Josephus problem can be solved using the power of 2. Given that there are total of 2^n people standing in a circle, the start position is 1 and skip is 1, then person at position 1 would always be the winner.
in other words: if we write n such that
n = 2^a + l where
l is smaller
2^a is the biggest power of
n; then the winner will be
Theorem: n = 2^a + l, where l < 2a W(n) = 2l + 1
n = 100 n = 64 + 36 (64 is the biggest power of n, and 36 is the reminder) w(n) = 2 * 36 + 1 = 73 (answer)