S2T2

Jim Randell 3:58 pm on 13 September 2019 Permalink Reply
Tags: by: Victor Bryant ( 108 )   

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  Jim Randell 4:06 pm on 13 September 2019 Permalink | Reply

  The two mystery numbers have a common divisor that is 3 digits, and also the have a common multiple that is 3 digits, so they are both 3 digit numbers themselves.

  Denoting the numbers UVW and XYZ we can use the [[ SubstitutedExpression() ]] solver from the enigma.py library to give a direct interpretation of the the puzzle.

  This run file executes in 989ms.

  Run: [ @repl.it ]

  #!/usr/bin/env python -m enigma -r
  
  SubstitutedExpression
  
  --distinct="CFHLM"
  --answer="(UVW, XYZ)"
  
  "gcd(UVW, XYZ) = HCF"
  "lcm(UVW, XYZ) = LCM"
  
  "all_different(UVW, XYZ, HCF, LCM)"
  
  "UVW < XYZ"
  

  Solution: The first two numbers are 278 and 417.

  hcf(278, 417) = 139
  lcm(278, 417) = 834

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  • 

    Jim Randell 5:14 pm on 13 September 2019 Permalink | Reply

    Here’s an alternative approach that uses a bit of analysis:

    Each of the mystery numbers must be some small (proper) multiple of HCF, say, A × HCF and B × HCF. And A and B must be co-prime.

    We can then use the fact that:

    hcf(p, q) × lcm(p, q) = p × q

    to give a faster alphametic approach.

    The following run file executes in 173ms.

    Run: [ @repl.it ]

    #!/usr/bin/env python -m enigma -r
    
    SubstitutedExpression
    
    --distinct="CFHLM"
    --answer="(A * HCF, B * HCF)"
    
    "1 < A < B"
    "gcd(A, B) = 1"
    
    "A * B * HCF = LCM"
    

    ---

    Actually it is clear that A × B < 10, so the only options are A = 2, B = 3, giving rise to the following one-line solution:

    % python -m enigma SubstitutedExpression "6 * HCF = LCM" --answer="(2 * HCF, 3 * HCF)"
    (6 * HCF = LCM) ((2 * HCF, 3 * HCF))
    (6 * 139 = 834) ((2 * 139, 3 * 139)) / C=3 F=9 H=1 L=8 M=4
    (2 * HCF, 3 * HCF) = (278, 417) [1 solution]
    

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  GeoffR 10:09 am on 17 September 2019 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 1..9:H; var 0..9:C; var 0..9:F;
  var 1..9:L; var 0..9:M;
  
  var 100..999:N1; var 100..999:N2;
  var 100..999:HCF = 100*H + 10*C + F;
  var 100..999:LCM = 100*L + 10*C + M;
  
  constraint all_different( [N1, N2, HCF, LCM] );
  constraint all_different( [H, C, F, L, M] );
  
  constraint LCM * HCF == N1 * N2;
  
  constraint N1 mod HCF == 0 /\ N2 mod HCF == 0;
  
  solve maximize(HCF);
  
  output [ " First number = " ++ show(N1) ++ "\n"
  ++ " Second number = " ++ show(N2) ++ "\n"
  ++ " Highest common factor = " ++ show(HCF) ++ "\n"
  ++ " Lowest common multiple = " ++ show(LCM) ]; 
  
  

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