Difference between revisions of "2017 AMC 10A Problems/Problem 16"
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==Solution== | ==Solution== | ||
− | + | 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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− | <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 18:49, 8 February 2017
Problem
There are horses, named Horse , Horse , . . . , Horse . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse runs one lap in exactly minutes. At time 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 , in minutes, at which all horses will again simultaneously be at the starting point is . Let 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
Solution
If we have horses, , 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 . Finally, .
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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