S2T2

Jim Randell 8:53 am on 21 February 2021 Permalink Reply
Tags: by: F. Moore ( 2 )   

• 

  Jim Randell 8:55 am on 21 February 2021 Permalink | Reply

  1s = 12d.

  Working in half-pennies (hp) the total amount to distribute is 100 d = 200 hp.

  And we are told:

  11 | (A + B)
  13 | (B + C)
  19 | (A + C)

  And the difference between two of the numbers is a square number of half pennies.

  The following Python program runs in 53ms.

  Run: [ @repl.it ]

  from enigma import irange, is_square, div, printf
  
  # solve the equations
  def solution(middle, AB, AC, BC):
    A = div(AB + AC - BC, 2)
    B = div(AB + BC - AC, 2)
    C = div(AC + BC - AB, 2)
    if not any(x is None or x < 0 for x in (A, B, C)):
      # output solution
      printf("A={A} hp, B={B} hp, C={C} hp; middle = {middle}")
  
  # A + C is some multiple of 19
  for AC in irange(19, 200, step=19):
  
    # B + C is some multiple of 13
    for BC in irange(13, 200, step=13):
  
      # A + B is some multiple of 11
      # and (A + B) + (B + C) + (A + C) = 2(A + B + C) = 400
      AB = 400 - (AC + BC)
      if AB % 11 > 0: continue
  
      # one of the differences is a square
      # B - C = (A + B) - (A + C)
      if is_square(abs(AB - AC)): solution('A', AB, AC, BC)
      # A - C = (A + B) - (B + C)
      if is_square(abs(AB - BC)): solution('B', AB, AC, BC)
      # A - B = (A + C) - (B + C)
      if is_square(abs(AC - BC)): solution('C', AB, AC, BC)
  

  Solution: The middle son was Clive. His share was 45d = (3s 9d).

  The shares are:

  A: 5 hp = 2.5 d
  B: 105 hp = 52.5 d (= 4s 4.5d)
  C: 90 hp = 45 d (= 3s 9d)

  And we see:

  A + B = 110 hp = 10× 11 hp
  B + C = 195 hp = 15× 13 hp
  A + C = 95 hp = 5× 19 hp

  And the difference between the eldest and the youngest is: B − A = 10² hp.

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• 

  Frits 11:53 am on 21 February 2021 Permalink | Reply

  from enigma import SubstitutedExpression, is_square, peek
  
  # the alphametic puzzle
  p = SubstitutedExpression(
    [# A = AST, B = BVW, C = CYZ
    
    # start with the most restrictive constraint
    
    # A + C is some multiple of 19
    "(AST + CYZ) % 19 == 0",
    
    # divide 200 half-pennies over 3 sons
    "200 - AST - CYZ = BVW", 
    
    # B + C is some multiple of 13
    "(BVW + CYZ) % 13 == 0",
    
    # A + B is some multiple of 11
    "(AST + BVW) % 11 == 0",
    
    # difference between the eldest and youngest was a perfect square
    "peek(2 - i for (i, (x, y)) in \
     enumerate(zip([AST, CYZ, BVW], [BVW, AST, CYZ])) \
     if is_square(abs(x - y))) = I",
    ],
    answer="I, [AST, BVW, CYZ][I]",
    d2i=dict([(2, "ABC")] + [(k, "ABCI") for k in range(3, 10)]),
    distinct="",
    reorder=0,
    verbose=0)   
  
  for (_, ans) in p.solve():
    print(f"{['Alan', 'Bob', 'Clive'][ans[0]]} with share {ans[1]} half-pennies")
  

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  • 

    Jim Randell 12:49 pm on 21 February 2021 Permalink | Reply

    Here’s an alternative solution using the [[ SubstitutedExpression() ]] solver:

    from enigma import (SubstitutedExpression, subsets, is_square, printf)
    
    # find candidate (A, B, C) values
    p = SubstitutedExpression(
      [ "A + B + C == 200", "(A + B) % 11 = 0", "(B + C) % 13 = 0", "(A + C) % 19 = 0" ],
      base=201,
      distinct=(),
      answer="(A, B, C)",
      template="",
    )
    
    # look at possible (A, B, C) values
    for (_, vs) in p.solve(verbose=0):
      # the values for the eldest and youngest differ by a square number
      for ((i, x), (j, y)) in subsets(enumerate(vs), size=2):
        if is_square(abs(x - y)):
          # identify the middle son
          m = "ABC"[3 - i - j]
          printf("A={vs[0]} hp, B={vs[1]} hp, C={vs[2]} hp; middle={m}")
    

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• 

  Hugh Casement 1:51 pm on 21 February 2021 Permalink | Reply

  Theoretically Alan and Clive could each have received 57 ha’pence and Bob 86.
  Zero is also a square number! However, I doubt that is the intended solution.

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  • 

    Jim Randell 2:02 pm on 21 February 2021 Permalink | Reply

    @Hugh: Well spotted!

    In my programs, both 0 and None evaluate to false, so using the return value from [[ is_square() ]] as a boolean value will reject 0. (It also doesn’t consider 0 to be allowed as a multiple of 11, 13, 19).

    You can explicitly check [[ is_square(...) is not None ]], or just import [[ is_square_p() ]] instead which returns a boolean.

    And if you do that you do indeed get two solutions:

    % python teaser23b.py
    A=5 hp, B=105 hp, C=90 hp; middle = C
    A=57 hp, B=86 hp, C=57 hp; middle = B
    

    The published solution is: “Clive, 3s 9d”.

    Perhaps the setter intended the shares to be three different values.

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• 

  John Crabtree 6:58 pm on 22 February 2021 Permalink | Reply

  Working in 1/2d, then A + B + C = 200
  A + B = 11p, p ≤ 18
  B + C = 13q, q ≤ 15
  A + C = 19r, r ≤ 10
  Summing the three gives 400 = 11p + 13q + 19r
  Using r as a variable as it has the fewest possible values:
  B = 200 – 19r, p = 8 + 3r mod 13, q = 3 + 7r mod 11 which leads to
  A = 31 + 52r mod 143
  C = 112 + 110r mod 143
  It is then straightforward to manually generate a table of values for A, B and C for each r. r = 1, 2 and 4 give A+B+C = 343. r = 3 and r = 5 to 10 give A+B+C = 200 and include the two solutions noted by Jim.

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