97. The Fibonacci sequence. Consider the Fibonacci sequence $1,1,2,3,5,13, \ldots$, whose terms $F_{n}$ (Fibonacci numbers) satisfy the relation $F_{n+2}=F_{n}+F_{n+1}$, $F_{1}=F_{2}=1$. Now consider the sequence of digits in the units place of the Fibonacci numbers. Will this sequence be cyclic, that is, can it be obtained by the unlimited repetition of the same finite set of digits, as is true, for example, in the case of the sequence $055055 \ldots$.
97. Yes, it is possible. Each member of the desired sequence $11235831 \ldots$ can be obtained by adding the previous two members and taking the digit in the units place of this sum.

In this sequence, two odd and one even members alternate. There are $5 \cdot 5=25$ ordered pairs composed of odd digits. Therefore, after no more than $3 \cdot 25=$

75 consecutive additions, one of these odd pairs will repeat, and a new cycle will begin. Since the sum (or difference) of two digits is unique, this repeating pair must coincide with the first odd pair (with which the sequence begins) rather than with any of the "internal" odd pairs. Indeed, it turns out that only 60 additions are required, and a new cycle of 60 digits will begin: 1123583145943707 74156178538190998752796516730336954 $932572910^{*} 11 \ldots$.

Similar reasoning will be valid in any number system, as well as not only for the Fibonacci sequence but also for any sequence that satisfies the recurrence relation.

$$
A_{n+k}=A_{n}+A_{n+1}+\cdots+A_{n+k-1}
$$