S2T2

Jim Randell 12:03 pm on 11 October 2020 Permalink Reply
Tags: by: J. V. Sharp   

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  Jim Randell 12:04 pm on 11 October 2020 Permalink | Reply

  Let’s suppose David intended to leave at m minutes past 8:00am (so m = 0 .. 59).

  At this time both the clock (which runs slow) and the watch (which runs fast) would both say the correct time:

  clock(m) = m
  watch(m) = m

  If the watch runs at a rate of w (> 1), and the clock at a rate of c (< 1) we have:

  clock(t) = m + (t − m)c
  watch(t) = m + (t − m)w

  Now, he actually left at n minutes past 8:00am (so n = 0 .. (m − 1)).

  And at this time the clock read 8:15 and the watch read 8:10:

  clock(n) = 15
  watch(n) = 10

  (The fact the clock is ahead of the watch tells us that the actual time is earlier than the intended time).

  So:

  m + (n − m)c = 15 ⇒ c = (m − 15) / (m − n)
  m + (n − m)w = 10 ⇒ w = (m − 10) / (m − n)

  And at the moment the train left (11:00am) the watch read 11:10:

  watch(180) = 190
  m + (180 − m)w = 190
  (180 − m)(m − 10) = (190 − m)(m − n)
  n = 1800 / (190 − m)
  m = 190 − 1800 / n

  And we need c < 1 and w > 1:

  m − 15 < m − n
  m − 10 > m − n
  10 < n < 15

  So there are only 4 values to check, which we can do manually or with a program.

  from enigma import (irange, div, printf)
  
  # choose a value for n (actual leave time)
  for n in irange(11, 14):
    # calculate m (intended leave time)
    m = div(190 * n - 1800, n)
    if m is None: continue
    printf("n={n} -> m={m}")
  

  Solution: David intended to leave at 8:40am. He actually left at 8:12am.

  The watch runs at a rate of 15/14 (approx. 107%). So it gains 2 minutes every 28 minutes.

  The clock runs at a rate of 25/28 (approx. 89%). So it loses 3 minutes every 28 minutes.

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  John Crabtree 6:38 pm on 13 October 2020 Permalink | Reply

  If not leaving at the intended time, Dave’s actual time of departure must be between those indicated on the clock and watch. Assume that he left x minutes after 8:10 where x = 1, 2, 3 or 4.
  The watch would be correct (170 – x) * x / (x + 10) = … = 180 – x – 1800 / (x + 10) minutes later, which is an integer.
  And so x = 2, and the solution follows.

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