S2T2

Jim Randell 8:47 am on 17 February 2026 Permalink Reply
Tags: by: C Higgins ( 36 ), by: H Bradley ( 36 )   

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  Jim Randell 8:48 am on 17 February 2026 Permalink | Reply

  If J starts with an amount X and is a (> 19) years old. L starts with an amount Y.

  Then in the actual scenario:

  J has X; L has Y
  J→L: X/2 + 19
  J has (X/2 − 19); L has (Y + X/2 + 19)
  L→J: (Y + X/2 + 19)/2 + a
  J has (X/2 − 19 + Y/2 + X/4 + 19/2 + a) = (3X/4 + Y/2 + a − 19/2)

  In the reverse scenario:

  J has X; L has Y
  L→J: Y/2 + a
  J has (X + Y/2 + a); L has (Y/2 − a)
  J→L: (X + Y/2 + a)/2 + 19
  J has (X + Y/2 + a − (X/2 + Y/4 + a/2 + 19)) = (X/2 + Y/4 + a/2 − 19)

  And so:

  (3X/4 + Y/2 + a − 19/2) is a multiple of (X/2 + Y/4 + a/2 − 19)

  i.e., for some positive integer k:

  k = (3X/4 + Y/2 + a − 19/2) / (X/2 + Y/4 + a/2 − 19)
  ⇒
  k = 2 + (114 − X)/(2X + Y + 2a − 76)

  And if we want a proper multiple, then k ≥ 2.

  Considering the fractional part of the expression:

  If X > 114, then the numerator is negative, and the denominator is positive, so this is no good.

  if X = 114, then the numerator is zero, and the denominator is positive, whatever the values of Y and a are.

  If X < 114, then we need to choose values of Y and a, such that Y ≥ 2a ≥ 40, and the denominator divides into the numerator to give a positive integer.

  Here is a Python program to examine the scenarios:

  from enigma import (irange, divisors_pairs, printf)
  
  # consider possible values for X
  # X/2 >= 19 => X >= 38
  # 114 - X >= 0 => X <= 114
  
  # consider possible X values
  for X in irange(38, 114):
    n = 114 - X
    if n == 0:
      # output scenario
      printf("X = {X}; a >= 20; Y >= 2n")
    else:
      # consider possible divisors of n
      for (d, x) in divisors_pairs(n, every=1):
        k = 2 + x
        # calculate z = (Y + 2a)
        z = d + 76 - 2*X
        # which must be >= 40 + 40 = 80
        if z < 80: continue
        # consider possible a values >= 20
        for a in irange(20, 90):
          Y = z - 2 * a
          if Y < 0: break
          if Y < 2 * a: continue
          # output scenario
          printf("X = {X} [n = {n}; d = {d}] -> k = {k}; a={a} Y={Y}")
  

  Solution: John started with £ 114.

  And it doesn’t matter how old he is (a), or Lauren’s initial amount (Y), as long as Y ≥ 2a ≥ 40, then John will end up with twice the amount in the reverse scenario.

  For example: X = 114, Y = 62, a = 25.

  actual:
  start: J=114, L=62
  (57+19) J→L: J=38, L=138
  (69+25) J→L: J=132, L=44

  reverse:
  start: J=114, L=62
  (31+25) L→J: J=170, L=6
  (85+19) J→L: J=66, L=110

  And 132 = 2×66.

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