2009 AMC 10B Problem 21 — Solution | LIVE by Po-Shen Loh

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2009 AMC 10B Problem 21

Below is the professionally curated solution for Problem 21 of the 2009 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B solutions, or check the answer key.

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Concepts:modular arithmetic geometric sequence pairing and grouping

Difficulty rating: 1420

21.

What is the remainder when $3^0+3^1+3^2+\cdots+3^{2009}$ is divided by $8?$

$0$

$1$

$2$

$4$

$6$

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 $3^2$ to $3^{2009}$ split into such blocks and contribute remainder $0.$ What remains is $3^0+3^1=4.$

Thus, the correct answer is D.

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