1. Pickmeupper!

X(A,B,C).......................................... ......................Y
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.

2. 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 $8\times369 = \boxed{2952\ \mathrm{miles}}$.

3. 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.