S2T2

Jim Randell 9:42 am on 6 May 2025 Permalink Reply
Tags: by: Danny Roth ( 84 )   

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  Jim Randell 9:42 am on 6 May 2025 Permalink | Reply

  If there are k houses on each side of the road, then opposite houses sum to (2k + 1).

  There are 2k houses on the road in total, and the number of George and Martha’s house n is a 2-digit number that is a multiple of the house opposite (which has number 2k + 1 − n).

  Hence:

  n > 2k + 1 − n
  ⇒
  k < (2n − 1) / 2

  And the maximum possible 2-digit value for n is 99, hence:

  k ≤ 98

  This gives us an upper bound for an exhaustive search.

  The following Python program runs in 56ms. (Internal runtime is 332µs).

  from enigma import (irange, group, ndigits, printf)
  
  # consider possible k values (number of houses on one side of the road)
  for k in irange(52, 98):
    t = 2 * k  # total number of houses
    # collect numbers that are a multiple of the house opposite
    ns = list(x for x in irange(k + 1, t) if x % (t + 1 - x) == 0)
    if len(ns) < 6: continue
    # group the numbers by digit count
    g = group(ns, by=ndigits)
    # ensure there is a 2-digit and 5 3-digits
    (n2s, n3s) = (g.get(2), g.get(3))
    if (not n2s) or (not n3s) or (len(n3s) < 5): continue
    # output solution
    printf("total houses = {t} [n2s={n2s} n3s={n3s}]")
  

  Solution: There are 134 houses in the road.

  And so the numbers of opposite houses sum to 135.

  George and Martha live at 90, which is opposite 45.

  And there are 6 houses that their 5 daughters could live at:

  108 (opposite 27)
  120 (opposite 15)
  126 (opposite 9)
  130 (opposite 5)
  132 (opposite 3)
  134 (opposite 1)

  ---

  If instead of an Avenue the road was a Close, with a house at the closed end (or several houses), then there would be further possible solutions.

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  Frits 12:39 pm on 8 May 2025 Permalink | Reply

  Using ideas of Jim and Brian.

  # the number of houses is even, say 2.k, so we want numbers 
  # n such that:
  # 
  #    n = m.(2.k + 1 - n)  ==>  (m + 1).n = m.(2.k + 1)
  #
  # this makes both n and m even (credit: B. Gladman)
  
  d3 = 5  # five 3-digit numbers
  mn = 49 + d3            # 2.k >= 100 + (d3 - 1) * 2 
  mx = (3 * 99 - 2) // 4  # n >= 2 * (2k + 1) - n
  
  # are there at least <a> house numbers with correct opposite house number?
  def find(k, a, strt, m=-1):
    c, x, k2 = 0, strt, k + k
    while c <= a and x >= m:
      c += (x % (k2 + 1 - x) == 0)
      x -= 2
    return c >= a  
  
  # consider possible k values (number of houses on one side of the road)
  for k in range(mn, mx + 1):
    # opposite housenumbers t.m / (m + 1) and t / (m + 1) with t = 2.k + 1
    # t.m / (m + 1) <= 98 or m <= 98 / (t - 98)  
    m = 98 // ((k2 := k + k) - 97) 
    # check for solutions of even n less than 100
    if any((k2 + 1) % x == 0 for x in range(3, m + 2, 2)):
      # check for at least <d3 - 1> solutions for even numbers 
      # in range 100 ... 2.k - 2
      if find(k, d3 - 1, k2 - 2, 100):
        print("answer:", k2)
  

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