You play a game with a friend with a pile of 24 stones. You take turns taking $2^n$ stones, where $n$ is any non-negative integer (including zero).

The player who takes the last stone wins!

If you go first, is there a strategy that guarantees you a win?

No
Yes

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No matter what you do, your opponent can always leave you with a number of stones that is divisible by three (which, of course, will never equal $2^n$ for any $n$ ), until you are down to only three stones.

At that point, you will take either one or two stones and he/she will take whatever is left (including the last stone!).