One hundred in a circle

Idea. With a power of 2 people, the person who starts with the sword survives. So remove people until the number left is a power of 2, and see who holds the sword then.

Step 1: powers of 2. With $2^m$ people and person 1 holding the sword, one trip around the circle eliminates every second person: $2,4,\dots ,2^m$. The last of them is eliminated by person $2^m-1$, who hands the sword back to person 1. Now $2^{m-1}$ people remain and person 1 starts again. Repeating, person 1 is the survivor.

Step 2: reach 64. The largest power of 2 below 100 is $64=100-36$. After 36 eliminations, those of persons $2,4,\dots ,72$, there are 64 people left and the sword is in the hands of person 73.

Step 3. From here the game is a circle of 64 people starting with person 73, so by Step 1 person 73 survives.

In general. Write $n=2^m+l$ with $0\le l<2^m$. The survivor is $2l+1$. For $n=100$ this gives $2\cdot 36+1=73$.