imc

To find the multiples of $3$ or $4$ but not $12$, you need to find the number of multiples of $3$ and $4$, and then subtract twice the number of multiples of $12$, because you overcount and do not want to include them. The multiples of $3$ are $\frac{2005}{3}=668R1.$ The multiples of $4$ are $\frac{2005}{4}=501R1$. The multiples of $12$ are $\frac{2005}{12}=167R1.$ So, the answer is $668+501-167-167=\boxed{835}$