1. What is the greatest common divisor of the numbers $11 n+4$ and $7 n+2$, where $n$ is a natural number?
II/1. The greatest common divisor of the numbers $11 n+4$ and $7 n+2$ also divides $7(11 n+4)-11(7 n+2)=6$, so it can be at most 6. If $n=4$, then $11 n+4=48$ and $7 n+2=30$, and the greatest common divisor of these two numbers is exactly 6. The answer is therefore 6.

2. method. We find the greatest common divisor of the numbers $11 n+4$ and $7 n+2$ using the Euclidean algorithm.

$$
\begin{aligned}
11 n+4 & =1 \cdot(7 n+2)+(4 n+2) \\
7 n+2 & =1 \cdot(4 n+2)+3 n \\
4 n+2 & =1 \cdot 3 n+(n+2) \\
3 n & =3 \cdot(n+2)-6
\end{aligned}
$$

The greatest common divisor of the above two numbers divides 6. If $n=4$, then $11 n+4=48$ and $7 n+2=30$, and the greatest common divisor of these two numbers is exactly 6. The answer is therefore 6.

Process (points may vary depending on the solution; points can be split)

(Points for the process can be split if the student does not reach the end: 1 point for starting the Euclidean algorithm, up to 2 points if there is a mistake but the process is correct and completed, and 1 point for the last line.)

Conclusion that the greatest common divisor divides 6, or that it is at most 6.

1 point Concrete example to achieve the upper bound, e.g., $n=4: \quad D(48,30)=6$ or generally $n=6 k-2, k \in \mathbb{N}$. (The greatest possible common divisor is therefore 6.)

2 points (An answer without an example earns 1 point.)