPLE % 3 = 0"
  "is_cube(EIGHT)"
  "NINE % 9 = 0"
  "is_prime(PRIME)"
  
  --answer="TIME"
  

  Solution: TIME = 3901.

  There are 4 solutions to the alphametic expressions, but the value for TIME is the same in all of them.

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• 

  Frits 12:39 pm on 23 March 2021 Permalink | Reply

  With a complicated formula for N.

  from enigma import is_prime
  
  alldiff = lambda seq: len(seq) == len(set(seq))
  
  # EIGHT is a cube
  for i in range(22, 47):
    s3 = str(i ** 3)
    E = int(s3[0])
    I = int(s3[1])
    T = int(s3[-1])
    if T == 0: continue             
    if E % 2 == 0: continue  # E may not be even (due to PRIME)
    if not alldiff(s3): continue
    
    # NINE is a multiple of 9
    # sumIE plus 2*N must be equal to k*9   (k is 1,2 or 3)
    sumIE = I + E
    N = (9 + 18 * (sumIE > 9) - sumIE) // 2 if sumIE % 2 else (18 - sumIE) // 2  
    if N == 0 or not alldiff(s3 + str(N)): continue   
    
    # TRIPLE is a multiple of 3
    LMPR = [x for x in range(10) if str(x) not in s3 + str(N)]
    candsM = [x for x in LMPR if (sumIE + T + sum(LMPR) - x) % 3 == 0]
     
    for M in candsM:
      LPR = [x for x in LMPR if x != M]
      for L in LPR:
        # PRIME may not be a multiple of 3
        if (sumIE + M + sum(LPR) - L) % 3 == 0: continue
        PR = [x for x in LPR if x != L]
        # consider permutations of PR
        for (P, R) in zip(PR, PR[::-1]):
          if P == 0: continue
          PRIME = 10000*P + 1000*R + 100*I + 10*M + E
          if not is_prime(PRIME): continue
          print(f"TIME={T}{I}{M}{E}")
  

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