Difference between revisions of "2017 AMC 12A Problems/Problem 12"
Thomas0115 (talk | contribs) (→See Also) |
m (→Solution) |
||
| Line 9: | Line 9: | ||
We know that Horse <math>k</math> will be at the starting point after <math>n</math> minutes iff <math>k|n</math>. Thus, we are looking for the smallest <math>n</math> such that at least <math>5</math> of the numbers <math>\{1,2,\cdots,10\}</math> divide <math>n</math>. Thus, <math>n</math> has at least <math>5</math> positive integer divisors. | We know that Horse <math>k</math> will be at the starting point after <math>n</math> minutes iff <math>k|n</math>. Thus, we are looking for the smallest <math>n</math> such that at least <math>5</math> of the numbers <math>\{1,2,\cdots,10\}</math> divide <math>n</math>. Thus, <math>n</math> has at least <math>5</math> positive integer divisors. | ||
| − | 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{(B) | + | 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>. |
== See Also == | == See Also == | ||
{{AMC12 box|year=2017|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2017|ab=A|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:09, 8 February 2017
Problem
There are 
horses, named Horse 1, Horse 2, 
, Horse 10. They get their names from ow many minutes it takes them to run one lap around a circular race track: Horse 
runs one lap in exactly 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 
, in minutes, at which all horses will gain simultaneously be at the starting point is 
. Let 
be the least time, in minutes, such that at least 
of the horses are again at the starting point. What is the sum of the digits of 
?
Solution
We know that Horse will be at the starting point after 
minutes iff 
. Thus, we are looking for the smallest such that at least of the numbers 
divide . Thus, has at least positive integer divisors.
We quickly see that 
is the smallest number with at least positive integer divisors, and that 
are each numbers of horses. Thus, our answer is 
.
See Also
| 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. 
