Teaser 2486: [Meet at the station]
From The Sunday Times, 16th May 2010 [link] [link]
Pat drove to the station at his regular speed to meet his brothers, John and Joe, whose train was due at 6pm. But John had caught an earlier train and, arriving half an hour early, decided to walk home at a steady pace. Pat waved as he passed John (a whole number of minutes before 6pm), but drove on to the station. Pat collected Joe on time and, a whole number of minutes later, they set off for home. Pat and Joe overtook John 13 minutes after Pat had waved to him.
At what time did Pat and Joe overtake John?
This puzzle was originally published with no title.
[teaser2486]
S2T2 
Jim Randell 4:23 pm on 28 August 2025 Permalink |
Suppose John walks at a speed of 1 unit per minute.
At a minutes before 6pm Pat passes John. At this time John has walked a distance of (30 − a) units from the station. And in 13 minutes time John will have walked an additional 13 distance units, so will be a distance (43 − a) units from the station.
After Pat passes John he travels the remaining distance (30 − a) to the station, in a time of a minutes (to arrive at 6pm).
So his velocity is: (30 − a)/a.
If the Pat and John leave the station at b minutes after 6pm, then the time taken to catch up with John is:
and:
(where a, b, c are all positive integer values).
So we can consider possible integer values for a and calculate the corresponding values for c and b.
This can be done manually or programatically.
This Python program runs in 75ms. (Internal runtime is 33µs).
from enigma import (irange, div, printf) for a in irange(1, 5): c = div((43 - a) * a, (30 - a)) if c is None: continue b = 13 - (a + c) if b > 0: printf("a={a} -> c={c} b={b}")Solution: Pat and Joe caught up with John at 6:09 pm.
Pat initially passed John at 5:56 pm, before arriving at the station 4 minutes later (at 6 pm) to pick up Joe.
They left the station at 6:03 pm and 6 minutes later (at 6:09 pm) passed John again.
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