National Competition

Combinatorics Difficulty 5.0 AIME, harder Prove it Austria

Consider $2016$ points arranged on a circle. We are allowed to jump ahead by $2$ or $3$ points in clockwise direction.
What is the minimum number of jumps required to visit all points and return to the starting point?

Solution

If the problem could be solved with $2016$ jumps, the total distance covered by these jumps would be strictly between $2 \cdot 2016$ and $3 \cdot 2016$ which makes a return to the original point impossible. Therefore, at least $2017$ jumps are required.
This is indeed possible, for example with the following sequence of points on the circle.
$0, 3, 6, \dots, 2013, 2015, 2, 5, \dots, 2012, 2014, 1, 4, \dots, 2011, 2013, 0.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.