What is the remainder of : $\sum_{i=0}^{2019} 3^i$ divided by $3^4$ ?
I know that $$\sum_{i=0}^{2019} 3^i = \frac{3^{2020}}{3-1} = \frac{3^{2020}}{2} $$
so $$\frac{\sum_{i=0}^{2019} 3^i}{3^4} = \frac{3^{2016}}{2}$$
how do I proceed from here? I tried running a program in python on it but I think the number is too big
On
The answer Shubham Johri provides is definitely the quickest approach and is the way we would think of that sum in terms of modular arithmetic (using mod $ \ 81 \ $ here). From where you were (once the correction is made to the geometric sum), you could write $$ \ \frac{3^{2020} - 1}{2 \ · \ 3^4} \ \ = \ \ \frac{(3^{2020} - 3^4) \ + \ (3^4 \ - \ 1)}{2 \ · \ 3^4} \ \ = \ \ \frac{3^{2016} - 1}{2} \ + \ \frac{80}{2 \ · \ 81} \ \ . $$ The numerator in the first term is a product of $ \ 2016 \ $ odd factors less $ \ 1 \ : \ $ the product is therefore odd and the numerator is even, making the first term an integer. The second term is the even numerator $ \ 80 \ = \ 2^4 · 5 \ $ divided by $ \ 2 \ $ and a product of four $ \ 3 $ 's, so this ratio is $ \ \frac{40}{81} \ $ "in lowest terms". The remainder in the specified division is thus $ \ 40 \ \ . $
Hint: Your sum is $3^0+3^1+3^2+3^3+3^4(3^0+3^1+...+3^{2015})$.