2017 AMC 12A Problem 12

Below is the professionally curated solution for Problem 12 of the 2017 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 12A solutions, or check the answer key.

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Concepts:least common multipledivisibility

Difficulty rating: 1630

12.

There are 1010 horses, named Horse 1,1, Horse 2,2, ,\ldots, 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\gt0, in minutes, at which all 1010 horses will again simultaneously be at the starting point is S=2520.S=2520. Let T>0T\gt0 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?

22

33

44

55

66

Solution:

Horse kk is at the starting point at time tt precisely when kt.k\mid t. So we want the smallest tt with at least 55 divisors among 1,2,,10.1,2,\ldots,10.

Checking small values, t=12t=12 is divisible by 1,2,3,4,1,2,3,4, and 6,6, giving exactly 55 such horses, and no smaller tt reaches 5.5. Thus T=12,T=12, and the sum of its digits is 1+2=3.1+2=3.

Thus, the correct answer is B.

Problem 12 in Other Years