2005 AMC 10A Problem 22

A number is in both sets only if it is a multiple of both $4$ and $6$. The smallest number that is a multiple of both is $12$, so the common numbers are multiples of $12$. The question is how far up they can go.

1. Where does each set stop?
- $S$ is $4, 8, 12, \ldots$, and its last number is $4 \times 2005 = 8020$.
- $T$ is $6, 12, 18, \ldots$, and its last number is $6 \times 2005 = 12030$.
2. Which multiples of $12$ are in both? A multiple of $12$ up to $8020$ is a multiple of $4$ no bigger than $8020$, so it is in $S$. It is also a multiple of $6$ no bigger than $12030$, so it is in $T$. A multiple of $12$ above $8020$ is not in $S$. So the smaller stopping point, $8020$, is the one that matters.
3. Count multiples of $12$ up to $8020$. $8020 \div 12 = 668$ with remainder $4$, since $12 \times 668 = 8016$. So there are $668$ of them.

Another way to see step 3: in $S$, the number $4k$ is a multiple of $12$ exactly when $k$ is a multiple of $3$. Among $k = 1, 2, \ldots, 2005$ there are $668$ multiples of $3$, because $3 \times 668 = 2004$.

The answer is $\boxed{\textbf{(D)}\ 668}$.

"The $2005$ smallest multiples of $4$" just means "multiples of $4$ from $4$ up to $8020$." Turn each set into a range with a clear top, then count the shared multiples (multiples of the least common multiple) up to the lower top. The lower top always decides the count.

A number lies in both sets only if it is a multiple of both $4$ and $6$, i.e. a multiple of $\operatorname{lcm}(4,6) = 12$.

Now find how far each set reaches. $S = \{4, 8, \dots, 4 \cdot 2005\}$ ends at $8020$; $T = \{6, 12, \dots, 6 \cdot 2005\}$ ends at $12030$. Every multiple of $12$ up to $8020$ is a multiple of $4$ below $S$'s cap and a multiple of $6$ below $T$'s cap, so it is in both sets; multiples of $12$ beyond $8020$ are missing from $S$.

So the common elements are the multiples of $12$ up to $8020$:
$$\left\lfloor \frac{8020}{12} \right\rfloor = 668 \quad (12 \cdot 668 = 8016).$$
Equivalently, every third element of $S$ ($4k$ with $3 \mid k$) is a multiple of $12$, and $\lfloor 2005/3 \rfloor = 668$.

The answer is $\boxed{\textbf{(D)}\ 668}$.