2005 AMC 10B Problem 13

1. Count each kind of multiple from $1$ to $2005$.
- Multiples of $3$: $2005 \div 3 = 668$ remainder $1$, so there are $668$.
- Multiples of $4$: $2005 \div 4 = 501$ remainder $1$, so there are $501$.
- Multiples of $12$: $2005 \div 12 = 167$ remainder $1$, so there are $167$.
2. Where are the multiples of $12$? A multiple of $12$, like $24$, is a multiple of $3$ and a multiple of $4$. So it was counted once in the $668$ and once more in the $501$.
3. Remove them. We want no multiples of $12$ at all. Each one was counted twice, so subtract $167$ twice:
$$668 + 501 - 167 - 167 = 1169 - 334 = 835.$$

The answer is $\boxed{\textbf{(C)}\ 835}$.

Picture two overlapping circles: multiples of $3$ and multiples of $4$. The overlap is the multiples of $12$. Adding the two circles counts the overlap twice. For "$3$ or $4$" you subtract it once, but for "$3$ or $4$ but not $12$" you want the overlap gone completely, so you subtract it twice.

Look at blocks of $12$. From $1$ to $12$, the good numbers are $3, 4, 6, 8, 9$: five of them ($12$ itself is left out). Every block of $12$ numbers repeats this pattern. There are $167$ full blocks up to $2004$, giving $167 \cdot 5 = 835$. The last number, $2005$, is not a multiple of $3$ or $4$, so it adds nothing.

Count each kind of multiple up to $2005$:
$$\left\lfloor \frac{2005}{3} \right\rfloor = 668, \qquad \left\lfloor \frac{2005}{4} \right\rfloor = 501, \qquad \left\lfloor \frac{2005}{12} \right\rfloor = 167.$$

A multiple of $12$ is a multiple of both $3$ and $4$, so it appears in both of the first two counts. We want such numbers removed completely, so we must take them out twice:
$$668 + 501 - 2 \cdot 167 = 1169 - 334 = 835.$$

The answer is $\boxed{\textbf{(C)}\ 835}$.