capecchi

Members
  • Content count

    6
  • Joined

  • Last visited

    Never

Community Reputation

0

About capecchi

  • Rank
    Newbie

capecchi's Activity

  1. capecchi added a post in a topic   

    Yeah but that's not an interesting question. If we look again at the moment there's a train leaving station 12 headed right and at the same moment a train arriving at station 12 from the left that means from station 12 to 1 and back to 12 there are 24 minutes spent at each station, then 22 minutes spent transiting between stations plus an extra unknown quantity for the loop at the end by station 1(call it L). So 46 minutes plus L on the left end of the track. Similarly, from station 12 to 24 and back to 12 we spend 24 minutes in the station, 24 minutes in transit between stations, plus some unknown quantity at the end by station 24 (call it R). So 48 minutes plus R on the right side of the track. What are we saying these days, 7 minutes between trains? It doesn't much matter since I have complete control now by being able to choose what R and L are. So if the time between trains is 7 minutes then 48+R has to be some multiple of 7. Why not make it 49, so R=1 and there are 7 trains on the right half of the track. Then 46+L has to be a multiple of 7, 49 is the closest again, so let R be 3 minutes and we have another 7 trains on the left side of the track. 14 trains.
    Unsatisfying though, since I could just pick whatever I wanted R and L to be to make it fit the model we are given.
    How bout we figure this out again: try this-

    There are 24 stations, we're at station 12. Every 8 minutes a train arrives from the left (for example, 10am=train arrives from the left, 10:08=train arrives from the left, etc) and TWO minutes after a train arrives from the left a train arrives from the right (so 10:02=train arrives from the right, 10:10=train arrives from the right, etc). If all transit times are equal including the time going around the loops at the end, what is the fewest number of trains on the track (excluding the case of a zero transit time between stations)? Recall, upon reaching station 1 (or station 24) a loop is made that returns you to station 1 (or 24) where you have to spend another minute at the station.

    Does this clear everything up?
    I have a solution to this. Anyone else? I changed it slightly so as to make everyone change their answers and think it through again.
    What if trains arrive every 4 minutes from the left (so 10:00=train from left, 10:02=train from right, 10:04=train from left, 10:06=train from right, etc)? An interesting question to ask yourself before you solve it is whether you think that this will require more or fewer trains than in the 8 minute case.
    • 0
  2. capecchi added a post in a topic   

    I'm getting frustrated. How many minutes are there between trains? I see support in this thread for 4,5,6, and 7 minutes. Another way to phrase the question: If I'm only considering trains traveling to the right and I see a train pass by, how long do I have to wait to see the next one?

    I've answered for 4,5, and 6 trains the answers being, respectively, 60, no solution, and 56.
    Now lets go through it again for 7 trains and someone tell me if I'm making a logical error somewhere. We take a snapshot of the situation one minute after the rightward moving train arrives at station 12. At this moment the rightward moving train (call it R) is just departing. Also a leftward moving train (call it L) is just arriving at station 12. Between L and R there are 24 stations (12,11,10,9,8,7,6,5,4,3,2,1,1,2,3,4,5,6,7,8,9,10,11,12) requiring 1 minute each or 24 minutes. There are also 23 transits that have to be made (12-11,11-10,10-9,9-8,8-7,7-6,6-5,5-4,4-3,3-2,2-1,1-1,1-2,2-3,3-4,4-5,5-6,6-7,7-8,8-9,9-10,10-11,11-12) each requiring some unknown amount of time x. So the total time between trains L and R is:
    23x+24
    This time must be some multiple of 7 (if we are to assume there are 7 minutes between trains and equal times between all stations as we should) so:
    23x+24=7n for some integer n
    Similarly there are 24 stations (13,14,15,16,17,18,19,20,21,22,23,24,24,23,22,21,20,19,18,17,16,15,14,13) between R and L for a total of 24 minutes, plus 25 transits (12-13,13-14,14-15,15-16,16-17,17-18,18-19,19-20,20-21,21-22,22-23,23-24,24-24,24-23,23-22,22-21,21-20,20-19,19-18,18-17,17-16,16-15,15-14,14-13,13-12) of x minutes each so that all transits including the ends have the same transit time gives a total for the time between R and L to be:
    25x+24
    again a multiple of 7 so:
    25x+24=7m for some integer m
    Since x need not be a whole number we can use the two equations to eliminate x and get an equation relating m,n. After algebra:
    175n-161m=48
    No such integers exist.
    To see for yourself, if you have access to something like Mathematica I used the code:
    FindInstance[25 7 n - 23 7 m == 48, {n, m}, Integers]
    Therefore I maintain if the gap is 4 minutes there are 60 trains.
    If the gap is 6 minutes there are 56 trains.
    If the gap is 5 or 7 minutes there is no solution which satisfies the given data (while requiring all transit times to be equal which I still maintain is the only way for this to be a meaningful thing to do). I think it's silly there's so much runaround on this forum. Anyone see an issue with my handling of this problem?
    Also, to reiterate what I've mentioned in previous posts, these answers are for the fewest trains possible given a non-zero transit time.

    Would like to hear comments about my work. Is it clear enough?
    I'd like some backers for my answers... anyone wanna support my math?
    • 0
  3. capecchi added a post in a topic   

    Alright then. If the new example is correct and minute by minute we go: train from left, train from right, no train, no train, no train, no train, train from left, etc. Then We need to solve:
    23x+24=6n and
    25x+24=6m
    This gives 23m=25n-8
    Which has a solution of m=n=4 but this again gives x=0 so if we want the first set of integers that give us a nonzero transit time we find:
    m=29,n=27. This gives x=6 minutes. So the track has 56 trains, each 6 minutes apart. For explanation of what I did here see my first post.
    • 0
  4. capecchi added a post in a topic   

    I agree there's something that needs to be cleared up. In the original post it sounds as if trains are arriving FROM THE LEFT every 4 minutes. The first example posted by wolfgang then makes it seem that trains arrive FROM THE LEFT every 5 minutes. Now this most recent example shows they arrive FROM THE LEFT every 7 minutes.
    So the question that needs to be clarified: How often do trains arrive FROM THE LEFT?
    • 0
  5. capecchi added a post in a topic   

    Correction to my last post. Due to an algebra error (24/4 is not 8) the equation to find solutions to is 23m=25n-12 for some integers m,n. The lowest pair satisfying this is m=n=6 which means that x=0. If we wish for a non-zero transit time the next pair of integers is n=29, m=31 giving x=4. Therefore there will be 60 trains.
    Also I'm confused by the example mentioned above. It implies there is actually a 5 minute gap between trains traveling in one direction. If this is the case then most of what I mentioned in the above post holds except we must solve the equations:
    23x+24=5n and
    25x+24=5m
    When we eliminate x we find the equation
    115m=125n-48
    which must be solved again, for some integers m,n. However, no such solution exists. (Easy enough to see when you consider that for any integers m,n 115m and 125n will have their last digit either 5 or 0, but when we subtract 48 from 125n we get the last digit is 2 or 7)
    So if trains to have 4 minute spacing, the answer (lowest number of trains with non-zero transit time) would be 60 and I suggest a new example of:
    10am... train arrives from the left
    10:01... train arrives from the right
    10:04... train arrives from the left
    10:05... train arrives from the right
    ...
    • 0
  6. capecchi added a post in a topic   

    64 trains. In order for this problem to remain an interesting logic puzzle I feel we need to assume things about transit time, ie that they are all equal. Without this assumption any amount of doctoring can yield a result, and many results besides, as we can see already in these posts. With this assumption lets call the transit time between stations 'x'. Lets consider the moment when the leftward moving train just arrives at station 12, and call this train L. This means that the rightward moving train that arrived a minute earlier is just leaving headed towards station 24, lets call this train R.
    Consider first train L. What we know is eventually train L will make it around the track to where train R is, and since trains arrive heading right every 4 minutes, there is an integer number of 4 minute intervals between trains R and L (heading through the end with station 1). With the convention that x is the transit time between stations (including the end 'turnaround' that takes a train from station 1 back to station 1 or 24 back to 24), then the time interval between L which just arrived at station 12 and R which is just leaving station 12 is:
    23x+24
    The 24 is from the fact that there are 24 stations that each require a stay of a minute (including station 12 twice since train L must wait a minute before it leaves as well as wait a minute when it returns before reaching the position where train R was. Since this must be a multiple of 4, we set:
    23x+24=4n
    A similar argument follows that there must be an integer number of 4 minute intervals between train R and train L as you go around station 24. This gives:
    25x+24=4m
    Combine these two equations to eliminate x and require that m and n be integers and the smallest pair of integers that offers a solution is n=31 and m=33. This means that there are 31 4 minute intervals from L to R and 33 4 minute intervals between R and L, or 64 total intervals of 4 minutes. 64 trains.
    • 0