Beer Signs on the Highway

Smith drove at a steady clip along the highway, his wife beside him.
"Have you noticed," he said, "that those annoying signs for Flatz beer seem to be regularly spaced along the road? I wonder how far apart they are."

Mrs. Smith glanced at her wrist watch, then counted the number of Flatz beer signs they passed in one minute.

"What an odd coincidence!" exclaimed Smith. "When you multiply that number by ten, it exactly equals the speed of our car in miles per hour."

Assuming that the car's speed is constant, that the signs are equally spaced and that Mrs. Smith's minute began and ended with the car midway between two signs, how far is it between one sign and the next?

(Source: My Best Mathematical and Logic Puzzles by Martin Gardner)

I solved this a little bit differently from how it's solved in the back of the book. Consider the equation:

\[ 10s = \frac{d}{t} \]

Where \(s\) is the number of signs encountered, \(d\) is the distance traveled (in miles), and \(t\) is the time elapsed (in hours). The right-hand side of the equation is simply derived from the old:

\[ d = rt \]

Or, distance equal rate times time, that one learned in middle school, solved for \(r\) and placed in the right-hand side of the first equation. All told, the first equation just says: the number of signs times ten is equal to the speed in miles per hour.

It's also known that one minute has passed, or one-sixtieth of an hour in the terms used here. With this fact, more can be figured out:

\begin{align*}

10s &= \frac{d}{\frac{1}{60}}\\
10s &= 60d \\
s &= 6d
\end{align*}

Conceptually, \(s\) can only have integer solutions, which constrains \(d\), and thus the speed of the car, but not uniquely. As stated by Gardner in the back of the book, the exact speed doesn't matter. But let's say the Smiths are going at a brisk clip of 60 mph for convenience. \(d\) is then one; \(s\), six. This means that the signs are spaced one-sixth of a mile apart. The same can be figured out from any rational value of \(d\) that leaves \(s\) an integer. (I understand this might also be constrained further by special relativity but there are still many possible solutions even then.)