Enigmatic Code

Programming Enigma Puzzles

Enigma 172: Enigmas galore

From New Scientist #1317, 5th August 1982 [link]

It’s easy to make up Enigmas: here are six letters-for-digits multiplication puzzles in one go! In each case, regarded as a separate problem, each letter stands consistently for a digit and different letters stand for different digits. Unfortunately I shall have to check if it’s possible to find numbers that fit and then discard the impossible puzzles (and if some numbers do fit I shall have to add additional clues to make the answer unique). How many of the following six Enigmas must I discard?

E.NIGMAS = ENIGMA
EN.IGMAS = ENIGMA
ENI.GMAS = ENIGMA
ENIG.MAS = ENIGMA
ENIGM.AS = ENIGMA
ENIGMA.S = ENIGMA

[enigma172]

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2 responses to “Enigma 172: Enigmas galore

  1. Jim Randell 27 February 2014 at 8:28 am

    We can arrive at the solution fairly easily by analysis. But this Python program checks all possible values for the letters. It runs in 3.8s (under PyPy).

    from itertools import permutations
    from enigma import irange, nconcat, printf
    
    # record how many solutions there are for each puzzle
    r = dict((i, 0) for i in irange(1, 6))
    
    # choose 7 digits for ENIGMAS
    for s in permutations(irange(0, 9), 7):
      ENIGMA = nconcat(s[:-1])
      # consider each puzzle
      for i in irange(1, 6):
        (a, b) = (nconcat(s[:i]), nconcat(s[i:]))
        if a * b == ENIGMA:
          r[i] += 1
          printf("[{i}] {a} x {b} = {ENIGMA}")
    
    for i in irange(1, 6):
      printf("[{i}] {n} solutions", n=r[i])
    

    Solution: Five of the equations must be discarded. Only the last one has solutions.

    • Jim Randell 27 February 2014 at 8:32 am

      Analytically we can see that:

      ENIGMA × S = ENIGMA

      simplifies to:

      S = 1

      and any assignment of the remaining digits to E, N, I, G, M, A will give a valid equation. So there are 9!/(9-6)! = 60480 possibilities (before additional constraints, such as disallowing leading zeros, are added).

      Similarly the remaining equations can be rewritten as:

      AS = 10 + A / ENIGM
      MAS = 100 + MA / ENIG
      GMAS = 1000 + GMA / ENI
      IGMAS = 10000 + IGMA / EN
      NIGMAS = 100000 + NIGMA / E

      The first and second equations are obviously not going to have solutions.

      For the third equation GMA/ENI would be a 1 or 2 digit number, making GMAS = 10xy, but then GMA = 10x and there is no possible assignment of ENI that would make this work.

      A similar argument eliminates the fourth and fifth equations from having solutions.

      Hence only the last equation given in the puzzle has solutions, and they are all trivial.

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