There are 3 numbers A, B, and C, such that $1001C - 2002A = 4004$, and $1001B + 3003A = 5005$. What is the average of A, B, and C?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E) }\text{Not uniquely determined}$
Notice that we don't need to find what $A, B,$ and $C$ actually are, just their average. In other words, if we can find $A+B+C$, we will be done. 
Adding up the equations gives $1001(A+B+C)=9009=1001(9)$ so $A+B+C=9$ and the average is $\frac{9}{3}=3$. Our answer is $\boxed{\textbf{(B) }3}$.