21. The remainder of $4^{0}+4^{1}+4^{2}+\cdots+4^{n}$ is 1 when it is divided by 7, then $n$ may be equal to $(\quad$.
(A) 214 .
(B) 215 .
(C) 216 .
(D) 217 .
（English-Chinese Mini Dictionary: remainder 余数）
Text: $4^{0}+4^{1}+4^{2}+\cdots+4^{n}$ when divided by 7 has a remainder of 1, then $n$ could be equal to ().
(A) 214 .
(B) 215 .
(C) 216 .
(D) 217 .
Answer: C.

Solution: Since
$$
4^{3 k}=\left(4^{3}\right)^{k}=64^{k}=(7 \times 9+1)^{k} \text {, }
$$

Therefore,
$4^{3 k}$ leaves a remainder of 1 when divided by 7,
$4^{3 k+1}=4 \cdot 4^{3 k}$ leaves a remainder of 4 when divided by 7,
$4^{3 k+2}=16 \cdot 4^{3 k}$ leaves a remainder of 2 when divided by 7,

Thus,
$4^{3 k}+4^{3 k+1}+4^{3 k+2}$ is divisible by 7.
To make $4^{0}+4^{1}+4^{2}+\cdots+4^{n}$ leave a remainder of 1 when divided by 7,

Then
$n$ must be a multiple of 3.

Therefore, the answer is C.