S2T2

Jim Randell 9:21 am on 1 May 2026 Permalink Reply
Tags: by: Victor Bryant ( 147 )   

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  Jim Randell 9:22 am on 1 May 2026 Permalink | Reply

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

  It runs in 106ms. (Internal runtime of the generated code is 21ms).

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  --distinct="ADENPRSTUY"
  --invalid="0,EPST"
  
  # exactly five of the six numbers are divisible by 9
  # so just 1 of them isn't, identify it as X = 1..6
  --invalid="0|7-9,X"
  
  "(X == 1) == (SETTER % 9 != 0)"
  "(X == 2) == (SENT % 9 != 0)"
  "(X == 3) == (PRETTY % 9 != 0)"
  "(X == 4) == (EASY % 9 != 0)"
  "(X == 5) == (SUNDAY % 9 != 0)"
  "(X == 6) == (TEASER % 9 != 0)"
  
  --answer="X"
  
  --template="(SETTER) (SENT) (PRETTY) (EASY) (SUNDAY) (TEASER) (X)"
  

  Solution: TEASER is not divisible by 9.

  There are 36 ways to assign the digits to the letters, but in all cases TEASER is the odd number out.

  ---

  If the puzzle had been set so that just one of the words is not divisibly by 29, then there is a single way to assign digits to the letters.

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  • 

    Ruud 11:07 am on 1 May 2026 Permalink | Reply

    Brute force reveals that there are 36 different solutions, all resulting in TEASER to be the one that’s not divisible by 9.

    import peek
    import istr
    
    count = 0
    for s, e, t, r, n, p, y, a, u, d in istr.permutations(range(10)):
        if s * p * e * t: # may not start with 0
            not_divisible_by_9 = ""
            for name in "setter sent pretty easy sunday teaser".split():
                if not istr(f":={name}").is_divisible_by(9):
                    if not_divisible_by_9:
                        break
                    not_divisible_by_9 = name
            else:
                if not_divisible_by_9:
                    count += 1
                    peek(count, not_divisible_by_9, setter, sent, pretty, easy, sunday, teaser)
    

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