3. If $x^{2}-13 x+1=0$, then the unit digit of $x^{4}+x^{-4}$ is
(A) 1 ;
(B) 3 ;
(C) 5 ;
(D) 7 .
3. (D)

Solution: From $x^{2}-13 x+1=0$, we know $x \neq 0$. Therefore, $x+x^{-1}=13$, $x^{2}+x^{-2}=13^{2}-2=167$. $x^{4}+x^{-4}=167^{2}-2$, thus the unit digit of $x^{4}+x^{-4}$ is $9-2=7$.