Here is a proof. Unfortunately, the proof is not constructive.The secret of
winning is "1", which is a fator of every integer.
If the first player(player A) can win by removing a number between 2 to n,
then our hypothesis holds. Or else, A can't win by removing any number
between 2 to n. We denote the situation after removing number i from [1, n]
by S(n, i), then for i = 2...n, S(n, i) is a winning situation. A can then
remove number 1 at the first step. No matter what B removes in the next
step, he will leave a situation S(n, i)(i is the number B removes), which is
a winning situation for the next player(A).
2009/10/1 nikhil <nikhilgar...@gmail.com>
>
> we have all the numbers written from 1- n. 2 players play
> alternatively. At any turn , a player removes a number and along with
> all its divisors present in the list. Player to remove last number
> wins.
>
> so given initial number n and player who is starting first , we are to
> find who wins if both play optimum.
>
>
> NOW , i have found that the the player who starts ALWAYS wins. Can
> anyone prove this or still better come up with a real strategy !
>
>
>
> cheers
> -
> nikhil
> Every single person has a slim shady lurking !
>
> >
>
--
此致
敬礼！
林夏祥
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