My Python 3.4.1 solution:

def numLeft(people,y):
    return 0 if people == 1 else (numLeft(people-1,y)+y)%people

print (numLeft(100,2) + 1)

The program returns 73 as the answer.

An intuitive, generalised form, that uses just maths, not Comp.Science

Let initial number of people be $n$ , and the power of $2$ just next to $n$ be $2^k$

(For example in this case, $n=100$ and $2^k=128$ )

Then the person who survives is $\biggl(n-(2^k-n)+1\biggr)$

See that in this case, it's $100-(128-100)+1=100-28+1=\boxed{73}$

It is depending on the power of 2 because , if there are $n_k$ people at the end of $k^{th}$ round ,then in $i^{th}$ round, $\lceil \dfrac{n_{i-1}}{2} \rceil$ people will survive. This means, if the number $n$ , if is itself a power of $2$ , then the person who survives at end will be $1$ . This you can check in mind. Everytime, half people die and sword comes to $1$

Thus, the formula holds.

U see that this question can be solved just by imagination. Let us start cutting the people . After one round , 2n-1 name of people lived. Now next round . Starting from 1 , here 4n-3 name people lived. So the one who was left should be of the form 4n-3 . 73 was that and so it was the answer . Now coming to real solution we will continue this process , everytime replacing n with next type of nos. Saved . So it will go next and finally the form will come in which we get 73 as only ans.

Might as well do it in c++.

#include <iostream>
main()
{
    int n, hPower2;
    std::cin >> n;
    hPower2 =1; //Highest power of two
    while( n > (hPower2 << 2))
        hPower2 <<= 2;
    std::cout << 2*(n-hPower2)+1 << "\n";
    return 0;
}

Look ma, no recursion

I did it by eliminating the options. First I eliminated option 98 as all the even numbered persons will die after the first round.
After the second go upto 100.
Then if we see the remaining numbers they form an arithmetic progression. 1,5,9,13,17,21......... Any term in this will be given as a+(n-1) d = nth term
Here a=1 and d=4, So 1+(n-1)4= number Analyzing it we see that the number which is in this A. P. will be a multiple of 4 after 1 is deducted from it. Only 73 is the option which meet this criteria, hence the answer is 73.

I don't know how to explain the mental math solution.

I guess you should write all of the odd numbers on a paper since it started with no.1 (which is an odd number.)

Then continue cancelling numbers accordingly.

Not sure what's the formula for this without programming..