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# geometric series

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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?

Jul 18, 2022

#1
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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 $$19208 (\text{mod} \ 19) = \color{brown}\boxed{18}$$

Jul 18, 2022

#1
+2666
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