S2T2

Jim Randell 9:54 am on 6 July 2021 Permalink Reply
Tags: by: J L Bowles ( 13 )   

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  Jim Randell 9:54 am on 6 July 2021 Permalink | Reply

  The three numbers used (when considered as words) must use 10 different letters between them, and the result is 5 letters long. So each of the numbers has at most 5 letters. But if one were to have 5 letters, that would only leave 5 letters for the remaining 2 numbers, and there are no numbers with 2 (or fewer) letters. So the numbers must have (4, 3, 3) letters.

  This Python program collects numbers with fewer than 5 letters, and finds 3 of them that between them use 10 different letters (including the letters of NONSO). We then use the [[ SubstitutedExpression ]] solver from the enigma.py library to solve the alphametic sum.

  It runs in 64ms.

  Run: [ @replit ]

  from enigma import irange, int2words, subsets, union, SubstitutedExpression, sprintf, printf
  
  # collect numbers (as words) with fewer than 5 letters
  words = list(x.upper() for x in map(int2words, irange(0, 10)) if len(x) < 5 and x.isalpha())
  
  # choose three words
  for ws in subsets(words, size=3):
    # check there are 10 different symbols (including NOS)
    ss = union(ws)
    if not(len(ss) == 10 and ss.issuperset('NOS')): continue
  
    # solve the alphametic puzzle
    exprs = list(sprintf("is_square({w})") for w in ws)
    t = sprintf("{ws[0]} + {ws[1]} + {ws[2]} = NONSO")
    exprs.append(t)
    p = SubstitutedExpression(exprs, verbose=0)
    for s in p.solve():
      printf("{t} | {s}", s=p.substitute(s, t))
  

  Solution: The three squares are: 9025, 784, 361.

  These correspond to:

  FOUR = 9025 (= 95²)
  SIX = 784 (= 28²)
  TEN = 361 (= 19²)

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  GeoffR 7:28 pm on 6 July 2021 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  % sets of three and four digit squares
  set of int: sq = {n*n | n in 10..99};
  
  % he used each of the digits 0 to 9 inclusive once only
  var 1..9:A; var 0..9:B; var 0..9:C; 
  var 1..9:D; var 0..9:E; var 0..9:F; 
  var 1..9:G; var 0..9:H; var 0..9:I; var 0..9:J; 
  
  constraint all_different([A, B, C, D, E, F, G, H, I, J]);
  
  % 2 no. 3-digit and 1 no. 4-digit squares - ex Jim's analysis
  var 100..999:ABC = 100*A + 10*B + C;
  var 100..999:DEF = 100*D + 10*E + F;
  var 1000..9999:GHIJ = 1000*G + 100*H + 10*I + J;
  
  var 10000..99999:NONSO; % the sum of the three squares
  
  constraint ABC in sq /\ DEF in sq /\ GHIJ in sq;
  
  constraint NONSO ==  ABC + DEF + GHIJ;
  
  % check digit pattern of NONSO for 'N' and 'O'
  constraint NONSO div 10000 == NONSO div 100 mod 10
  /\ NONSO div 1000 mod 10 == NONSO mod 10;
  
  solve satisfy;
  
  output ["Three squares are " ++ show(ABC) ++ ", " ++ show(DEF) 
  ++ ",  " ++ show(GHIJ) ++ "\n" ++ 
  "NONSO, the 5-digit sum = " ++ show(NONSO)];
  
  % Three squares are 361, 784,  9025
  % NONSO, the 5-digit sum = 10170
  % ----------
  % Three squares are 784, 361,  9025
  % NONSO, the 5-digit sum = 10170
  % ----------
  % ==========
  
  
  
  

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  • 

    Frits 6:08 pm on 7 July 2021 Permalink | Reply

    @GeoffR,

    It is not directly stated but I think N, O and S are meant to be different digits. It doesn’t matter for the solution.

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  John Crabtree 10:31 am on 9 July 2021 Permalink | Reply

  FIVE and NINE only allow one 3-digit number (TWO), and so FOUR is the four-digit number and must contain 0, and, as F = 8 or 9, so must be 9025, 9604 or 9801.
  The sum of the digits of NONSO = 0 mod 9. N = 1. If O = 8, S = 0 which is invalid. If O = 6, S = 4, which is invalid. And so FOUR = 9025, S = 7, ie SIX = 784, which leaves TEN = 361.

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  GeoffR 12:22 pm on 9 July 2021 Permalink | Reply

  # three/four digit squares list
  sq = [x * x for x in range(10, 100)]
  
  digits = set('1234567890')
  
  from itertools import permutations
  
  # 1st stage permutation
  for p1 in permutations('1234567890', 3):
    A, B, C = p1
    if A == '0':
      continue
    ABC = int(A + B + C)
    if ABC not in sq:
      continue
    
    # 2nd stage permutation
    q1 = digits.difference(p1)
    for p2 in permutations(q1, 3):
      X, Y, Z = p2
      if X == '0':
        continue
      XYZ = int(X + Y + Z)
      if XYZ not in sq:
        continue
  
      # 3rd stage permutation
      q2 = q1.difference(p2)
      for p3 in permutations(q2):
        P, Q, R, S = p3
        if P == '0':
          continue
        PQRS = int(P + Q + R + S)
        if PQRS not in sq:
          continue
  
        # form sum of three squares
        NONSO = ABC + XYZ + PQRS
        # check digit pattern of NONSO
        if 10000 < NONSO < 100000:
          if NONSO // 1000 % 10 == NONSO % 10:  # 'O'
            if NONSO // 10000 == NONSO // 100 % 10: # 'N' 
              print(f"Three Squares are {ABC}, {XYZ}, {PQRS}")
            
  # Three Squares are 361, 784, 9025
  
  
  

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