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 $a^2 \equiv 0, 1 \pmod 3$, 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 $b=3$.