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Thread: Pickmeupper!

  1. #1
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    Pickmeupper!

    X(A,B,C).......................................... ......................Y
    XY is a straight road.
    A, B and C are together at X, and leave at the same time.
    A's speed is 1 mph, B's speed is 2 mph, C's speed is 8 mph.

    C travels directly to Y, arrives after u hours,
    then comes back and meets A after v hours.
    He picks up A and returns towards Y (speed still 8 mph).
    He drops off A after w hours; A continues.

    C returns, and meets B somewhere, after x hours.
    He picks up B and returns towards Y (speed still 8 mph).
    C and B arrive at Y at same time as A does, after y hours.

    The hours u,v,w,x,y are "total hours so far".
    And these hours are all integers.

    What is the minimum distance from X to Y?

    My solution: a 4digit number, digits adding to 18.
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  2. #2
    Newbie Olinguito's Avatar
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    Re: Pickmeupper!

    Spoiler:
    The distance from X to Y is 8u miles.

    At the moment C first arrives at Y, A has travelled u miles from X so the separation between them is 7u miles. When C heads towards A, their relative velocity is 8 + 1 = 9 mph; so C meets A in \tfrac{7u}9 hours: i.e.

    v-u=\frac{7u}9\ \Rightarrow\ v=\frac{16u}9

    The point where they meet is v miles from X (distance covered by A in v hours).

    At time w hours, when C drops off A, B has travelled 2w miles from X; C is v+8(w-v) = 8w-7v miles from X. Their separation is therefore 6w-7v miles. When C heads towards B, their relative velocity is 8 + 2 = 10 mph, so C meets B in \tfrac{6w-7v}{10} hours: i.e.

    x-w = \frac{6w-7v}{10}\ \Rightarrow\ x=\frac{16w-7v}{10} = \frac{16w-7\left(\frac{16u}9\right)}{10}=\frac{72w-56u}{45}

    The point where they meet is 2x miles from X (distance covered by B in x hours).

    At this moment, A has travelled x-w miles towards Y from where it was dropped off by C, i.e. (8w-7v)+(x-w)=x+7w-7v miles from X: i.e.

    8u+7v-7w-x = 8u+7\left(\frac{16u}9\right)-7w-\frac{72w-56u}{45} = \frac{976u-387w}{45}

    miles from Y. C and B themselves are

    8u-2x = 8u-\frac{144w-112u}{45} = \frac{472u-144w}{45}

    miles from Y. As C travels 8 times faster than A,

    8(976u-387w)=472u-144w\ \Rightarrow\ w=\frac{917u}{369}

    Since w must be an integer and gcd(369,917) = 1, the smallest value that u can take is 369. This makes v = 656, w = 917, x = 224, which are all integers. We check that

    y-x = \frac{976u-387w}{45} = \frac{976(369)-387(917)}{45} = 117

    so y = 341 is also an integer. Hence the minimum distance between X and Y is 8\times369 = \boxed{2952\ \mathrm{miles}}.
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  3. #3
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    Re: Pickmeupper!

    YES! Nicely done.

    In case this makes the problem clearer:
    Code:
    1:[X]C@8.............................................................>u[Y]  
    
    2:[X]A@1------------>v<..............................................@8[Y]  
                         
    3:[X]           @8(A,C)...............................>w(A)---------->y[Y]  
    
    4:[X]B@2*****************************>x<..............@8(C)            [Y]
    
    5:[X]                            @8(B,C).............................>y[Y]
    1:C travels to Y in u hours
    2:C comes back and meets A after v hours (v includes the u hours)
    3:A travels with C at 8mph, drops off A after w hours; A continues, C goes back.
    4:C meets B after x hours
    5:B travels with C at 8mph and arrive at Y after y hours, which is at same time as A.
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