S2T2

Jim Randell 8:10 am on 29 October 2024 Permalink Reply
Tags: by: Danny Roth ( 84 )   

• 

  Jim Randell 8:11 am on 29 October 2024 Permalink | Reply

  I assumed that the rail fare does not have a trailing zero, so that the room rate does not have a leading zero.

  This Python program finds the first solution where the rail fare and room rate are less than £100.

  It runs in 80ms. (Internal runtime is 188µs).

  from enigma import (irange, nrev, div, printf)
  
  # consider possible rail fares
  for F in irange(1, 99):
    # disallow trailing zeros
    if F % 10 == 0: continue
  
    # the room rate is the reverse of this
    R = nrev(F)
  
    # cost for 30 nights and 33 nights
    t30 = 4*F + 30*R
    t33 = 4*F + 33*R
  
    # t33 is a k per cent increase on t30
    r = div(100 * t33, t30)
    if r is None: continue
  
    printf("F={F} R={R} r={r}%")
    break
  

  Solution: The cost of the room was £ 54 per night.

  And so the rail fares were £ 45 per person (single).

  The cost for 30 days is:

  4× 45 + 30× 54 = 1800

  And the cost for 33 days:

  4× 45 + 33× 54 = 1962

  And we have:

  1962 / 1800 = 1.09

  So the increase is 9%.

  ---

  There are larger solutions, for example:

  F=45, R=54
  F=495, R=594
  F=4545, R=5454
  F=4995, R=5994
  F=45045, R=54054
  F=49995, R=59994
  F=450045, R=540054
  F=454545, R=545454
  F=495495, R=594594
  F=499995, R=599994
  …

  But the room is described as “inexpensive”, so only the first of these is a viable solution.

  However, they all involve a 9% increase.

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• 

  GeoffR 12:34 pm on 29 October 2024 Permalink | Reply

  
  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 1..9:A; var 1..9:B; 
  var 1..99: per_cent;
  
  % Assume room cost < £100
  var 12..98: Room == 10*A + B; 
  var 12..98: Fare == 10*B + A;
  
  % Two people with outgoing and return trips = 4 fares
  var 600..6000: Cost30 == 30 * Room + 4 * Fare;
  var 660..6060: Cost33 == 33 * Room + 4 * Fare;
  
  % Calculate percentage increased room cost
  constraint (Cost33 - Cost30) * 100 ==  Cost30 * per_cent;
  
  solve satisfy;
  
  output ["Room cost = " ++ show(Room) ++ " pounds per night."
  ++ "\n" ++ "Cost percentage increase = " ++ show(per_cent) ++ " %"];
  
  % Room cost = 54 pounds per night.
  % Cost percentage increase = 9 %
  % ----------
  % ==========
  

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• 

  Ruud 4:15 am on 30 October 2024 Permalink | Reply

  from istr import istr
  
  
  istr.repr_mode("int")
  for train in istr(range(10, 100)):
      room = train[::-1]
      if room[0] != 0:
          total30 = int(train * 4 + room * 30)
          total33 = int(train * 4 + room * 33)
          increase = ((total33 - total30) / total30) * 100
          if increase % 1 == 0:
              print(f"{train=} {room=} {total30=} {total33=} {increase=:.0f}%")
  

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• 

  Frits 5:43 am on 3 November 2024 Permalink | Reply

  Objective was to use fewer iterations.

  from fractions import Fraction as RF
  
  # fare = 10 * f + g, room cost = 10 * g + f
  
  # t30 = 4*F + 30*R = 70*f + 304*g  
  # t33 = 4*F + 33*R = 73*f + 334*g
  
  # 100 + p = 100 * (73*f + 334*g) / (70*f + 304*g)     
  
  # integer percentages less than 10
  for p in range(9, 0, -1):
    # (7300*f + 33400*g) - (100 + p) * (70*f + 304*g) = 0
    # ((300 - p*70)*f + (3000 - 304*p)*g) = 0
    
    # calculate - f / g  (g will not be zero)
    r = RF(3000 - 304*p, 300 - p*70)
    if r > 0: continue  
    f, g = abs(r.numerator), abs(r.denominator)
    if not (0 < f < 10 and g < 10): continue 
    
    # double check
    if 100 * (73*f + 334*g) / (70*f + 304*g) - 100 != p: continue
    print(f"answer: {10 * g + f} pounds per night") 
  

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