9. The sequence of Fibonacci numbers $F_{1}, F_{2}, \ldots, F_{n}, \ldots$ is such that $F_{1}=F_{2}=1$ and that the n-th term (with $n \geq 3$) is the sum of the two preceding ones (the first terms of the sequence are therefore $F_{1}=1, F_{2}=1, F_{3}=2=1+1, F_{4}=3=2+1, F_{5}=5=3+2$). How many Fibonacci numbers have exactly 2016 digits in their decimal representation?
(A) At least 2 and at most 3
(B) At least 4 and at most 5
(D) At least 8 and at most 9
(E) 10 or more
(C) At least 6 and at most 7
9. The answer is $\mathbf{( B )}$. First, observe that (for $n \geq 2$) the relation $F_{n+1}=F_{n}+F_{n-1}$, together with the fact that $F_{n-1} \leq F_{n}$, implies that $F_{n}$ is at least half of $F_{n+1}$. Now let $k \geq 1$ and $F_{n}$ be the smallest Fibonacci number with $k+1$ decimal digits (so $F_{n} \geq 10^{k}$). We have $F_{n-1} \geq \frac{1}{2} \cdot 10^{k}$, and thus $F_{n+1}=F_{n}+F_{n-1} \geq\left(1+\frac{1}{2}\right) \cdot 10^{k}$. Proceeding in the same way, we obtain the inequalities $F_{n+2} \geq\left(\frac{3}{2}+1\right) \cdot 10^{k}, F_{n+3} \geq\left(\frac{5}{2}+\frac{3}{2}\right) 10^{k}, F_{n+4} \geq\left(4+\frac{5}{2}\right) \cdot 10^{k}$, and $F_{n+5} \geq\left(\frac{13}{2}+4\right) \cdot 10^{k}>10^{k+1}$, from which we see that $F_{n+5}$ has at least $k+2$ decimal digits.

On the other hand, by definition we have $F_{n-2} \leq F_{n-1}<10^{k}$, from which we successively obtain $F_{n}<2 \cdot 10^{k}, F_{n+1}<3 \cdot 10^{k}, F_{n+2}<5 \cdot 10^{k}, F_{n+3}<8 \cdot 10^{k}$, so $F_{n+3}$ still has $k+1$ decimal digits. In conclusion, for any number $k \geq 2$, there are at least 4 and at most 5 Fibonacci numbers with $k$ digits.