$2^n$ Stones

Logic Level 3

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

1 solution

Geoff Pilling
Jun 12, 2017

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!).

I can start with only 1,2 or 4.If I start 16 or 8 then opponent will take 8 or 16 respectively i.e. oppoent will win.

Kushal Bose - 3 years, 12 months ago

I think no matter what you take, you lose eventually, no?

e.g.

If you take 1, your opponent takes 2 and you are once again left with a number that is divisible by 3.

If you take 2, your opponent takes 1 and you are once again left with a number that is divisible by 3.

If you take 4, your opponent takes 2 and you are once again left with a number that is divisible by 3.

If you take 8, your opponent takes 16 and wins!

If you take 16, your opponent takes 8 and wins!

Similarly, all the way down to 0, where he has taken the last one and won!

Geoff Pilling - 3 years, 12 months ago

yes I am saying that we can not take 8 or 16 then first player will be easily defeated in two steps

Kushal Bose - 3 years, 12 months ago

2 n 2^n 2 n Stones

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$2^n$ Stones