Last one alive x

We can prove this by induction. First, $n=0$ , $X=1$ , clearly, as the number one is the only person, the proccess will never take place.

Now, let's suppose that for $n$ the survivor is the number 1. That is, when initially there are $X=2^n$ people, the one who's left at the final is the $1^\text{st}$ person.

If there are $X=2^{n+1}$ people at the beginning, after one move, all the $2k^\text{th}$ persons will die as each $(2k+1)^\text{th}$ person kills the one who's next. Therefore, after one turn only the half of the initial people will survive, that is $2^n$ people, and the sword will end in $1^\text{st}$ 's hands. We apply our induction hypothesis, thus, $1^\text{st}$ person will be the last one alive.

Note that if there are $X=2^{n}+1$ , the person that survives is the $3^\text{rd}$ one.

Great question.. Its easy to say that in each iteration the number of people standing are reduced to half.. So if the total number of people are 2^n then definitely the 1st man will be standing at the end.. But what if its not a perfect 2^n? This is complex.. Assuming x people standing.. X>2^n for some n.. For every extra man (x-2^n) the last standing man will kill the first one.. So the man standing will definitely be ((x-2^n)*2)+1 Hard to visualise this.. I tried to be clear but logic is one thing that cannot be made clear that easily.. Try the formula for different numbers and check.. All the best(y)