S2T2

Jim Randell 4:56 pm on 3 December 2021 Permalink Reply
Tags: by: Colin Vout ( 36 )   

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  Jim Randell 5:08 pm on 3 December 2021 Permalink | Reply

  If we suppose a major unit consists of x minor units (x ≤ 10), then the side of the square can be expressed as a major units plus b minor units (= ax + b minor units) where a < b < x.

  The following Python program runs in 45ms.

  Run: [ @replit ]

  from enigma import (irange, subsets, div, sq, printf)
  
  # consider x minor units in a major unit (no more than 10)
  # consider a major + b minor units (a < b < x)
  for (a, b, x) in subsets(irange(1, 10), size=3):
    # the side of the square is (ax + b) (minor units)
    s = a * x + b
    # can s^2 be expressed as k x^2 + 15?
    k = div(sq(s) - 15, sq(x))
    if k is not None:
      printf("x={x} a={a} b={b}; s={s} (= {a}x + {b}); s^2 = {k}x^2 + 15")
  

  Solution: The area is a square with each side measuring 41 minor units.

  So the entire area measures 41 × 41 = 1681 minor units.

  There are 7 minor units in a major unit (so the side of the square is 5 major + 6 minor units).

  And the area of the square can be expressed as 34 square major + 15 square minor units.

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  • 

    Jim Randell 11:51 am on 4 December 2021 Permalink | Reply

    Or using the [[ SubstitutedExpression ]] solver from the enigma.py library:

    #! python3 -m enigma -r
    
    SubstitutedExpression
    
    --base=11
    
    "A < B < X"
    
    "div(sq(A * X + B) - 15, sq(X))"
    
    --answer="A * X + B"
    

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  Frits 3:28 pm on 4 December 2021 Permalink | Reply

  I have the feeling X might be logically determined as the Wolframalpha site for this equation only shows solutions for a specific positive X.

    
  #!/usr/bin/env python3 -m enigma -r
   
  SubstitutedExpression
   
  --base=11
   
  "B < X"
  
  # if X is even then A * X + B must be odd so B must be odd as well
  "X % 2 or B % 2"
  
  # if X is 10 then sq(A * X + B) ends on a 5 so B must be 5
  "X < 10 or B == 5"
  
  "A < B"
   
  "div(2 * A * B * X + B * B - 15, sq(X))"
   
  --answer="A * X + B"
  # if X is 5 then sq(A * X + B) ends on 0 or 5 so B must be 0 or 5, impossible
  --invalid="0|1|2|3|5,X"
  --invalid="0|1|10,B"
  # A can't be 8 as if X is 10 then B must be 5
  --invalid="0|8|9|10,A"
  #--verbose=256   # print generated code
  

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  GeoffR 9:29 am on 5 December 2021 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  % a major units, x minor units in one major unit and b minor units
  var 1..9:a; var 1..9:b; var 1..9:x;
  % S = side of perfect square
  var 1..50:S;
  
  constraint a < b  /\ b < x;
  constraint S == a * x + b;  
  
  % the area is a certain number of square major units
  % plus a remainder of 15 square minor units
  constraint S * S mod (x*x) == 15;
  
  solve satisfy;
  
  output["Side in minor units = " ++ show(S) ++ "  " 
  ++ "\n" ++ "a = " ++ show(a)  ++ ", b = " ++ show(b) ++ ", x = " ++ show(x)];
  
  
  

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  • 

    Jim Randell 9:54 am on 5 December 2021 Permalink | Reply

    @GeoffR: x is “no more than 10”, so [[ var 1..10: x; ]] would be more appropriate.

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  GeoffR 10:19 am on 5 December 2021 Permalink | Reply

  @Jim: Yes, correct. I was assuming x < 10.
  Fortunately, it does not affect the answer.

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