$\textbf{Circular Killing}$
One hundred prisoners on the death row are ordered to stand in a circle and are numbered from $1$ to $100$ in sequence. The king then gives a sword to No $1$. No 1 kills the No $2$ and passes the sword to No $3$. The No $3$ then kills No $4$ and passes the sword to the next person alive, i.e. No $5$. All people continuously does the same until only one person survives. Who is the last survivor?
$\textbf{Answer}$
The person numbered $73$.
$\textbf{Analysis}$
An initial thought is that one has to be always positioned as an odd number in order to survive. For example, in the first round, all even numbered prisoners are killed. Afterwards, some odd numbered prisoners become even number positioned, such as the original No $3$ now stands on the $2^{nd}$ position, and No $7$ stands on the $4^{th}$ position, etc. These people will be killed in the second round. Afterwards, in the third round, the original No $5$ will be killed because he now stands on the $2^{nd}$ position.
After two rounds of killing, there are $25$ prisoners left which is an odd number. Hence, the last one will survive and the original No $1$ will be killed in the next round. This situation will make our tracking difficult because the numbering system shifts as a result of No $1$'s death.
This leads us to investigate under which condition the numbering system will stay the same, i.e. the No $1$ will never be killed which can always serve as the anchor for numbering. More importantly, in such a case, No $1$ will be the last survivor!
In order for the No $1$ to always survive, the number of surviving prisoners must always be even after each round of killing. This means that the number of living people must be a power of $2$. We can quickly verify that when the initial number of prisoners are $2$, $4$, $8$, $16$, etc, the first person always survives and will be the last survivor.
Now, given the initial number is $100$ which is not a power of $2$, we need to find out when the number of surviving person is a power of $2$. The largest power of $2$, not exceeding $100$, is $64$. Hence, the moment when $100-64=36$ people have been killed is a magic moment. Because there are $64=2^6$ people left, if we reset the numbers and start counting at this moment, every round of killing will leave an even number of survivors $$64\rightarrow 32\rightarrow 16\rightarrow 8\rightarrow 4\rightarrow 2\rightarrow 1$$
Now it is clear whoever stands as the new No $1$ at that moment is the last survivor. By the rule, the $36^{th}$ victim must be No $72$. Hence, the last survivor is No $73$.