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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 $T$?

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

Solution 1

Since the least common multiple of $4$ and $6$ is $12$, the elements that are common to $S$ and $T$ are all multiples of $12$. Moreover, as the largest element of $S$ is $4 \cdot 2005$, while that of $T$ is $6 \cdot 2005$, which is larger, several of the multiples of $12$ that are in $T$ will not be in $S$, whereas all the multiples of $12$ that are in $S$ will be in $T$.

Thus we only need to find the number of multiples of $12$ that are in $S$, and so we observe that as $4 \cdot 3 = 12$, these multiples of $12$ are precisely every $3$rd element of $S$. It follows that there are $\left\lfloor\frac{2005}{3}\right \rfloor = \boxed{\textbf{(D) } 668}$ such elements.

Solution 2

As in Solution $1$, we find that the elements common to $S$ and $T$ are precisely the multiples of $12$. As $S$ has exactly $2005$ elements, these must range from $4 \cdot 1 = 4$ to $4 \cdot 2005 = 8020$, and similarly the elements of $T$ range from $6 \cdot 1 = 6$ to $6 \cdot 2005 = 12030$. This means any element $n \in S \cap T$ must satisfy both $4 \leq n \leq 8020$ and $6 \leq n \leq 12030$, which reduces to simply $6 \leq n \leq 8020$.

Accordingly, as $12 \cdot 0 = 0 < 6 < 12 \cdot 1 = 12$ and $12 \cdot 668 = 8016 < 8020 < 12 \cdot 669 = 8028$, the multiples of $12$ in the required interval range from $12 \cdot 1$ to $12 \cdot 668$, so there are precisely $\boxed{\textbf{(D) } 668}$ of them.

Video Solution

https://youtu.be/D6tjMlXd_0U

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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