$29 \cdot 27$ For every odd number $p>1$, we have  
(A) $(p-1)^{\frac{1}{2}(p-1)}-1$ is divisible by $p-2$.  
(B) $(p-1)^{\frac{1}{2}(p-1)}+1$ is divisible by $p$.  
(C) $(p-1)^{\frac{1}{2}(p-1)}$ is divisible by $p$.  
(D) $(p-1)^{\frac{1}{2}(p-1)}+1$ is divisible by $p+1$.  
(E) $(p-1)^{\frac{1}{2}(p-1)}-1$ is divisible by $p-1$.  
(18th American High School Mathematics Examination, 1967)
[Solution] Let $p>1$ be an odd number, then $p-1$ is even, and $\frac{1}{2}(p-1)=n$ is a positive integer. Therefore,
$$
\begin{aligned}
(p-1)^{\frac{p-1}{2}}-1 & =(2 n)^{n}-1 \\
& =(2 n-1)\left[(2 n)^{n-1}+(2 n)^{n-2}+\cdots+1\right]
\end{aligned}
$$

we know it contains the factor $2 n-1=p-2$. Thus, (A) is correct.
When $p=5$, (B), (C), (D), and (E) are all incorrect.
Therefore, the answer is (A).