Winter Mathematical Competition

Consider a band of length $n$. It is clear that no move is possible for $n=1,2,3$. For $4\le n\le 7=3+2+2$ Alexander has a winning move. For $n=8$ and $9$ after the first move of Alexander the largest length is between $4$ and $7$ and then Denitza has a winning move. Similarly, for $10\le n\le 9+8+8=25$, Alexander has a winning strategy since he can cut the band in such a way that the largest length is either $8$ or $9$ and so on. It follows by induction that Denitza has a winning strategy if and only if $n=3^k$ or $n=3^k-1$ for some integer $k>1$.

The numbers $3^k$ and $3^2-1=2^3$ obviously have the desired form. We shall prove that there are no other solutions. Assume that $a^b=3^k-1$. Since $a^2\equiv 0,1(\mathrm{mod}3)$, the number $b$ is even. Then $(a+1)(a^{b-1}+\cdots +a+1)=3^k$ and we

have $a+1=3^i$, and $3^{k-i}=a^{b-1}+\cdots +a+1=A(a+1)+b$, for some integers $i$ and $k$, $0<i<k$. Hence 3 divides $b$ and setting $c=a^{b/3}$ we have $3^k=(c+1)((c+1{)}^2-3c)$. Then $c+1=3^j$ and $(c+1{)}^2-3c=3^{k-j}$, $0<j<k$. In particular, 9 divides $(c+1{)}^2$, but does not divide $3c$. We consecutively find $k-j=1$, $c=2$, $a=k=2$ and $b=3$.