The distance from X to Y is 8

*u* miles.

At the moment C first arrives at Y, A has travelled

*u* miles from X so the separation between them is 7

*u* miles. When C heads towards A, their relative velocity is 8 + 1 = 9 mph; so C meets A in $\displaystyle \tfrac{7u}9$ hours: i.e.

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

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 2

*w* miles from X; C is $\displaystyle v+8(w-v) = 8w-7v$ miles from X. Their separation is therefore $\displaystyle 6w-7v$ miles. When C heads towards B, their relative velocity is 8 + 2 = 10 mph, so C meets B in $\displaystyle \tfrac{6w-7v}{10}$ hours: i.e.

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

The point where they meet is 2

*x* miles from X (distance covered by B in

*x* hours).

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

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

miles from Y. C and B themselves are

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

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

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

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

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

so

*y* = 341 is also an integer. Hence the minimum distance between X and Y is $\displaystyle 8\times369 = \boxed{2952\ \mathrm{miles}}$.