Let N be the sum of the geometric series 8^1 + 8^2 + ... + 8^2021. What is the remainder when N is divided by 19?

Guest Jul 18, 2022

#1**0 **

Note the remainders for each term can be written in separate "cycles": 8, 7, 18, 11, 12, 1

There are 336 complete "cycles" (note there are 2021 terms), each with a sum of 57 The last cycle has 5 terms, meaning it has a sum of 56.

This means that the sum of their remainders is \(57 \times 336 + 56 = 19208\), meaning the remainder is \(19208 (\text{mod} \ 19) = \color{brown}\boxed{18}\)

BuilderBoi Jul 18, 2022

#1**0 **

Best Answer

Note the remainders for each term can be written in separate "cycles": 8, 7, 18, 11, 12, 1

There are 336 complete "cycles" (note there are 2021 terms), each with a sum of 57 The last cycle has 5 terms, meaning it has a sum of 56.

This means that the sum of their remainders is \(57 \times 336 + 56 = 19208\), meaning the remainder is \(19208 (\text{mod} \ 19) = \color{brown}\boxed{18}\)

BuilderBoi Jul 18, 2022