2005 AMC 10A Problem 22

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 $\boxed{\textbf{(D)}\ 668}$.

"The $N$ smallest multiples of $m$" is the interval $[m, mN]$, so intersecting two such sets means intersecting two intervals of multiples; the shorter interval decides the count. Translate "first $2005$" into an explicit upper bound before counting, then count multiples of the lcm below that bound.

• Counting multiples of 12 up to 12030 (the top of T) to get 1002, or using 2005/6 = 334 from the wrong ratio.
• Dividing $2005$ by $12$ to get $167$ (choice (A) is nearby), as if the sets contained the integers $1$ to $2005$.