At each of the given $n$ points on a circle, either $+1$ or $-1$ is written. The following operation is performed: between any two consecutive numbers on the circle their product is written, and the initial $n$ numbers are deleted. Suppose that, for any initial arrangement of $+1$ and $-1$ on the circle, after finitely many operations all the numbers on the circle will be equal to $+1$. Prove that $n$ is a power of two.
Assume $f(n)$ to be the last remaining number .
Note that if all the numbers on the circle will be $1$ then $ f(n)=1.$
If there exists only $1$ negative $1$ and $(n-1)1's$ then $f(n)=-1$
The question demands $f(n)=1 $ iff $n=2^r$ for any natural number $r$
Now if $N=2^k \cdot r$ for $2^k||n$, then let the numbers on the circle be $[1,-1,1...,-1]$
And after transformations we obtain: $[1,-1,1...,-1]->[-1,-1..]->[1,1,1..1]->....->[-1,1,1,1]..->[-1]$ so here $f(n)=-1 \neq1$
Hence n will have to be a power of 2.(one small error:2 doesn't work since [1,-1]=-1 so the question should read "At each of the given n points on a circle, either +1 or -1 is written. The following operation is performed: between any two consecutive numbers on the circle their product is written, and the initial n numbers are deleted. Suppose that, for any initial arrangement of +1 and -1 on the circle, after finitely many operations all the numbers on the circle will be equal to +1. Prove that n is a power of two greater than 2)