= Solution
At $T_{\tau_{k-1}}^{(k)}$, part b and the strong-stationarity assumption make $X^{(k)}/2$ uniform on $\mathbb Z_{2^{k-1}}$ and independent of the stopping index. One further lazy step either stays or moves by one, each parity occurring with probability one half; conditional on the even residue already selected, this chooses uniformly between its two lifts to $\mathbb Z_{2^k}$. Thus
$$
\tau_k=T_{\tau_{k-1}}^{(k)}+1
$$
has a uniform terminal state independent of $\tau_k$, and is a <strong stationary time>.
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