Difference between revisions of "2017 AMC 12A Problems/Problem 12"
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<math>\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6</math> | <math>\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6</math> | ||
| − | ==Solution== | + | ==Solution 1== |
We know that Horse <math>k</math> will be at the starting point after <math>n</math> minutes if <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 if <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 | + | 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>. |
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| + | ==Solution 2== | ||
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| + | In order for at least <math>5</math> horses to finish simultaneously, the current time needs to have at least <math>5</math> divisors. Thus the number must have a form of either <math>p^4</math> or <math>p^2*q</math>, which have <math>5</math> and <math>6</math> factors respectively. The smallest number of the first form is <math>16</math>, and the smallest number of the second form is <math>12.</math> Thus, our answer is <math>1+2=\boxed{\textbf{(B) } 3}</math>. | ||
== See Also == | == See Also == | ||
Latest revision as of 13:21, 19 December 2020
Contents
Problem
There are 
horses, named Horse 1, Horse 2, 
, Horse 10. They get their names from how 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 again 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 1
We know that Horse will be at the starting point after 
minutes if 
. 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 
.
Solution 2
In order for at least horses to finish simultaneously, the current time needs to have at least divisors. Thus the number must have a form of either 
or 
, which have and 
factors respectively. The smallest number of the first form is 
, and the smallest number of the second form is 
Thus, our answer is .
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. 
