Two bikers, Bill and Sal, simultaneously set off from one end of a straight road. Neither biker moves at a constant rate, but each continues biking until he reaches one end of the road, at which he instantaneously turns around. When they meet at the opposite end from where they started, Bill has traveled the length of the road eleven times and Sal seven times. Find the number of times the bikers passed each other moving in opposite directions.
Define a pass $(\mathrm{P})$ to be an time when Bill and Sal pass one another moving in opposite directions and a turn (T) to be a time when one of the bikers turns around. If both bikers turn around simultaneously, we may alter their speeds slightly, causing one turn to happen before the other, without affecting the number of passes. We distinguish three states:

A. The bikers are moving in the same direction.

B. The bikers are moving toward each other.

C. The bikers are moving away from one another.

We notice that the bikers can only change state by turning or passing. Moreover, there are only three possible state changes:

- From state A, one biker reaches the end and turns around, creating state B.
- From state B, the bikers pass one another, creating state C.
- From state $\mathrm{C}$, one biker reaches the end and turns around, creating state A.

Since there are no other possibilities and we begin at state A, the states must follow the sequence ABCABC $\cdots \mathrm{ABCA}$ and the turns and passes TPTTPT ... TPT. We are given that Bill makes ten turns and Sal six, or 16 in all. Consequently there are eight passes.