Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution)
Line 9: Line 9:
 
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>.  
  
−
Since <math>4\cdot3=12</math> every <math>3</math>rd element of <math>S</math> will be a multiple of <math>12</math>
+
Since <math>4\cdot3=12</math> every <math>3</math>rd element of <math>S</math> will be a multiple of <math>12</math>.
  
 
Therefore the answer is <math>\left \lfloor\frac{2005}{3} \right \rfloor=\boxed{\textbf{(D)}  668}</math>
 
Therefore the answer is <math>\left \lfloor\frac{2005}{3} \right \rfloor=\boxed{\textbf{(D)}  668}</math>

Revision as of 14:09, 26 January 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 $T$?

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

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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_22&oldid=143317"