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Jim Randell 10:08 am on 20 July 2021 Permalink Reply
Tags: by: Eric Emmet ( 10 )   

• 

  Jim Randell 10:09 am on 20 July 2021 Permalink | Reply

  Working in units of 10p means the wages are in the range 100 – 200.

  This Python program runs in 54ms.

  Run: [ @replit ]

  from enigma import irange, div, subsets, int2base, printf
  
  # possible values
  values = irange(100, 200)
  
  # find possible (x, f(x)) pairs
  def pairs(f):
    for x in values:
      y = f(x)
      if y is not None and y in values:
        yield (x, y)
  
  # consider BW/B pairs (BW = 1.05 * 0.9 * B)
  for (BW, B) in pairs(lambda x: div(200 * x, 21 * 9)):
  
    # consider DKP/E pairs (E = 1.2 * (DKP - 10))
    for (DKP, E) in pairs(lambda x: div(6 * (x - 10), 5)):
  
      # consider D values
      for D in (E - 10, E + 10):
        if D not in values: continue
  
        # consider A values (A > D)
        for A in values:
          if not(A > D): continue
          C = div(11 * min(B, D, E), 10)
          if C is None or C not in values: continue
          # the set of wages
          ws = set([A, B, C, D, E])
          if len(ws) != 5: continue
          # remaining jobs
          js = ws.difference([BW, DKP])
          if len(js) != 3: continue
          for (DS, DO, WO) in subsets(js, size=len, select="P"):
            # check DS, DO values
            if not(DO % 2 == 1 and DS == max([A, B, C, D, E])): continue
            # jobs: map: wage -> job
            jobs = dict(zip([BW, DKP, DS, DO, WO], "BW DKP DS DO WO".split()))
  
            # output solution
            fmt = lambda x: int2base(x * 10, group=2, sep='.')
            printf("A={A} ({j})", A=fmt(A), j=jobs[A])
            printf("B={B} ({j})", B=fmt(B), j=jobs[B])
            printf("C={C} ({j})", C=fmt(C), j=jobs[C])
            printf("D={D} ({j})", D=fmt(D), j=jobs[D])
            printf("E={E} ({j})", E=fmt(E), j=jobs[E])
            printf()
  

  Solution: The wages per week are: Alf = £18.90; Bert = £20.00; Charlie = £12.10; Duggie = £11.00; Ernie = £12.00.

  And their jobs:

  Alf = Bottle-Washer
  Bert = Door-Shutter
  Charlie = Door-Opener
  Duggie = Door-Knob-Polisher
  Ernie = Welfare Officer

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• 

  Frits 4:00 pm on 21 July 2021 Permalink | Reply

  Manual solution.

  # Consider amounts in 10p.
  
  # Ernie = 1.2 * (Door-Knob-Polisher - 10)                              (rule 6)
  # Ernie = 108 + 6 * YZ,     YZ  less than 16  
  # Ernie must be even so also Duggie must be even                       (rule 4)
  
  # Bottle-Washer = 1.05 * 0.9 * Bert, 189 * Bert = 200 * Bottle-Washer  (rule 3)
  # 189 and 200 are coprime so Bottle-Washer = 189, Bert = 200
  # Bottle-Washer is odd so not Bert, Duggie or Ernie
  # Charlie has to be a multiple of 11                                  (rule 11)
  # so --- Bottle-Washer must be Alf ---, Alf = 189
  # so --- Door-Shutter must be Bert ---                                 (rule 7)
  
  # Door-Opener's wages are to be an odd multiple of 10p                 (rule 5)
  # Alf and Bert have diffeent jobs, Duggie and Ernie are even
  # so --- Charlie is the Door-Opener ---
  
  # Ernie is to get 20% more than £1 less than the Door-Knob-Polisher    (rule 6)
  # three jobs already have been assigned
  # so --- Duggie is the Door-Knob-Polisher ---
  # so --- Ernie is the Welfare Officer ---
  # Ernie is paid more than Door-Knob-Polisher
  # so worst paid has to be Duggie 
  # so Ernie = Duggie + 10                                               (rule 4)
  # Door-Knob-Polisher = Ernie / 1.2 + 10 = 5 / 6 * (108 + 6 * YZ) + 10 =
  # 5 * YZ + 100 so YZ must be even (as Duggie is even)
  
  # Charlie = 1.1 * Duggie = 1.1 * (Ernie - 10)                    (rule 1 and 4)
  # so Ernie must be a multiple of 10
  # so YZ must be 2 or 12    (Ernie = 108 + 6 * YZ and YZ is even)
  # so (Charlie, Duggie, Ernie) must be resp. (121, 110, 120) or (187, 170, 180)
  # only in first case rule 6 is obeyed
  
  # so Alf = £18.90, Bert = £20, Charlie = £12.10, Duggie = £11 and Ernie = £12
  

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