1. A model car is tested on some closed circuits that are 600 meters long, consisting of flat sections and uphill or downhill sections. All the uphill and downhill sections have the same gradient. The tests highlight some curious facts:

(a) the speed of the model car depends only on whether the car is on an uphill, flat, or downhill section; calling these three speeds $v_{s}, v_{p}$, and $v_{d}$, respectively, we have $v_{s}<v_{p}<v_{d}$;

(b) $v_{s}, v_{p}$, and $v_{d}$, expressed in meters per second, are integers;

(c) regardless of the composition of the circuit (with more or fewer uphill and downhill sections), the travel time is always 50 seconds.

Find all possible values of $v_{s}, v_{p}$, and $v_{d}$.
Solution: We indicate all lengths in meters and all times in seconds, omitting the units of measurement.

The possible triples of speeds are $\left(v_{s}, v_{p}, v_{d}\right)=(10,12,15),(9,12,18),(8,12,24)$ and $(7,12,42)$.

First, consider a completely flat circuit 600 long. Hypothesis (c) then gives $50=\frac{600}{v_{p}}$, from which $v_{p}=\frac{600}{50}=12$.

Next, consider a generic circuit of length 600, where the uphill section is $s$ long, the flat section is $p$ long, and the downhill section is $d$ long.

By construction, $s+p+d=600$; moreover, for the circuit to close, the elevation gain from the uphill sections must equal the elevation loss from the downhill sections: since the slopes are the same for all non-flat sections, there must be as much uphill as downhill, i.e., $s=d$.

The flat section is then $p=600-s-d=600-2 s$, and the time taken by the car to complete the circuit is $\frac{s}{v_{s}}+\frac{p}{v_{p}}+\frac{d}{v_{d}}=\frac{s}{v_{s}}+\frac{600-2 s}{12}+\frac{s}{v_{d}}=s\left(\frac{1}{v_{s}}+\frac{1}{v_{d}}-\frac{1}{6}\right)+\frac{600}{12}$, which equals $50=\frac{600}{12}$ if and only if $s\left(\frac{1}{v_{s}}+\frac{1}{v_{d}}-\frac{1}{6}\right)=0$.

The speeds then satisfy condition (c) of the problem if and only if the preceding relation holds for any closed circuit, i.e., for any choice of $s$ (between 0 and 300).

A necessary and sufficient condition on the speeds is therefore $\frac{1}{v_{s}}+\frac{1}{v_{d}}=\frac{1}{6}$; $v_{s}$ and $v_{d}$ are both non-zero, so we can rewrite this equation as $v_{s} v_{d}=6\left(v_{s}+v_{d}\right)$, or $\left(v_{s}-6\right)\left(v_{d}-6\right)=36$.

$v_{d}$ is an integer greater than $v_{p}=12$, so $v_{d}-6$ is a divisor of 36 strictly greater than 6, i.e., it is one of 9, 12, 18, or 36 (which correspond respectively to the values 15, 18, 24, 42 for $v_{d}$).

Solving the equation for $v_{s}$, we get $v_{s}=6+\frac{36}{v_{d}-6}$, or $v_{s}=10,9,8,7$.

SECOND SOLUTION: Once we have the equation $\frac{1}{v_{s}}+\frac{1}{v_{d}}=\frac{1}{6}$, it is sufficient to observe that $v_{s}<v_{p}=12$, and on the other hand, we cannot have $v_{s} \leq 6$, because otherwise $\frac{1}{v_{s}} \geq \frac{1}{6}$, from which $\frac{1}{v_{d}} \leq 0$, which is clearly absurd. It follows that the only possibilities for $v_{s}$ are 7, 8, 9, 10, 11, and it is easy to verify which of these lead to integer values for $v_{d}$.