What is the remainder when 3^0+3^1+3^2+…+3^2009 is divided by 8?

2009 AMC 10B Problem 21

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21.

What is the remainder when

3^0+3^1+3^2+\cdots+3^{2009}

is divided by

8?

0

1

2

4

6

3^0+3^1+3^2+3^3=40

is a multiple of

8

Group the

2010

terms into blocks of four and handle the leftover

Any four consecutive powers of

3

sum to a multiple of

3^0+3^1+3^2+3^3=40,

which is divisible by

8.

The terms from

3^2

3^{2009}

split into such blocks and contribute remainder

0.

What remains is

3^0+3^1=4.

Thus, the correct answer is D.