1000-digit Fibonacci Number

This problem can be solved with a scientific calculator!

We're looking at Fibonacci numbers again, after last seeing them in problem 2. Our goal is to find the index of the first 1000 digit number in the sequence. Mathematically, we can state this is as the lowest $n$ such that $$1+\log{F_n} \geq 1000$$ (Here, $\log$ is the base 10 logarithm).

To assist in finding $n$, we can employ Binet's formula: $$F_n = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}$$ Here, $\phi$ is the golden ratio.

Binet's formula is exact, but we can make our work a little easier by approximating. $$F_n \approx \frac{\phi^n}{\sqrt{5}}$$ This approximation works since $(-\phi)^{-n}$ approaches 0 as $n$ increases. This also means this approximation gets more accurate as $n$ increases, which is especially convenient since we're looking for a large $n$. Substituting this approximation into the above inequality, we have $$1 + \log{\frac{\phi^n}{\sqrt{5}}} \geq 1000$$

Now with a little bit of algebra, we can solve for $n$. $$n \geq 999\log_{\phi}10 + \log_{\phi}\sqrt{5}$$ If you're trying to solve this with a scientific calculator, you probably don't have a $\log_{\phi}$ button (please let me know if you do), but we can just use the logarithmic change of base formula. This ends up getting us $$n \geq 4781.85927\ldots$$ Therefore, we want $n=4782$.

Relevant sequences

• Fibonacci numbers: A000045

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