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Difference between revisions of "2005 AMC 10A Problems/Problem 22"

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==Problem==
 
==Problem==
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Let <math>S</math> be the [[set]] of the <math>2005</math> smallest positive multiples of <math>4</math>, and let <math>T</math> be the set of the <math>2005</math> smallest positive multiples of <math>6</math>. How many elements are common to <math>S</math> and <math>T</math>?
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Let <math>S</math> be the set of the <math>2005</math> smallest positive multiples of <math>4</math>, and let <math>T</math> be the set of the <math>2005</math> smallest positive multiples of <math>6</math>. How many elements are common to <math>S</math> and <math>T</math>?
  
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<math> \mathrm{(A) \ } 166\qquad \mathrm{(B) \ } 333\qquad \mathrm{(C) \ } 500\qquad \mathrm{(D) \ } 668\qquad \mathrm{(E) \ } 1001 </math>
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<math>
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\textbf{(A) } 166\qquad \textbf{(B) } 333\qquad \textbf{(C) } 500\qquad \textbf{(D) } 668\qquad \textbf{(E) } 1001
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</math>
  
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==Solution==
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==Solution 1==
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Since the [[least common multiple]] <math>\mathrm{lcm}(4,6)=12</math>, the [[element]]s that are common to <math>S</math> and <math>T</math> must be [[multiple]]s of <math>12</math>.  
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Since the least common multiple of <math>4</math> and <math>6</math> is <math>12</math>, the elements that are common to <math>S</math> and <math>T</math> are all multiples of <math>12</math>. Moreover, as the largest element of <math>S</math> is <math>4 \cdot 2005</math>, while that of <math>T</math> is <math>6 \cdot 2005</math>, which is larger, several of the multiples of <math>12</math> that are in <math>T</math> will not be in <math>S</math>, whereas all the multiples of <math>12</math> that are in <math>S</math> will be in <math>T</math>.
  
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Since <math>4\cdot2005=8020</math> and <math>6\cdot2005=12030</math>, several multiples of <math>12</math> that are in <math>T</math> won't be in <math>S</math>, but all multiples of <math>12</math> that are in <math>S</math> will be in <math>T</math>. So we just need to find the number of multiples of <math>12</math> that are in <math>S</math>.  
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Thus we only need to find the number of multiples of <math>12</math> that are in <math>S</math>, and so we observe that as <math>4 \cdot 3 = 12</math>, these multiples of <math>12</math> are precisely every <math>3</math>rd element of <math>S</math>. It follows that there are <math>\left\lfloor\frac{2005}{3}\right \rfloor = \boxed{\textbf{(D) } 668}</math> such elements.
  
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Since <math>4\cdot3=12</math> every <math>3</math>rd element of <math>S</math> will be a multiple of <math>12</math>
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==Solution 2==
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As in Solution <math>1</math>, we find that the elements common to <math>S</math> and <math>T</math> are precisely the multiples of <math>12</math>. As <math>S</math> has exactly <math>2005</math> elements, these must range from <math>4 \cdot 1 = 4</math> to <math>4 \cdot 2005 = 8020</math>, and similarly the elements of <math>T</math> range from <math>6 \cdot 1 = 6</math> to <math>6 \cdot 2005 = 12030</math>. This means any element <math>n \in S \cap T</math> must satisfy both <math>4 \leq n \leq 8020</math> and <math>6 \leq n \leq 12030</math>, which reduces to simply <math>6 \leq n \leq 8020</math>.
  
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Therefore the answer is <math>\lfloor\frac{2005}{3}\rfloor=\boxed{\textbf{(D)} 668}</math>
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Accordingly, as <math>12 \cdot 0 = 0 < 6 < 12 \cdot 1 = 12</math> and <math>12 \cdot 668 = 8016 < 8020 < 12 \cdot 669 = 8028</math>, the multiples of <math>12</math> in the required interval range from <math>12 \cdot 1</math> to <math>12 \cdot 668</math>, so there are precisely <math>\boxed{\textbf{(D) } 668}</math> of them.
  
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~ LaTeX edits by dolphin7
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==Video Solution==
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https://youtu.be/D6tjMlXd_0U
  
 
==See Also==
 
==See Also==

Latest revision as of 02:19, 2 July 2025

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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