33-11 Given that $a$ is a root of the equation $x^{2}-5 x+1=0$, then the last digit of $a^{4}+a^{-4}$ is
(A) 3.
(B) 5.
(C) 7.
(D) 9.
(6th "Jinyun Cup" Junior High School Mathematics Invitational, 1989)
[Solution] Let $\alpha$ be a root of the equation, then $\alpha^{2}-5 \alpha+1=0$. That is, $\alpha^{2}+1=5 \alpha$, or $\frac{\alpha^{2}+1}{\alpha}=5$.
Also, $\alpha^{4}+\alpha^{-4}=\left(\alpha^{2}+\alpha^{-2}\right)^{2}-2$
\[
\begin{array}{l}
=\left[\left(\alpha+\alpha^{-1}\right)^{2}-2\right]^{2}-2 \\
=\left[\left(\frac{\alpha^{2}+1}{\alpha}\right)^{2}-2\right]^{2}-2 \\
\quad=(25-2)^{2}-2=23^{2}-2,
\end{array}
\]
The last digit of this is 7.
Therefore, the answer is $(C)$.