56. (7) Given a sequence of natural numbers $u_{0}, u_{1}, \ldots$, where $u_{0}=1$. It turns out that there exists a natural number $k$ such that the equality $u_{n+1} u_{n-1} = k u_{n}$ holds for all natural numbers $n$. It is also known that $u_{2013}=100$. Find $k$.
- Answer: $k=10$. Let $u_{1}$ be denoted by $a$. Simple calculations show that $u_{2}=k a, u_{3}=k^{2}, u_{4}=k^{2} / a, u_{5}=k / a, u_{6}=1, u_{7}=a$, which means the sequence

![](https://cdn.mathpix.com/cropped/2024_06_04_fb90edf90df37d91faedg-27.jpg?height=928&width=951&top_left_y=310&top_left_x=558)

Fig. 12: to problem 48.

![](https://cdn.mathpix.com/cropped/2024_06_04_fb90edf90df37d91faedg-27.jpg?height=691&width=943&top_left_y=1805&top_left_x=565)

Fig. 13: to problem 54.
is periodic with a period of 6. Therefore, $u_{2013}=u_{3}=k^{2}$, from which $k=10$.

(British Olympiad 1995)