S2T2

Jim Randell 6:41 am on 6 July 2025 Permalink Reply
Tags: by: Howard Williams ( 44 )   

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  Jim Randell 6:48 am on 6 July 2025 Permalink | Reply

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

  It runs in 78ms. (Internal runtime of the generated program is 279µs).

  Run: [ @codepad ]

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  # sqrt(ABCDEF) = XYZ
  --distinct="ABCDEF"
  --invalid="0,ABCDEFXZ"
  --answer="XYZ"
  
  # XYZ is the square root of ABCDEF
  "sq(XYZ) = ABCDEF"
  
  # the sum of the digits of the square is 5 times the
  # sum of the first and last digit
  "A + B + C + D + E + F == 5 * (A + F)"
  
  # the final three digits are in descending order
  "D > E > F"
  
  # [optional] performance tweaks
  --reorder="[1]"
  --invalid="1|2,X"
  

  Solution: The square root is 661.

  And so the 6-digit number is: 661^2 = 436921.

  And we have:

  4 + 3 + 6 + 9 + 2 + 1 = 25 = 5 × (4 + 1)
  9 > 2 > 1

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• 

  GeoffR 10:17 am on 6 July 2025 Permalink | Reply

  
  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 1..9:a; var 1..9:b; var 1..9:c; var 1..9:d;
  var 1..9:e; var 1..9:f; 
  var 316..999:ans; % square root of abcdef
  
  var 100000..999999: abcdef == 100000*a + 10000*b 
  + 1000*c + 100*d + 10*e + f;
  
  constraint all_different([a, b, c, d, e, f]);
  
  constraint a + b + c + d + e + f == 5 * (a + f);
  constraint d > e /\ e  > f;
  
  predicate is_sq(var int: c) =
     let {
        var 0..ceil(pow(int2float(ub(c)),(1/2))): z
     } in z*z = c;
  
  constraint ans * ans == abcdef;
  constraint is_sq(abcdef);
  
  solve satisfy;
  
  output["Square root of the six digit number = " ++  show(ans)];
  
  
  

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• 

  Frits 3:03 pm on 6 July 2025 Permalink | Reply

  # last 2 digits y and z  (abcdef = xyz^2)
  last2 = set(i for i in range(11, 100) 
              if i % 10 and 9 < (m := (i * i) % 100) < 90 and m // 10 > m % 10)
  
  for yz in last2:
    # A + F <= 7 so X < 9
    for x in range(3, 9):
      xyz = 100 * x + yz
      abcdef = xyz * xyz
      if len(set(s := str(abcdef))) != 6: continue
      if s[3] <= s[4] or "0" in s: continue
      dgts6 = [int(x) for x in s]
      if sum(dgts6) != 5 * (dgts6[0] + dgts6[-1]): continue
  
      print("answer:", xyz)
  

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  • 

    Frits 5:57 pm on 6 July 2025 Permalink | Reply

    Processing six-digit numbers to find a valid square root.

    from enigma import is_square
    
    # last 2 digits e and f  (abcdef = xyz^2)
    last2 = set(divmod(m, 10) for i in range(11, 100) 
                if i % 10 and 9 < (m := (i * i) % 100) < 90 and m // 10 > m % 10)
    for (e, f) in last2:
      ef = 10 * e + f
      for d in range(e + 1, 10):
        _def = 100 * d + ef
        # 21 <= 5 * (a + f) <= 39  or  4 < a + f < 8
        for a in set(range(max(1, 5 - f), 8 - f))- {d, e, f}:
          # remaining for b + c
          if not (3 <= (bc := 5 * (a + f) - (a + d + e + f)) <= 17): continue
          adef = 10**5 * a + _def
          # determine missing numbers x and y (x + y = bc with x < y)
          for x in set(range(max(1, bc - 9), (bc + 1) // 2)) - {a, d, e, f}:
            if (y := bc - x) in {a, x, d, e, f}: continue
            for b, c in ((x, y), (y, x)):
              n = adef + 10**4 * b + 10**3 * c
              if (xyz := is_square(n)):
                print("answer:", xyz)
    

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• 

  Ruud 4:22 pm on 6 July 2025 Permalink | Reply

  import math
  import istr
  
  for i in range(int(math.sqrt(100_000)), int(math.sqrt(1_000_000))):
      if (sum((sq := istr(i * i))) == 5 * (sq[0] + sq[-1])) and sq.all_distinct() and "0" not in sq and sq[-3] > sq[-2] > sq[-1]:
          print(sq, i)
  

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  GeoffR 9:17 am on 9 July 2025 Permalink | Reply

  sq6 = [ n**2 for n in range(317, 1000)    
          if len(set(str(n**2))) == 6 ]
  
  for num in sq6:
    ds = str(num)
    if '0' in ds: continue
    dig = [int(x) for  x in ds]
    if (dig [3] > dig[4] > dig[5]):
      if sum(dig[:6]) == 5 * (dig[0] + dig[5]):
        print (f"Square root is {int(num ** 0.5)}.")
  
  

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