, 2022

Solution:

Suppose that Candice starts driving at $n$ miles per hour. Then she slows down and drives $(n-1)$ mph, $(n-2)$ mph, and so on, with her last speed being $(m+1)$ mph.

Then the total distance traveled is
$$\begin{aligned} \frac{n}{60} + \frac{n-1}{60} + \cdots + \frac{m+1}{60} &= \frac{1}{60}\left(\frac{n(n+1)}{2} - \frac{m(m+1)}{2}\right) \\ &= \frac{n^2 + n - m^2 - m}{120} \\ &= \frac{(n+m+1)(n-m)}{120} \end{aligned}$$
Since the total distance travelled is $2/3$, we have $(n+m+1)(n-m) = 120 \cdot 2/3 = 80$. We know $m$ is nonnegative since Candice's speed is always positive, so $n+m+1 > n-m$. Thus, $n+m+1$ and $n-m$ are a factor pair of $80$, with $n+m+1$ greater and $n-m$ smaller. Since one is even and one is odd, this means we either have $(n+m+1, n-m) = (80, 1)$ or $(16, 5)$. The first case is impossible since it gives $n-m=1$, which would imply that Candice drives at $n$ mph the whole way home. Therefore, $(n+m+1, n-m) = (16, 5)$. Since $n-m=5$, she gets home at 5:05 pm.