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

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2009 AMC 10B Problem 21

Attempt Problem 21 of the 2009 AMC 10B below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

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

$0$

$1$

$2$

$4$

$6$

Answer: D

Concepts:modular arithmetic geometric sequence pairing and grouping

Difficulty rating: 1420

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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