Number of Fibonacci Numbers with Same Number of Decimal Digits

Theorem
Let $n$ be an integer such that $n > 1$.

When expressed in decimal notation, there are either $4$ or $5$ Fibonacci numbers with $n$ digits.

Proof
Let $F_m$ be an $n$-digit Fibonacci number.

Then $F_m \ge 10^{n - 1} \ge 10$.

We have:

Thus:

So $F_{m + 5}$ is a number of at least $n + 1$ digits.

Therefore there is at most $5$ Fibonacci numbers with $n$ digits: $F_m, F_{m + 1}, F_{m + 2}, F_{m + 3}, F_{m + 4}$.

Similarly, $10 \le F_m < 10^n$.

Suppose $F_m$ is the smallest Fibonacci numbers with $n$ digits.

Then $F_{m - 1} < 10^{n - 1}$.

Thus:

So there is at least $4$ Fibonacci numbers with $n$ digits: $F_m, F_{m + 1}, F_{m + 2}, F_{m + 3}$.

Combining the arguments above, there are either $4$ or $5$ Fibonacci numbers with $n$ digits.