Who survives at the end ?

We see that when $n$ is even the sword goes back to the Person 1 after 1 iteration When $n$ is odd the sword goes to the Person n after 1 iteration. In general, at any stage, if number of person is $n$ and sword is with Person $p$ then if $n$ is even, sword will remain with Person $p$ after 1 iteration and number of person will reduce to $\large\frac{n}{2}$ and if $n$ is odd, sword will goes to Person $p-1$ after 1 iteration and number of person reduces to $\large\frac{n+1}{2}$ .

On this basis, we conclude that at any stage, if the number of persons is a power of 2 and sword is with Person $p$ , then at the end Person $p$ will survive.

Solving in general,

If $n$ are persons are there, we have to find the value of integral $k$ such that $2^k \leq n < 2^{k+1}$ . Now we have to find $r$ such that $n = 2^k + r$ where $r$ will range from 0 to $2^k - 1$

$\hspace{15pt} r = n - 2^k\newline \Rightarrow 2r = 2n - 2^{k+1} \newline\Rightarrow 2r = n + n - 2^{k+1} \leq n-1\hspace{20pt} [\because n < 2^{k+1} \Rightarrow n - 2^{k+1} < 0 \Rightarrow n - 2^{k+1} \leq -1]$

This last equation assures that it is possible to kill $r$ persons in $1^{st}$ iteration starting form Person 1 to get the number of person as $2^k$ that is the power of 2.

Since Person 1 kills Person 2, Person 3 kills Person 4 and so on, so $r^{th}$ person to die is the Person $2r$ who is kiiled by the Person $2r - 1$ and this Person after killing Person $2r$ gives the sword to Person $2r + 1\hspace{20pt} [$ Person $2r + 1$ exists because $2r + 1 \leq n]$

Now, sword is with Person $2r + 1$ and there are $2^k$ persons left. Hence at the end Person $2r + 1$ will survive.

Now finding the value of $r$ in terms of $n$

We have,

$\hspace{13pt} 2^k \leq n < 2^{k+1} \newline\Rightarrow k \leq \text{log}_2 n < k + 1\newline \Rightarrow k = \lfloor \text{log}_2 n \rfloor$

Now $r = n - 2^{\lfloor \text{log}_2 n \rfloor}$

Hence at the end, Person $2(n - 2^{\lfloor \text{log}_2 n \rfloor}) + 1$ will survive.

In the question, $n$ is given $10052020$ .

$\Rightarrow \text{log}_2 n = \text{log}_2 10052020 = 23.261 \newline \Rightarrow \lfloor \text{log}_2 n \rfloor = 23 \newline \Rightarrow 2^{\lfloor \text{log}_2 n \rfloor} = 2^{23} = 8388608 \newline \Rightarrow n - 2^{\lfloor \text{log}_2 n \rfloor} = 10052020 - 8388608 = 1663412 \newline \Rightarrow 2(n - 2^{\lfloor \text{log}_2 n \rfloor}) = 2\cdot 1663412 = 3326824 \newline \Rightarrow 2(n - 2^{\lfloor \text{log}_2 n \rfloor}) + 1 = 3326824 + 1 = 3326825$

Hence at the end Person 3326825 will survive.