2005 AMC 10B Problem 13

Between $1$ and $2005$ there are $668$ multiples of $3,$ $501$ multiples of $4,$ and $167$ multiples of $12.$ Every multiple of $12$ is both a multiple of $3$ and of $4,$ so removing them from each group gives $$\begin{aligned} &(668 - 167) \\ &\quad {}+ (501 - 167) = 835 \end{aligned}$$ numbers that are multiples of $3$ or $4$ but not $12.$ Thus, C is the correct answer.