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 "2017 AMC 12A Problems/Problem 12"

m (→‎Problem)
(→‎Solution)
Line 10: Line 10:
  
 
We quickly see that <math>12</math> is the smallest number with at least <math>5</math> positive integer divisors, and that <math>1,2,3,4,6</math> are each numbers of horses. Thus, our answer is <math>1+2=\boxed{\textbf{(B) } 3}</math>.
 
We quickly see that <math>12</math> is the smallest number with at least <math>5</math> positive integer divisors, and that <math>1,2,3,4,6</math> are each numbers of horses. Thus, our answer is <math>1+2=\boxed{\textbf{(B) } 3}</math>.
 +
 +
 +
==Solution 2==
 +
 +
In order for at least 5 horses to finish simultaneously, the current time needs to have at least 5 divisors. Thus the number must have form of either <math>p^4</math> or <math>p^2*q</math>, which have 5 and 6 factors respectively. The smallest number of the first form is <math>16</math>, and the smallest number of the second from is <math>12.</math>  Thus, our answer is <math>1+2=\boxed{\textbf{(B) } 3}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 00:19, 25 January 2019

Problem

There are $10$ horses, named Horse 1, Horse 2, $\ldots$, Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$, in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S = 2520$. Let $T>0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T$?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution

We know that Horse $k$ will be at the starting point after $n$ minutes if $k|n$. Thus, we are looking for the smallest $n$ such that at least $5$ of the numbers $\{1,2,\cdots,10\}$ divide $n$. Thus, $n$ has at least $5$ positive integer divisors.

We quickly see that $12$ is the smallest number with at least $5$ positive integer divisors, and that $1,2,3,4,6$ are each numbers of horses. Thus, our answer is $1+2=\boxed{\textbf{(B) } 3}$.


Solution 2

In order for at least 5 horses to finish simultaneously, the current time needs to have at least 5 divisors. Thus the number must have form of either $p^4$ or $p^2*q$, which have 5 and 6 factors respectively. The smallest number of the first form is $16$, and the smallest number of the second from is $12.$ Thus, our answer is $1+2=\boxed{\textbf{(B) } 3}$.

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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
2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 12 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=2017_AMC_12A_Problems/Problem_12&oldid=100818"