There are 2016 points arranged on a circle. We are allowed to jump over 2 or 3 points in a clockwise direction as we please.

How many jumps must one make at a minimum to reach all points and return to the starting point?

(Gerd Baron)
We first show indirectly that we cannot manage if we use all points exactly once.

Since 2016 is divisible by 2 and 3, it is not possible with jumps of a single length alone. Therefore, both must occur. Due to the cyclical nature, we may assume without loss of generality a position with the jump sequence 2, 3. Let $A, B, C, D, E, F$ be six consecutive points on the circle, and we assume that $A, C$, and $F$ are jumped to in succession. From a point before $A$, we can only reach $B$ as a new point. From there, we can only reach $D$ or $E$. Therefore, it is not possible in 2016 jumps. We must therefore jump to $A, B$, or $C$ twice and pass through this block a total of three times. Hence, we choose series of jumps of length 3.

![](https://cdn.mathpix.com/cropped/2024_04_17_23ba34befa521d651d71g-05.jpg?height=899&width=1062&top_left_y=227&top_left_x=497)

Figure 3: Problem 2, Solution 3

![](https://cdn.mathpix.com/cropped/2024_04_17_23ba34befa521d651d71g-05.jpg?height=934&width=1060&top_left_y=1549&top_left_x=498)

Figure 4: Problem 2, Solution 4

With 1, 4, 7, ..., 2014, 2016, 3, 6, ..., 2013, 2015, 2, 5, ..., 2015, 1, we have completed the round in 2017 jumps (only 2015 is visited twice). The answer we are looking for is therefore: 2017 jumps.

(Gerd Baron)