Good job !!
just to add , i think the exact winning strategy would be based on
distribution of factors of numbers and that would depend on
distribution of primes...
So it occurs to me that there might not be a "closed form" winning
strategy possible !

Still thanks for the proof !


cheers

-
nikhil
Every single person has a slim shady lurking.

On Oct 2, 7:08 am, saltycookie <saltycoo...@gmail.com> wrote:
> 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).
>
> On 10月1日, 上午2时53分, nikhil <nikhilgar...@gmail.com> wrote:
>
> > 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 !

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