S2T2

Jim Randell 8:04 am on 13 October 2022 Permalink Reply
Tags: by: Danny Roth ( 85 )   

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  Jim Randell 8:05 am on 13 October 2022 Permalink | Reply

  None of the individual digits can be 0, as then the product of the digits would also be 0.

  We can use the [[ SubstitutedExpression ]] solver from the enigma.py library to solve this puzzle.

  The following run file executes in 65ms. (Internal runtime of the generated code is 1.1ms)

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  --distinct=""
  --digits="1-9"
  --answer="ABCD"
  
  # The following multiplied together give the value of the PIN
  # (i) the sum of the digits;
  # (ii) the mean of the digits (= (i) / 4);
  # (iii) the product of the digits
  "sq(A + B + C + D) * (A * B * C * D) == 4 * ABCD"
  
  # the PIN is odd
  "D % 2 = 1"
  

  Solution: The PIN is: 1715.

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• 

  GeoffR 12:16 pm on 13 October 2022 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;
  constraint D mod 2 == 1; % PIN number is odd
  
  var 1000..9999:PIN = 1000*A + 100*B + 10*C + D;
  
  % Num 1the sum of the four digits in the PIN
  var 4..36:Num1 = A + B + C + D;
  
  var 4..36: Num2; %  average = Num2 / 4
  constraint Num2 == A + B + C + D;
  
  % Num3 was the product of the four digits
  var 1..6561: Num3 == A * B * C * D ; % UB = 9^4
  
  % Martha multiplied these three numbers together
  constraint 4 * PIN == Num1 * Num2 * Num3;
  
  solve satisfy;
  
  output[ "PIN = " ++ show(PIN) ];
  
  % PIN = 1715
  % ----------
  % ==========
  
  
  

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