S2T2

Jim Randell 8:13 am on 20 February 2026 Permalink Reply
Tags: by: ??? ( 4 )   

• 

  Jim Randell 8:14 am on 20 February 2026 Permalink | Reply

  This Python program runs in 69ms. (Internal runtime is 1.6ms)

  from enigma import (irange, decompose, tuples, csum, multiply, is_square_p, printf)
  
  # the sequence of digits
  digits = list(irange(1, 9))
  n = len(digits)
  
  # divide the digits into k groups (min groups = 1, max groups = n)
  for k in irange(1, n):
    for js in decompose(len(digits), k, increasing=0, sep=0, min_v=1):
      # split the digits into the groups
      gs = list(digits[i:j] for (i, j) in tuples(csum(js, empty=1), 2))
      # calculate the result of the process
      r = multiply(sum(ns) for ns in gs)
      # both B and J found a square number
      if is_square_p(r):
        # B's was odd
        if r % 2 == 1:
          printf("Billy: {gs} -> {r}")
        # J's groups each have more than one digit
        if all(len(ns) > 1 for ns in gs):
          printf("Jilly: {gs} -> {r}")
  

  Solution: Billy = 12-3-456-78-9; Jilly = 12-3456-789.

  So:

  Billy = (1+2)×(3)×(4+5+6)×(7+8)×(9) = 18225 = 135^2
  Jilly = (1+2)×(3+4+5+6)×(7+8+9) = 1296 = 36^2

  Here is a complete list of square numbers that can be formed this way:

  12345678-9 → 324 = 18^2
  12-3456-789 → 1296 = 36^2 (only one with multiple digits per group)
  123-4-5-6789 → 3600 = 60^2
  1-2-34567-8-9 → 3600 = 60^2
  1-2-3-4-5-6789 → 3600 = 60^2
  12-345-6-789 → 5184 = 72^2
  12-3-45-6-789 → 11664 = 108^2
  1-234-567-8-9 → 11664 = 108^2
  1-23-4-5-6-789 → 14400 = 120^2
  12-3-456-78-9 → 18225 = 135^2 (only odd square)
  12-3-4-567-8-9 → 46656 = 216^2

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• 

  ruudvanderham 12:07 pm on 20 February 2026 Permalink | Reply

  I make expressions by putting either )*( or + between the digits 1 – 8, like
  (1)*(2+3+4+5)*(6+7)*(8)*(9)
  and then evaluate that with eval.
  For Jilly, I then also check whether the lengths of all parts split by ‘)*(‘ is greater than 2.

  import itertools
  import math
  
  for ops in itertools.product(("+", ")*("), repeat=8):
      exp = "(1" + "".join(f"{op}{i}" for i, op in enumerate(ops, 2)) + ")"
      if math.isqrt(val := eval(exp)) ** 2 == val:
          if val % 2:
              print("Billy", exp, "=", val)
          else:
              if all(len(part) > 2 for part in exp.split(")*(")):
                  print("Jilly", exp, "=", val)
  

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  • 

    Ruud 3:41 pm on 20 February 2026 Permalink | Reply

    With a more elegant check for no groups of one digit:

    import itertools
    import math
    
    for ops in itertools.product(("+", ")*("), repeat=8):
        exp = "(1" + "".join(f"{op}{i}" for i, op in enumerate(ops, 2)) + ")"
        if math.isqrt(val := eval(exp)) ** 2 == val:
            if val % 2:
                print("Billy", exp, "=", val)
            else:
                if all("+" in part for part in exp.split("*")):
                    print("Jilly", exp, "=", val)
    

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