TITLE: Find the last three digits of the $2020^{th}$ fibonacci number.
QUESTION [6 upvotes]: Consider the Fibonacci sequence $1, 1, 2, 3, 5, 8, 13, ...$ What are the last three digits (from left to right) of the $2020^{\text{th}}$ term?

The problem is taken from here. I have no idea how to approach this problem. Maybe the general term of the Fibonacci sequence will be of some use?
$$F_n=\frac{(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n}{\sqrt{5}}$$
Hints are welcomed. Thanks in advance.

REPLY [4 votes]: By Chinese remainder theorem, it suffices to find the result mod $8$ and mod $125$.
I assume that you know how to do it mod $8$ (by finding period).
For mod $125$, we write $$2^nF_n = \frac 1 {\sqrt 5} \left(\sum_{k = 1}^n\binom n k \sqrt 5^k - \sum_{k = 1}^n\binom n k (-\sqrt 5)^k\right) = 2\sum_{k = 0\\2\nmid k}^n\binom n k 5^\frac{k - 1}2.$$ Now we only have to compute the sum for $k = 1, 3, 5$, as all other terms are obviously multiples of $125$.
This, together with a simple calculation of $2^{2020} \mod 125$, will give you the result.