5. The divisors of a natural number $n$ (including $n$ and 1), which has more than three divisors, were listed in ascending order: $1=d_{1}<d_{2} \ldots<d_{k}=n$. The differences $u_{1}=d_{2}-d_{1}, u_{2}=d_{3}-d_{2}, \ldots, u_{k-1}=d_{k}-d_{k-1}$ turned out to be such that $u_{2}-u_{1}=u_{3}-u_{2}=\ldots=u_{k-1}-u_{k-2}$. Find all such $n$. (S. Berlov)
Answer: 10. Solution: Let $n$ be odd. Then $u_{k-1}=d_{k}-d_{k-1} \geq n-n / 3=2 n / 3$. In this case, $u_{k-2}=d_{k-1}-d_{k-2}n / 3$, but $u_{k-1}-u_{k-2}=u_{k-2}-u_{k-3}<u_{k-2}<n / 3$ - a contradiction. In the case of even $n$, we get $u_{k-1}=d_{k}-d_{k-1}=n / 2, u_{k-2}=d_{k-1}-d_{k-2}=n / 2-d_{k-2}$, so $u_{k-1}-u_{k-2}=d_{k-2}$. But $u_{2}-u_{1}=d_{3}-2 d_{2}+d_{1}=d_{3}-3$, which can only equal $d_{k-2}$ when $k=4$. Then $d_{k-2}=d_{2}=2, d_{3}=n / 2=d_{2}+3=5$ and $n=10$.