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Difference between revisions of "2017 AMC 10A Problems/Problem 16"
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==Solution==
 
==Solution==
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If we have horses, <math>a_1, a_2, \ldots, a_n</math>, then any number that is a multiple of the all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that <math>\text{lcm}(1,2,3,2\cdot2,2\cdot3) = 12</math>. Finally, <math>1+2 = \boxed{\textbf{(B)}\ 3}</math>.
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If we have horses, <math>a_1, a_2, \ldots, a_n</math>, then any number that is a multiple of the all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that <math>\text{LCM}(1,2,3,2\cdot2,2\cdot3) = 12</math>. Finally, <math>1+2 = \boxed{\textbf{(B)}\ 3}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=15|num-a=17}}
 
{{AMC10 box|year=2017|ab=A|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:15, 8 February 2017

Problem

There are $10$ 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 $k$ 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

If we have horses, $a_1, a_2, \ldots, a_n$, then any number that is a multiple of the all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that $\text{LCM}(1,2,3,2\cdot2,2\cdot3) = 12$. Finally, $1+2 = \boxed{\textbf{(B)}\ 3}$.

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

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