$33 \cdot 12$ If $x^{2}-13 x+1=0$, then the units digit of $x^{4}+x^{-4}$ is
(A) 1.
(B) 3.
(C) 5.
(D) 7.
(China Junior High School Mathematics League, 1992)
[Solution] From $x^{2}-13 x+1=0$, we know $x \neq 0$, and the product of its two roots is 1, meaning they are reciprocals of each other, so we have $\quad x+x^{-1}=13$.
Therefore, $\quad x^{2}+x^{-2}=167$.
Thus, $\quad x^{4}+x^{-4}=167^{2}-2$.
Hence, the unit digit of $x^{4}+x^{-4}$ is 7.
Therefore, the answer is $(D)$.