# 2005 AMC 10A Problems/Problem 22

## Problem

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 $T$? $\mathrm{(A) \ } 166\qquad \mathrm{(B) \ } 333\qquad \mathrm{(C) \ } 500\qquad \mathrm{(D) \ } 668\qquad \mathrm{(E) \ } 1001$

## Solution

Since the least common multiple $\mathrm{lcm}(4,6)=12$, the elements that are common to $S$ and $T$ must be multiples of $12$.

Since $4\cdot2005=8020$ and $6\cdot2005=12030$, several multiples of $12$ that are in $T$ won't be in $S$, but all multiples of $12$ that are in $S$ will be in $T$. So we just need to find the number of multiples of $12$ that are in $S$.

Since $4\cdot3=12$ every $3$rd element of $S$ will be a multiple of $12$

Therefore the answer is $\left \lfloor\frac{2005}{3} \right \rfloor=\boxed{\textbf{(D)} 668}$

## Video Solution

CHECK OUT Video Solution: https://youtu.be/D6tjMlXd_0U

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 