# Difference between revisions of "2017 AMC 10A Problems/Problem 16"

## 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 1

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, $1+2 = \boxed{\textbf{(B)}\ 3}$.

## Solution 2

We are trying to find the smallest number that has $5$ one-digit divisors. Therefore we try to find the LCM for smaller digits, such as $1$, $2$, $3$, or $4$. We quickly consider $12$ since it is the smallest number that is the LCM of $1$, $2$, $3$ and $4$. Since $12$ has $5$ single-digit divisors, namely $1$, $2$, $3$, $4$, and $6$, our answer is $1+2 = \boxed{\textbf{(B)}\ 3}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 