Difference between revisions of "2005 AMC 10A Problems/Problem 22"
Dairyqueenxd (talk | contribs) (→Problem) |
Dairyqueenxd (talk | contribs) (→Solution) |
||
| Line 11: | Line 11: | ||
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> |
==Video Solution== | ==Video Solution== | ||
Latest revision as of 11:55, 14 December 2021
Contents
Problem
Let 
be the set of the 
smallest positive multiples of 
, and let 
be the set of the smallest positive multiples of 
. How many elements are common to and ?
Solution
Since the least common multiple 
, the elements that are common to and must be multiples of 
.
Since 
and 
, several multiples of that are in won't be in , but all multiples of that are in will be in . So we just need to find the number of multiples of that are in .
Since 
, every 
rd element of will be a multiple of .
Therefore the answer is 
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. 
