Question 219, Find the number of subsets B of the set $\{1,2, \ldots, 2005\}$, where B satisfies that the sum of its elements has a remainder of 2006 when divided by 2048.
Question 219, Answer: Let $A=\left\{2^{0}, 2^{1}, 2^{2}, \ldots, 2^{10}\right\}$. Note that, the sum of elements of the subsets of $A$ exactly traverses: 1, 2, ... 2048, which means it exactly traverses the complete residue system modulo 2048.

Let $S=\{1,2, \ldots, 2005\}$, then for any $C \subseteq M \backslash A$, there exists a unique $D \subseteq A$ such that the sum of elements of $C \cup D$ is congruent to 2006 modulo 2048. Therefore, the number of $B$ that satisfy the given condition is $2^{|M \backslash A|}=2^{1994}$.