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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> \textbf{(A) } 166\qquad \textbf{(B) } 333\qquad \textbf{(C) } 500\qquad \textbf{(D) } 668\qquad \textbf{(E) } 1001 </math>
  
 
==Solution==
 
==Solution==
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Since <math>lcm(4,6)=12</math>, the elements that are common to <math>S</math> and <math>T</math> must be multiples of <math>12</math>.  
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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>.  
  
 
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>.  
 
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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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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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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Therefore the answer is <math>\lfloor\frac{2005}{3}\rfloor=668\Rightarrow D</math>
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Therefore the answer is <math>\left \lfloor\frac{2005}{3} \right \rfloor=\boxed{\textbf{(D) }  668}</math>
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==Video Solution==
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CHECK OUT Video Solution: https://youtu.be/D6tjMlXd_0U
  
 
==See Also==
 
==See Also==
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*[[2005 AMC 10A Problems]]
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{{AMC10 box|year=2005|ab=A|num-b=21|num-a=23}}
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*[[2005 AMC 10A Problems/Problem 21|Previous Problem]]
 
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*[[2005 AMC 10A Problems/Problem 23|Next Problem]]
 
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 11:55, 14 December 2021

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

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

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AMC 10 Problems and Solutions

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