Question

当 $3^0+3^1+3^2+\cdots+3^{2009}$ 除以 $8$ 时，余数是多少？

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

$0$

$1$

$2$

$4$

$6$

$3^0+3^1+3^2+3^3=40$ 是 $8$ 的倍数。

$3^0+3^1+3^2+3^3=40$ is a multiple of $8$

将 $2010$ 项分成每四项一组，并处理剩余项。

Group the $2010$ terms into blocks of four and handle the leftover

任意连续四个 $3$ 的幂之和都是 $3^0+3^1+3^2+3^3=40$ 的倍数，而这个数能被 $8$ 整除。

从 $3^2$ 到 $3^{2009}$ 的项可以分成这样的四项组，对余数贡献为 $0$ 。剩下的是 $3^0+3^1=4$ 。

所以正确答案是 D。

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.