S2T2

Jim Randell 8:48 am on 9 December 2025 Permalink Reply
Tags: by: Andrew Skidmore ( 85 )   

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  Jim Randell 8:49 am on 9 December 2025 Permalink | Reply

  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.

  The following run file executes in 98ms. (Internal runtime of the generated code is 14ms).

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  # suppose the numbers are: AB CD EF (in order)
  --distinct=""
  --invalid="5|6|7|8|9,ACE"
  "A <= C" "C <= E"
  "0 < AB < CD < EF < 50"
  
  # the two 6-digit numbers are: ABCDEF, EFCDAB
  # and 3 times the difference divided by the sum of the original numbers
  # is the same as the product of two of the original numbers
  "div(abs(ABCDEF - EFCDAB) * 3, AB + CD + EF) in { AB * CD, AB * EF, CD * EF }"
  
  # [optional] tidy up output
  --template="(AB CD EF)"
  --solution=""
  --verbose="1-H"
  

  Solution: The three winning numbers were: 32, 33, 36.

  These are assembled into the two 6-digit numbers:

  ascending: 323336
  descending: 363332

  And the dividing 3 times the difference of these by the sum of the original numbers gives:

  3 × (363332 − 323336) / (32 + 33 + 36) = 1188

  And 1188 is the product of two of the original numbers:

  1188 = 33 × 36

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  Ruud 10:33 am on 9 December 2025 Permalink | Reply

  import istr
  
  for numbers in istr.combinations(range(1, 50), 3):
      n_descending = istr("").join(sorted(numbers, reverse=True))
      n_ascending = istr("").join(numbers)
      n_difference = n_descending - n_ascending
      if (n_difference * 3).is_divisible_by(sum(numbers)) and any(n0 * n1 == n_difference * 3 / sum(numbers) for n0, n1 in istr.combinations(numbers, 2)):
          print(numbers)
  

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  Frits 7:58 pm on 10 December 2025 Permalink | Reply

  The sum of the winning numbers must be 101.

  See Teaser 2509

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  • 

    Jim Randell 10:24 am on 11 December 2025 Permalink | Reply

    This analysis allows a nice compact program:

    # for numbers (x, y, z) we have:
    #
    #   3.9999.(z - x) = (x + y + z).{ x.y | x.z | y.z }
    #
    # 101 is a prime factor of 9999 that can only appear in (x + y + z)
    # and since (x + y + z) in [6, 144] it follows:
    #
    #   (x + y + z) = 101
    #
    # leaving:
    #
    #   3.99.(z - x) = { x.y | x.z | y.z }
    
    from enigma import (decompose, printf)
    
    for (x, y, z) in decompose(101, 3, increasing=1, sep=1, min_v=1, max_v=49):
      if 297 * (z - x) in { x * y, x * z, y * z }:
        printf("({x}, {y}, {z})")
    

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