9,10,11 |

Authors: Agakhanov N.Kh., Bogdanov I.I.

In how many ways can the numbers $2^{0}, 2^{1}, 2^{2}, \ldots, 2^{2005}$ be divided into two non-empty sets $A$ and $B$ such that the equation $x^{2}-S(A) x+S(B)=0$, where $S(M)$ is the sum of the numbers in set $M$, has an integer root?
If $x_{1} \leq x_{2}$ are the roots of the equation, then $x_{1}, x_{2} \in \mathbf{N}$ and $x_{1}+x_{2}=S(A), x_{1} x_{2}=S(B)$, therefore $\left(x_{1}+1\right)\left(x_{2}+1\right)=S(B)+$ $S(A)+1=1+2+4+\ldots+2^{2005}+1=2^{2006}$. Hence, $x_{1}+1=2^{k}, x_{2}+1=2^{2006-k}$, where $k$ can take values $1,2, \ldots, 1003$.

Conversely, let $x_{1}, x_{2}$ be numbers of this form. Then they are the roots of the equation $x^{2}-p x+q=0$, where $p=2^{k}+$ $2^{2006-k}-2, q=2^{2006}-1-p$.

However, the number $p$ has a unique decomposition into a sum of different powers of two, and in this decomposition, the powers of two do not exceed $2^{2005}$, while the binary representation of $q$ contains 1s in the positions where the number $p$ has zeros, since $p$ $+q=2^{2006}-1$. Thus, for each $k$ from 1 to 2003, there is a unique partition ( $A, B$ ) that gives the specified roots $x_{1}$ and $x_{2}$.

## Answer

1003 ways.