## Task A-4.5.

On a circle, 3000 points are marked. One of these points is where a jumper is located. The jumper makes each jump by skipping one or two marked points in the clockwise direction and landing on the next marked point. Determine the minimum number of jumps the jumper has made if it has landed on each marked point at least once and returned to the point from which it started.
## Solution.

We claim that the answer is 3001 jumps.

Label the points in the clockwise direction with numbers $1,2, \ldots, 3000$, starting from the point where the jumper is initially located. The jumper can visit each point at least once and return to the starting point in 3001 jumps as follows:

$1 \xrightarrow{3} 4 \xrightarrow{3} 7 \xrightarrow{3} \ldots \xrightarrow{3} 2998 \xrightarrow{2} 3000 \xrightarrow{3} 3 \xrightarrow{3} \ldots \xrightarrow{3} 2097 \xrightarrow{2} 2999 \xrightarrow{3} 2 \xrightarrow{3} \ldots \xrightarrow{3} 2999 \xrightarrow{2} 1 . \quad 2$ points

We will show that the jumper cannot do this in fewer jumps. Let $k$ be the number of jumps of length 2, and $\ell$ be the number of jumps of length 3. Since the jumper must return to the starting point, i.e., must complete a whole number of full circles, there exists a natural number $n$ such that

$$
2 k+3 \ell=3000 n
$$

Since the jumper must visit each labeled point at least once, we have

$$
k+\ell \geqslant 3000
$$

If $k+\ell=3000$, then from

$$
3000 n=2 k+3 \ell=2(k+\ell)+\ell=6000+\ell
$$

it would follow that 3000 divides $\ell$. Since $k, \ell \geqslant 0$, it would have to be $\ell=0$ or $\ell=3000$. The case $\ell=3000$ would mean that all jumps are of length 3, and the jumper would not visit every labeled point.

Therefore, it is not possible that $k+\ell=3000$.