$10 \cdot 75$ If $n$ is a natural number, and the sum $1981^{n}+1982^{n}+1983^{n}+1984^{n}$ is not divisible by 10, then what condition must $n$ satisfy? Try to explain the reason.
(China Beijing Junior High School Grade 2 Mathematics Competition, 1983)
[Solution] Since the unit digit of $a^{n}$ cycles with a period of 4, we can discuss the cases for $n=4k+1$, $4k+2$, $4k+3$, and $4k$ respectively.
Let $\quad M=1981^{n}+1982^{n}+1983^{n}+1984^{n}$.
When $n=4k+1$ (where $k$ is a non-negative integer), the unit digit of $M$ is the same as the unit digit of $1+2+3+4=10$, which means the unit digit of $M$ is 0, so $10 \mid M$.

When $n=4k+2$ (where $k$ is a non-negative integer), the unit digit of $M$ is the same as the unit digit of $1+4+9+6=20$, which means the unit digit of $M$ is 0, so $10 \mid M$.

When $n=4k+3$ (where $k$ is a non-negative integer), the unit digit of $M$ is the same as the unit digit of $1+8+7+4=20$, which means the unit digit of $M$ is 0, so $10 \mid M$.

When $n=4k$ (where $k$ is a positive integer), the unit digit of $M$ is the same as the unit digit of $1+6+1+6=14$, which means the unit digit of $M$ is 4, so $10 \nmid M$.
Therefore, $M$ is not divisible by 10 if and only if $n$ is a multiple of 4.