2005 AMC 10A Problem 22

Below is the professionally curated solution for Problem 22 of the 2005 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10A solutions, or check the answer key.

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Concepts:least common multiple multiple counting integers in a range

Difficulty rating: 1690

22.

Let $S$ be the set of the $2005$ smallest positive multiples of $4,$ and let $T$ be the set of the $2005$ smallest positive multiples of $6.$ How many elements are common to $S$ and $T?$

$166$

$333$

$500$

$668$

$1001$

Solution:

The elements common to $S$ and $T$ are the multiples of $\operatorname{lcm}(4,6) = 12.$ Now $S$ contains multiples of $4$ up to $4 \cdot 2005 = 8020,$ while $T$ reaches up to $6 \cdot 2005 = 12{,}030,$ so the common elements are the multiples of $12$ not exceeding $8020.$ There are $\left\lfloor \dfrac{8020}{12} \right\rfloor = 668$ of them.

Thus, the correct answer is D.

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