What would happen in the variant of the Josephus problem in which people in odd-numbered positions are killed in each round?
Position $2^{\lfloor \log_2 N \rfloor}$ is always safe since it is the position that maximizes the number of trailing zeros in its binary representation.
Claim: The person originally sitting in position $n$ is killed in the $i$-th round iff the binary representation of $n$ has exactly $i-1$ trailing zeroes.
Proof: By strong induction on $i$. If $i=1$ the claim is trivially true since $n$ has $0$ trailing zeroes iff $n$ is odd.
If $i &gt; 1$, by induction hypothesis we have that all remaining people were originally sitting in positions whose binary representation has $i$ or more trailing zeros, i.e, those positions are exactly the integers in $S = \{1, \dots, N\}$ that are multiples of $2^i$.
If $n=2^i k$ then there are exactly $k$ positions in $S$ smaller than or equal to $n$, i.e., $n$ is killed iff $k$ is odd.