S2T2

Jim Randell 4:34 pm on 5 November 2021 Permalink Reply
Tags: by: Victor Bryant ( 139 )   

• 

  Jim Randell 4:44 pm on 5 November 2021 Permalink | Reply

  The following Python program runs in 51ms.

  Run: [ @replit ]

  from enigma import irange, inf, nsplit, flatten, printf
  
  # consider lucky numbers, n
  for n in irange(2, inf):
    # and the initial multiple
    for k in irange(2, inf):
      # the three numbers
      ns = list(k * n ** x for x in (1, 2, 3))
      # collect the digits in the numbers
      ds = flatten(nsplit(n) for n in ns)
      if len(ds) > 10: break
      if len(ds) == 10 and len(set(ds)) == 10:
        printf("ns = {ns} [n = {n}; k = {k}]")
    if k == 2: break
  

  Solution: The three numbers are: 306, 918, 2754.

  The lucky number is 3.

  So we have:

  102 × 3 = 306
  306 × 3 = 918
  918 × 3 = 2754

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• 

  GeoffR 5:34 pm on 5 November 2021 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 2..10: LN; % my lucky number
  
  var 1..9: A; var 0..9: B; var 0..9: C;
  var 1..9: D; var 0..9: E; var 0..9: F;
  var 1..9: G; var 0..9: H; var 0..9: I; var 0..9: J;
  
  % the three numbers written down have different digits
  constraint all_different ([A, B, C, D, E, F, G, H, I, J]);
  
  % My three numbers
  var 100..999: ABC = 100*A + 10*B + C;
  var 100..999: DEF = 100*D + 10*E + F;
  var 1000..9999: GHIJ = 1000*G + 100*H + 10*I + J;
  
  % First number is a multiple of my lucky number
  constraint ABC mod LN == 0;
  
  % Second and third lucky numbers
  constraint ABC * LN == DEF /\ DEF * LN == GHIJ;
  
  solve satisfy;
  
  output ["My three numbers are " ++show(ABC) ++ ", " ++ show(DEF) ++ 
  " and " ++ show(GHIJ)
  ++ "\n" ++ "Lucky number is " ++ show(LN) ];
  
  
  

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• 

  NigelR 5:51 pm on 5 November 2021 Permalink | Reply

  First number must be less than 1000 or the three numbers would have more than 10 digits between them. Code below runs in 39ms:

  for lucky in range(1,100):
      for first in range(1,1000):
          if first%lucky!=0:
              continue
          second = first*lucky
          third = second*lucky
          res=str(first)+str(second)+str(third)
          if len(set(res)) == 10 and len(res) == 10:
              print (‘first: ‘, first,’  second: ‘, second,’  third:’, third,’  lucky:’, lucky)
              exit()
  

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• 

  GeoffR 8:52 am on 6 November 2021 Permalink | Reply

  # Assume a digit split of 3:3:4 for 10 digits for three numbers.
  # The multiplier is less than 10 between two increasing 3-digit numbers
  # The three numbers are ABC, DEF and GHIJ and the lucky number is LN
  
  from enigma import nsplit, all_different
  
  from enigma import Timer
  timer = Timer('ST3085')
  timer.start()
  
  for LN in range(2, 10):
    for ABC in range(102, 987):
      if ABC % LN != 0:
        continue
      A, B, C = nsplit(ABC)
      if not all_different(A, B, C):
        continue
      DEF = ABC * LN
      if DEF < 100 or DEF > 999:
        continue
      D, E, F = nsplit(DEF)
      if not all_different(A, B, C, D, E, F):
        continue
      GHIJ = DEF * LN
      if GHIJ < 1000 or GHIJ > 9999:
        continue
      G, H, I, J = nsplit(GHIJ)
      if not all_different(A, B, C, D, E, F, G, H, I, J):
        continue
      print(f"My three numbers are {ABC}, {DEF} and {GHIJ}")
      timer.stop()
      timer.report()  # 4.98ms (I9 processor)
      
  
  
  
  

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  • 

    Frits 10:33 am on 9 November 2021 Permalink | Reply

    @GeoffR, a digit split of 2:3:5 is also possible.

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  GeoffR 11:09 am on 9 November 2021 Permalink | Reply

  @Frits:
  I decided to try a 3:3:4 digit split first as this seemed a more likely candidate for a solution. As this digit split gave a single answer, I did not try a 2:3:5 digit split.

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