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 $\lfloor\frac{2005}{3}\rfloor=668\Longrightarrow \mathrm{(D)}$

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2005 AMC 10A Problems/Problem 22

Problem

Solution

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