Enigmatic Code

Programming Enigma Puzzles

Enigma 1630: Powering up

From New Scientist #2796, 22nd January 2011 [link]

ENIGMA is a number in which the digits are different and non-zero.

ENIGMA is divisible without remainder by E, by the square NI, and by the cube GMA.

What is ENIGMA?

[enigma1630]

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2 responses to “Enigma 1630: Powering up

  1. jimrandell 13 December 2011 at 11:06 pm

    This Python program runs in 30ms.

    from enigma import irange, split, is_distinct, concat, printf
    
    # three digit cubes
    for i in irange(5, 9):
      GMA = i ** 3
      (G, M, A) = split(GMA, int)
      if not is_distinct(M, A): continue
      if not is_distinct(G, M, A): continue
    
      for j in irange(4, 9):
        NI = j ** 2
        (N, I) = split(NI, int)
        if not is_distinct(N, G, M, A): continue
        if not is_distinct(I, N, G, M, A): continue
    
        for E in irange(1, 9):
          if not is_distinct(E, N, I, G, M, A): continue
          ENIGMA = int(concat(E, N, I, G, M, A))
    
          if ENIGMA % E > 0: continue
          if ENIGMA % NI > 0: continue
          if ENIGMA % GMA > 0: continue
    
          printf("ENIGMA={ENIGMA} [GMA={GMA} ({i}^3) NI={NI} ({j}^2) E={E}]")
    

    Solution: ENIGMA = 349125.

  2. geoffrounce 31 January 2015 at 8:31 am

    Referring to Hugh’s latest comment on Enigma 254, the description and solution of this puzzle does seem to show a better pattern for the letters used. It also finds the solution in only 17 loops and is based on Jim’s original solution.

     
    for E in range(1, 10):
      for j in range(4, 10):
        NI = j**2
        if len(set(str(E) + str(NI))) != 3: continue
        for i in range(5, 10):
          GMA = i**3
          if len(set(str(E)+ str(NI)+ str(GMA)))!= 6: continue
          ENIGMA = E*100000 + NI*1000 + GMA
          if all(ENIGMA % Z == 0 for Z in (E,NI,GMA)):
            print('E={}, NI={}, GMA={}, ENIGMA={}'.format(E,NI,GMA,ENIGMA))
    

    Solution: E=3, NI=49, GMA=125, ENIGMA=349125

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