S2T2

Jim Randell 3:07 pm on 11 February 2022 Permalink Reply
Tags: by: Andrew Skidmore ( 83 )   

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  Jim Randell 3:42 pm on 11 February 2022 Permalink | Reply

  We can use some handy routines from the enigma.py library to help in solving this puzzle. [[ recurring() ]] finds the recurring part of the decimal representation of a fraction, and [[ filter_unique() ]] can be used to find sets that are unique by the count of multiples of 10 involved.

  The following Python program runs in 91ms.

  Run: [ @replit ]

  from enigma import (defaultdict, irange, union, recurring, join, subsets, filter_unique, printf)
  
  # generate sets of possible numbers
  def generate():
    # choose the denominator (2-digits, one of them a 7)
    for d in union([irange(17, 97, step=10), irange(70, 79)]):
  
      # consider the smaller numbers, n/d = 0.(xyz)...
      # and record them by the digits that make up the repetend
      r = defaultdict(list)
      for n in irange(1, d - 1):
        (_, nr, rr) = recurring(n, d)
        if (not nr) and len(rr) == 3:
          r[join(sorted(rr))].append(n)
  
      # find sets of 3 numerators
      for ns in r.values():
        for ss in subsets(ns, size=3, fn=list):
          yield tuple(ss + [d])
  
  # find sets unique by the count of multiples of 10 involved
  fn = lambda ns: sum(n % 10 == 0 for n in ns)
  for ns in filter_unique(generate(), fn).unique:
    printf("{ns}")
  

  Solution: Liam’s numbers are: 4, 30, 40, 74.

  So we have:

  4/74 = 0.(054)…
  30/74 = 0.(405)…
  40/74 = 0.(540)…

  And this the only set of 4 numbers where 2 of them are divisible by 10.

  In all there are 30 possible sets of 4 numbers. 12 sets use a denominator of 37, and the numbers in these sets can be multiplied by 2 to give 12 further sets (that give the same fractions) using a denominator of 74. The remaining 6 sets have a denominator of 27.

  Of these there are 19 sets with none of the 4 numbers divisible by 10, and 10 sets with one of the 4 numbers divisible by 10. The remaining set has two of the numbers divisible by 10, and so gives the answer.

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  GeoffR 3:09 pm on 15 February 2022 Permalink | Reply

  Not very elegant code, but it finds the answer, and runs in less than 100ms.
  I used two Python dictionaries in this solution.

  from enigma import recurring, join
  from collections import defaultdict
  
  # for potential number lists
  Nums = defaultdict(list)
  
  # for counting values in Nums divisible by 10
  D10 = defaultdict(list) 
  
  # using list of possible denominators including digit 7
  # .. to search for repeating decimal patterns
  for n in (17,27,37,47,57,67,70,71,72,73,74,75,76,77,78,79,87,97):
    (_, nr, rr) = recurring(1, n)
    if not(nr) and len(rr) == 3:
      print(f"Number = {n}, reciprocal = {1/n}")
  
  # only 27 and 37 were found as denominators with repeating patterns
  # ... add multiples 54 and 74 as extra 2-digit numbers
  # ...in search for all 3-digit repeating patterns
  for n in range(1, 80):
    for d in (27, 37, 54, 74):
      if n < d:
        (_, nr, rr) = recurring(n, d)
        if not(nr) and len(rr) == 3:
          Nums[join(sorted(rr)) + " " + str(d)].append(n)
  
  # add denominator (from the key) to values            
  for k, v in Nums.items():  
    v = v + [int(k[4:6])]
  
  # form a new dictionary D10, based on the Nums dict values
  # ... which are divisible by 10
  for k, v in Nums.items():
    tot10 = 0
    # count numbers divisible by 10
    for x in v:
      q, r = divmod(int(x), 10)
      if q and r == 0:
        tot10 += 1
      # add denominator to D10 values
      D10[tot10].append(v + [int(k[4:6])] )
  
  # look for a unique set of four numbers
  # ... based on the number of values divisible by 10
  for k, v in D10.items():
    if len(v) == 1:
      print(f"Four numbers are {v}")
  
  
  

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  A.T.Surherland 9:40 am on 16 February 2022 Permalink | Reply

  I have obtained an interesting answer :-
  The sum of the 3 lowest numbers = the greatest number.
  ie the sum of the fractions = 1.
  Is there a reason for this condition?.

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  • 

    Jim Randell 12:01 pm on 16 February 2022 Permalink | Reply

    There are 12 sets of numbers that work with a denominator of 37. Of these 6 have the fractions sum to 1, and 6 have the fractions sum to 2. And we can obviously multiply all the numbers up by a factor of 2 to get another 12 sets that behave the same.

    The remaining 6 sets of numbers (with denominator 27) give fractions that don’t sum to a whole number (but are all multiples of 1/9).

    This is presumably due to the 3 different digits in the repetends lining up, so that the repetend of the sum is a single digit, and in the case of .(9)… this gives a whole number.

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  • 

    Brian Gladman 12:29 pm on 16 February 2022 Permalink | Reply

    If we add 0.abc… + 0.bca… + 0.cab… we will get 0.111… x (a + b + c) = (a + b + c) / 9, so sum(numerators) / denominator = (a + b + c) / 9. So this happens when a + b + c == 9, which it does in our case. The sum can be 2 when a + b + c equals 18 but it is not necessarily an integer.

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  GeoffR 11:07 am on 16 February 2022 Permalink | Reply

  Yes, This is my understanding of the reason.

  A repeating fraction of the format 0.abcabcabc.. can be expressed exactly as a rational fraction abc/999 – see number theory elsewhere.

  Each of the three numerators in the answer divided by the denominator, gives a 3-digit repeating pattern.

  Without displaying the answer publicly, as this is a current Sunday Times Teaser,
  the three repeating fractions for this teaser can be expressed as:

  0.abcabcabc   0.cabcabcab   0.bcabcabca – since the three fractions use the same digits as a stated teaser condition.

  The rational fractions for this teaser are then abc/999, cab/999 and bca/999.

  Also, abc + cab + bca = 999.

  It can be seen that abc/999 + cab/999 + bca/999 = 1.

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