Enigmatic Code

Programming Enigma Puzzles

Enigma 1609: I remember it well

From New Scientist #2774, 21st August 2010 [link]

I am a bit vague about my bank security number –

It is a four figure number (or possibly five);

It uses four different digits (or possibly five);

The first digit is four (or possibly five);

The second digit is four (or possibly five) and none of the other digits is smaller;

The number is a prime multiplied by four (or possibly five);

If you reverse the order of the digits you get a number which is the product of four (or possibly five) different primes;

In four (or possibly five) of these statements my first guess is correct.

What is my bank security number?

[enigma1609]

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One response to “Enigma 1609: I remember it well

  1. jimrandell 2 January 2012 at 11:46 pm

    The following Python program runs in 128ms.

    from enigma import irange, is_prime, factor
    
    for n in irange(4444, 55999):
    
      # A. is a 4 (or 5) digit number
      s = str(n)
      g1 = len(s)
    
      # B. uses 4 (or 5) different digits
      g2 = len(set(s))
      if g2 not in (4, 5): continue
    
      # C. starts with 4 (or 5)
      g3 = s[0]
      if g3 not in ('4', '5'): continue
    
      # D. second digit is 4 (or 5)...
      g4 = s[1]
      if g4 not in ('4', '5'): continue
      # ... and none of the other digits are smaller
      if list(filter(lambda x: x < s[1], [s[0]] + list(s[2:]))): continue
    
      # E. prime * 4 (or 5)
      g5 = ''
      if n % 4 == 0 and is_prime(n // 4): g5 += '4'
      if n % 5 == 0 and is_prime(n // 5): g5 += '5'
      if not g5: continue
    
      # F. if you reverse the number, it is the product of 4 (or 5) different primes
      r = int(s[::-1])
      f = factor(r)
      g6 = len(set(f))
      if g6 != len(f): continue
      if g6 not in (4, 5): continue
    
      # G. 4 (or 5) of these statements yield 4.
      # so, if there are 3 statements yielding 4 then the last one can yield 4 making 4 of them altogether.
      # if there are 4 statements yielding 4 the last statement cannot be satisfied (G4 => G5)
      # if there are 5 statements yielding 4 the last can yield 5 making 5 of them altogether
      l = list(map(int, (g1, g2, g3, g4, g5, g6)))
      g = l.count(4)
      if g not in (3, 5): continue
      print(n, l + [g], g)
    

    Solution: The security number is 5468.

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