2009 AMC 10B Problem 21

What is the remainder when

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

is divided by

8?

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Solution

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.

Concepts: modular arithmetic · geometric sequence · pairing and grouping

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.