Maths Olympiad Prep

Track / Stage 3 / 100 of 260 #100 of 1964

Problem 100

AMC 10/12, early questions
Number theory Difficulty 3.4 Multiple choice

There are 1010 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 kk runs one lap in exactly kk 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>0S > 0, in minutes, at which all 1010 horses will again simultaneously be at the starting point is S=2520S = 2520. Let T>0T>0 be the least time, in minutes, such that at least 55 of the horses are again at the starting point. What is the sum of the digits of TT?

Pick one

Official solution

We know that Horse kk will be at the starting point after nn minutes if knk|n. Thus, we are looking for the smallest nn such that at least 55 of the numbers {1,2,,10}\{1,2,\cdots,10\} divide nn. Thus, nn has at least 55 positive integer divisors.
We quickly see that 1212 is the smallest number with at least 55 positive integer divisors and that 1,2,3,4,61,2,3,4,6 are each numbers of horses. Thus, our answer is 1+2=(B) 31+2=\boxed{\textbf{(B) } 3}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.