S2T2

Jim Randell 9:48 am on 23 January 2026 Permalink Reply
Tags: by: Victor Bryant ( 140 )   

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  Jim Randell 9:49 am on 23 January 2026 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 14.4ms).

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  # no leading zeros (ONE, TWO, SIX)
  --invalid="0,OTS"
  
  # collect primes < 1000
  --code="primes.expand(999)"
  
  # ONE is not prime, but it is an odd perfect square
  "ONE not in primes"
  "E % 2 = 1"
  "is_square(ONE)"
  
  # TWO is divisible by 2, but not 4
  "O % 2 = 0"
  "WO % 4 != 0"
  
  # the number of odd primes less than SIX is TWO
  "icount(primes.between(3, SIX - 1)) = TWO"
  
  --template="(ONE) (TWO) (SIX)"
  --answer="NOW"
  

  Solution: NOW = 820.

  There are two possible assignments of digits to letters:

  ONE = 289, TWO = 102, SIX = 564
  ONE = 289, TWO = 102, SIX = 567

  The next prime after 563 is 569, so we count 102 odd primes if X is 4 or 7.

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  ruudvanderham 11:24 am on 23 January 2026 Permalink | Reply

  import peek
  import istr
  
  for o, n, e, t, w, s, i, x in istr.permutations(range(10), 8):
      if o * t * s != 0:
          if not (one := istr("=one")).is_prime() and one.is_odd() and one.is_square():
              if (two := istr("=two")).is_divisible_by(2) and not two.is_divisible_by(4):
                  if len(istr.primes(3, (six := istr("=six")))) == two:
                      peek(one, two, six)
  

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