［Solution］From the given, $x$ can be written as $x=10^{2 n+4}+10^{n+3}+50$. Then
$$
x-25=10^{2(n+2)}+10 \cdot 10^{n+2}+5^{2}=\left(10^{n+2}+5\right)^{2} \text {, }
$$

we know that it is a perfect square.
Therefore, the answer is $(C)$.
$30 \cdot 32$ Given the number $x=1 \overbrace{00 \cdots 001}^{n 0} \stackrel{n+0 \uparrow 0}{00 \cdots 0050}$. Then
(A) $x$ is a perfect square.
(B) $(x-50)$ is a perfect square.
(C) $(x-25)$ is a perfect square.
(D) $(x+50)$ is a perfect square.
(1st "Hope Cup" National Mathematics Invitational, 1990)