5. A5 (BUL) Find all positive integers $a_{1}, a_{2}, \ldots, a_{n}$ such that 
$$ \frac{99}{100}=\frac{a_{0}}{a_{1}}+\frac{a_{1}}{a_{2}}+\cdots+\frac{a_{n-1}}{a_{n}} $$
where $a_{0}=1$ and $\left(a_{k+1}-1\right) a_{k-1} \geq a_{k}^{2}\left(a_{k}-1\right)$ for $k=1,2, \ldots, n-1$.
5. Let $a_{1}, a_{2}, \ldots, a_{n}$ satisfy the conditions of the problem. Then $a_{k}>a_{k-1}$, and hence $a_{k} \geq 2$ for $k=1, \ldots, n$. The inequality $\left(a_{k+1}-1\right) a_{k-1} \geq$ $a_{k}^{2}\left(a_{k}-1\right)$ can be rewritten as  $$ \frac{a_{k-1}}{a_{k}}+\frac{a_{k}}{a_{k+1}-1} \leq \frac{a_{k-1}}{a_{k}-1} . $$  Summing these inequalities for $k=i+1, \ldots, n-1$ and using the obvious inequality $\frac{a_{n-1}}{a_{n}}<\frac{a_{n-1}}{a_{n}-1}$, we obtain $\frac{a_{i}}{a_{i+1}}+\cdots+\frac{a_{n-1}}{a_{n}}<\frac{a_{i}}{a_{i+1}-1}$. Therefore  $$ \frac{a_{i}}{a_{i+1}} \leq \frac{99}{100}-\frac{a_{0}}{a_{1}}-\cdots-\frac{a_{i-1}}{a_{i}}<\frac{a_{i}}{a_{i+1}-1} \quad \text { for } i=1,2, \ldots, n-1 $$  Consequently, given $a_{0}, a_{1}, \ldots, a_{i}$, there is at most one possibility for $a_{i+1}$. In our case, (1) yields $a_{1}=2, a_{2}=5, a_{3}=56, a_{4}=280^{2}=78400$. These values satisfy the conditions of the problem, so that this is a unique solution.