Test Question $A-1$. How many evil numbers have a 10-adic representation that is exactly 0,1 alternating and ends with 1?
Let $P_{\mathrm{k}}=1010 \cdots 101$ ( $k$ zeros) be a prime number in decimal notation, $k \geqslant 1$. Clearly, $P_{1}=101$ is a prime number. When $k>2$, if $k$ is odd, then 101 clearly divides $P_{\times 3}$. If $k$ is even, then
$$
\begin{array}{l}
11 P_{k}=11 \cdots 1(2 k+2 \text { ones}) \stackrel{\text { denote as }}{=} Q_{2 k+2}, \\
\text { hence } 11 P_{k}=Q_{k+1} \cdot R,
\end{array}
$$
$R$ is $100 \cdots 01(k$ zeros$)$, and 11 divides $R$, so $Q_{\mathrm{x}+1}$ divides $P_{\mathrm{k}}$.

Therefore, only the prime number (101) satisfies the problem's requirements.