AMC10 2009 B

AMC10 2009 B · Q21

AMC10 2009 B · Q21. It mainly tests Remainders & modular arithmetic, Powers & residues.

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

将 $3^0 + 3^1 + 3^2 + \dots + 3^{2009}$ 除以 8 所得的余数是多少？

(A) 0 0

(B) 1 1

(D) 4 4

(E) 6 6

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The sum of any four consecutive powers of 3 is divisible by $3^0+3^1+3^2+3^3=40$ and hence is divisible by 8. Therefore $$(3^2+3^3+3^4+3^5)+\cdots+(3^{2006}+3^{2007}+3^{2008}+3^{2009})$$ is divisible by 8. So the required remainder is $3^0+3^1=4$.

答案（D）：任意连续四个 3 的幂之和可被 $3^0+3^1+3^2+3^3=40$ 整除，因此也可被 8 整除。于是 $$(3^2+3^3+3^4+3^5)+\cdots+(3^{2006}+3^{2007}+3^{2008}+3^{2009})$$ 可被 8 整除。因此所求余数为 $3^0+3^1=4$。

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