10.2. A proper divisor of a natural number is any of its divisors, different from one and the number itself. Find all natural numbers that have at least two distinct proper divisors and are divisible by the difference of any two of them.
Answer: $6,8,12$.

Solution. Let the required number be ${ }^{n}$ and show that it must be even. Otherwise, all its divisors must be odd, and their difference must be even. But an even number cannot divide an odd number - a contradiction.

Due to the evenness of ${ }^{n}$, it must have proper divisors 2 and $\frac{n}{2}$, so ${ }^{n}$ must divide their difference, that is, the number $\frac{n}{\frac{n}{2}-2}=\frac{2 n}{n-4}=2+\frac{8}{n-4}$ must be an integer.

This is only possible when $\frac{8}{n-4}$ is an integer, that is, when $n-4$ is an even divisor of the number 8. From this, we get all possible $n=6,8,12$. We check each of them, and they all satisfy the condition of the problem.

Grading criteria. It is proven that $n$ is even: 1 point. Divisibility by $\frac{n}{2}-2$ is considered: 2 points. From this, all possible $n=6,8,12$ are found: 3 points. Verification is performed for all $n=6,8,12$: 1 point.