= Solution
Write the clock states modulo $M$. The <detailed balance equations> are $p_m\lambda_m=\nu p_{m-1}$. The product condition makes these equations consistent around the circle, so the <stationary distribution> is
$$
\boxed{p_m=\frac{w_m}{\sum_{k=0}^{M-1}w_k},\qquad w_0=1,\qquad w_m=\prod_{j=1}^{m}\frac{\nu}{\lambda_j}.}
$$
Indeed, the rate of the <time reversal of a continuous-time Markov chain> along a clockwise edge is $p_{m+1}\lambda_{m+1}/p_m=\nu$, while its anticlockwise rate is $p_{m-1}\nu/p_m=\lambda_m$. Thus the clock is a <reversible Markov chain>.
The reversed clockwise transitions can be driven by a single independent rate-$\nu$ <Poisson process>, since their rate does not depend on the clock state. They correspond to the original anticlockwise transitions. Consequently \b[the anticlockwise event times form a stationary Poisson process of rate $\nu$]. Its events before a given time are independent of the clock state at that time, which is the relevant <quasireversibility> property. If clockwise and anticlockwise transitions have the same destination, their direction is retained as an event mark when reversing the process.
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