If $a = 1,~ b = 10, ~c = 100$, and $d = 1000$, then $(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)$ is equal to
$\textbf{(A) }1111\qquad \textbf{(B) }2222\qquad \textbf{(C) }3333\qquad \textbf{(D) }1212\qquad  \textbf{(E) }4242$
Adding all four of the equations up, we can see that it equals 
\[3(a+b+c+d)\]
This is equal to $3(1111) = \boxed{\textbf{(C) }3333}$ ~awin