S2T2

Jim Randell 9:29 am on 27 December 2022 Permalink Reply
Tags: by: Rex Gooch ( 3 )   

• 

  Jim Randell 9:31 am on 27 December 2022 Permalink | Reply

  See also: Teaser 1040, Teaser 3053.

  This Python program uses the [[ SubstitutedExpression ]] solver from the enigma.py library to find answers to the embedded puzzle. It then calculates the GCD of the answers, and looks for 3-digit multiples of this value that do not occur.

  It runs in 62ms.

  Run: [ @replit ]

  from enigma import (SubstitutedExpression, mgcd, irange, nsplit, printf)
  
  # suppose the number is: ABCDEFGHIJ
  p = SubstitutedExpression(
    [
      "IJ % 2 = 0",
      "HIJ % 3 = 0",
      "GHIJ % 4 = 0",
      "FGHIJ % 5 = 0",
      "EFGHIJ % 6 = 0",
      "DEFGHIJ % 7 = 0",
      "CDEFGHIJ % 8 = 0",
      "BCDEFGHIJ % 9 = 0",
      "ABCDEFGHIJ % 10 = 0", # or: "0 = J"
    ],
    d2i=dict(),
    answer="HIJ",
  )
  
  # solve the puzzle and accumulate the answers
  ns = set(p.answers(verbose=0))
  # calculate the gcd of the answers
  g = mgcd(*ns)
  
  # look for 3-digit multiples of g
  for m in irange.round(100, 999, step=g, rnd='I'):
    # does not occur in an answer
    if m in ns: continue
    # has 3 different digits
    if len(set(nsplit(m, 3))) != 3: continue
    # output solution
    printf("multiple = {m} [gcd = {g}]")
  

  Solution: The 3-digit multiple that does not occur is: 960.

  There are 202 answers to the embedded puzzle. And the final 3 digits of each answer form a multiple of 120.

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• 

  GeoffR 9:41 am on 28 December 2022 Permalink | Reply

  This program finds the maximum number of solutions and sets of digits for HIJ values.

  From the output values in the HIJ set, it can be readily seen that the gcd is 120 and the missing multiples of this gcd are 600 and 960. 600 can be ruled out as a solution as it contains repeating digits, so the missing gcd multiple is 960 (8 x 120).

  from functools import reduce
  
  # convert digit sequence to number
  d2n = lambda s: reduce(lambda x,  y: 10 * x + y, s)
  
  J = 0   # 10th last digit
  cnt = 0  # counter for number of solutions
  dig3 = set()  # set for all different values of HIJ
  
  for I in range(1, 10):
    IJ = d2n([I, J])
    if IJ % 2 != 0:continue
    
    for H in range(1, 10):
      if H in (I, J): continue
      HIJ = d2n([H, I, J])
      if HIJ % 3 != 0: continue
      
      for G in range(1, 10):
        GHIJ = d2n([G, H, I, J])
        if G in (H, I, J): continue
        if GHIJ % 4 != 0:continue
         
        for F in range(1, 10):
          if F in (G, H, I, J): continue
          FGHIJ = d2n([F, G, H, I, J])
          if FGHIJ % 5 != 0: continue
          
          for E in range(1, 10):
            if E in (F, G, H, I, J):continue
            EFGHIJ = d2n([E, F, G, H, I, J])
            if EFGHIJ % 6 != 0:continue
            
            for D in range(1, 10):
              if D in (E, F, G, H, I, J):continue
              DEFGHIJ = d2n([D, E, F, G, H, I, J])
              if DEFGHIJ % 7 != 0:continue
              
              for C in range(1, 10):
                if C in (D, E, F, G, H, I, J):continue
                CDEFGHIJ = d2n([C, D, E, F, G, H, I, J])
                if CDEFGHIJ % 8 != 0: continue
                
                for B in range(1, 10):
                  if B in (C, D, E, F, G, H, I, J):continue
                  BCDEFGHIJ = d2n([B, C, D, E, F, G, H, I, J])
                  if BCDEFGHIJ % 9 != 0:continue
                  
                  for A in range(1, 10):
                    if A in (B, C, D, E, F, G, H, I, J):continue
                    ABCDEFGHIJ = d2n([A, B, C, D, E, F, G, H, I, J])
                    if ABCDEFGHIJ % 10 != 0:continue
                    # update HIJ digit set and number of solutions
                    dig3.add(HIJ)
                    cnt += 1
  
  print('Number of solutions = ', cnt)  
  print('HIJ set = ', dig3)  
  
  # Number of solutions =  202
  # HIJ set =  {480, 840, 360, 240, 720, 120}
  
  
  
  
  

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