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2017 AMC 10A Problem 16

Problem 16 of 25IntermediateNumber Theory

There are 1010 horses, named Horse 1,1, Horse 2,2, and so on through Horse 10.10. 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 00 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,S > 0, in minutes, at which all 1010 horses will again simultaneously be at the starting point is S=2520.S=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 T?T?

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Solution

Horse kk is back at the starting point after tt minutes exactly when tt is divisible by kk. Thus we need the least positive tt that is divisible by at least five of the integers 1,2,,101,2,\ldots,10. Checking upward, no number below 1212 has five divisors from this list: for example, 66 has 1,2,3,61,2,3,6, 88 has 1,2,4,81,2,4,8, 99 has 1,3,91,3,9, and 1010 has 1,2,5,101,2,5,10. The number 1212 is divisible by 1,2,3,4,1,2,3,4, and 66, so the least possible time is T=12T=12. The sum of its digits is 1+2=31+2=3. Thus, B is the correct answer.

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Concepts: divisibility · least common multiple

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.