S2T2

Jim Randell 5:01 pm on 26 March 2021 Permalink Reply
Tags: by: Danny Roth ( 84 )   

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  Jim Randell 5:04 pm on 26 March 2021 Permalink | Reply

  See also: Enigma 339b, Enigma 1643, Enigma 1667, and the recently posed Puzzle #102.

  Here’s a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library. It runs in 96ms.

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  --digits="1-9"
  
  "ABCDEFGH % 8 = 0"
  "ABCDEFG % 7 = 0"
  "ABCDEF % 6 = 0"
  "ABCDE % 5 = 0"
  "ABCD % 4 = 0"
  "ABC % 3 = 0"
  "AB % 2 = 0"
  
  --answer="ABCDEFGHI"
  

  Solution: The number is 381654729.

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  Frits 10:40 pm on 26 March 2021 Permalink | Reply

  Tested in a Python online editor.

  from itertools import permutations
  
  print(*["".join(x) for x in permutations("123456789")
        if all(int("".join(x[:i])) % i == 0 for i in range(2, 9))])
  

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    Tony Brooke-Taylor 9:02 pm on 28 March 2021 Permalink | Reply

    I used a few shortcuts with the aim of minimising wasted iterations. Running my code on Replit takes about 3.5ms, compared with 1.13s for Frits’ sledgehammer approach. On the other hand, mine takes a few more lines!

    Hope I manage to paste this in an appropriate format:

    
    evens=list(range(2,9,2))
    Xs=list(range(2,9,4))
    odds=list(range(1,10,2))
    odds.remove(5)
    
    #Exploit the fact that the three-digit and six-digit numbers must be divisible by 3, and that the 6-digit number can be expressed as the sum of ABC*1000 and XYZ, so the digits of both ABC and XYZ must sum separately to 3. Let the last three digits be RST. 
    
    #For a four-digit number to be divisible by four, the last two digits must make a number divisible by four. Therefore if the third digit is odd, the fourth must be an odd multiple of 2
    for X in Xs:
      vec=[None for i in range(9)]
      vec[4]=5#no zeros allowed, and the five-digit number divisible by 5
      vec[3]=X
      Z=10-X#consequence of sum(X,Y,Z) divisible by three
      vec[5]=Z
      Bs=[i for i in evens if i not in vec]#remaining evens go to 2nd and 8th
      for B in Bs:
        vec[1]=B
        S=Bs[(Bs.index(B)-1)%2]
        vec[7]=S  
    
    #Try each pair of odd numbers (other than 5) in positions 1 and 3, subject to requiring that ABC divisible by 3 (requiring A+B+C divisible by 3)  
        for A in odds:
          Cs=odds.copy()
          Cs.remove(A)
          for C in Cs:
            if sum([A,B,C])%3==0:
              vec[0]=A
              vec[2]=C
    
    #Complementary odds then fall into positions 7 and 9          
              Rs=Cs.copy()
              Rs.remove(C)
              for R in Rs:
                vec[6]=R
                if sum([i*10**(6-vec.index(i)) for i in vec[:7]])%7==0 and sum([i*10**(7-vec.index(i)) for i in vec[:8]])%8==0:
                  Ts=Rs.copy()
                  Ts.remove(R)
                  for T in Ts:
                    vec[8]=T
                    print("The telephone number is",sum([i*10**(8-vec.index(i)) for i in vec[:9]]))
    
    
    
    
    

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    • 

      Frits 10:57 am on 29 March 2021 Permalink | Reply

      Hi Tony,

      A couple of recommendations:

      index() is expensive so line 35 may be written to :

       
      if sum([x * 10**(2 - i) for i, x in enumerate(vec[5:8])]) % 8 == 0 and \
         sum([x * 10**(6 - i) for i, x in enumerate(vec[:7])]) % 7 == 0:
      

      You can also use something like “dgts2nbr” as in puzzle [[ https://enigmaticcode.wordpress.com/2017/04/10/enigma-1120-assorted-numbers/ ]]

      – line 16-19 can be rewritten to:

       
      for j, B in enumerate(Bs):
        vec[1] = B
        vec[7] = Bs[1 - j]
      

      or

       
      for j in range(2):
        vec[1] = Bs[j]
        vec[7] = Bs[1 - j]
      

      or

       
      for (B, S) in permutations(Bs):
        vec[1] = B
        vec[7] = S 
      

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        Tony 6:01 pm on 29 March 2021 Permalink | Reply

        Thanks Frits: now I know about enumerate and reduce. As I’m getting the hang of looping over iterables I had forgotten it is still possible to loop with a counter, so I would probably choose your second approach to lines 16-19. I was trying to avoid using itertools once I realised I only needed to deal with pairs.

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  Hugh Casement 12:51 pm on 27 March 2021 Permalink | Reply

  Isn’t this effectively the same as Brainteaser 1040 (4 July 1982)?

  That one at least spared us George and Martha, who get dragged in again and again for no good reason.

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    Jim Randell 3:36 pm on 27 March 2021 Permalink | Reply

    @Hugh: It does seem to be. I suppose it is hard to come up with over 3000 puzzles without some overlap. Or to check that there is no overlap!

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      Hugh Casement 1:14 pm on 30 March 2021 Permalink | Reply

      And the recent Puzzle 102 in New Scientist. I call that plagiarism!

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  GeoffR 1:39 pm on 20 May 2021 Permalink | Reply

  from itertools import permutations
  from enigma import join
  
  # the number 45 is divisible by nine
  i = 9
  
  for p in permutations('123456789', 8):
    a, b, c, d, e, f, g, h = p
  
    # A geometric pattern to this code  
  
    if int(a + b + c + d + e + f + g + h) % 8 == 0:
      if int(a + b + c + d + e + f + g) % 7 == 0:
        if int(a + b + c + d + e + f) % 6 == 0:
          if int(a + b + c + d + e) % 5 == 0:
            if int(a + b + c + d) % 4 == 0:
              if int(a + b + c) % 3 == 0:
                if int(a + b) % 2 == 0:
                  tel_num = join((a, b, \
                  c, d, e, f, g, h, i))
                  print('No. is',tel_num)
                  # No. is 381654729
  
  

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  Dirk 9:43 am on 9 July 2021 Permalink | Reply

  Teaser 1882 (11/10/1998): Multiple Solutions by Rex Gooch is also linked with brainteaser 1040/3053.

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