3.12 Let \( K \) be the set of all natural numbers, each of which is formed by alternating 0s and 1s, with the first and last digits being 1. How many primes are there in \( K \)?
(50th Putnam Mathematical Competition, 1989)
[Solution] 1 is not a prime number, while 101 is a prime number.
Let $101010 \cdots 01$ contain 3 or more 1s.
Such a number can always be expressed as
$$
1+100+100^{2}+\cdots+100^{n}=\frac{100^{n+1}-1}{100-1} \quad(n \geqslant 2) .
$$

Since
$$
\begin{aligned}
\frac{100^{n+1}-1}{100-1} & =\frac{10^{2 n+2}-1}{10^{2}-1} \\
& =\frac{\left(10^{n+1}+1\right)\left(10^{n+1}-1\right)}{(10+1)(10-1)} \\
& =\frac{\left(10^{n+1}+1\right) \cdot 99 \cdots 9}{(10+1) \cdot 9} \\
& =\frac{\left(10^{n+1}+1\right)}{(10+1)} \cdot 11 \cdots 1 .
\end{aligned}
$$

If $n$ is odd, then $n+1$ is even, in which case $\frac{11 \cdots 11}{11}$ is an integer greater than 1.
Thus, $\frac{11 \cdots 11}{11} \cdot\left(10^{n+1}+1\right)$ is the product of two integers greater than 1, meaning $\frac{100^{n+1}-1}{100-1}$ is a composite number.

If $n$ is even, then $n+1$ is odd, in which case $10+1$ divides $10^{n+1}+1$, and the quotient is greater than 1.

Thus, $\frac{10^{n+1}+1}{10+1} \cdot 11 \cdots 11$ is the product of two integers greater than 1, meaning $\frac{100^{n+1}-1}{100-1}$ is a composite number.
This shows that when $n \geqslant 2$, such a number is always composite.
Therefore, the only prime number in $K$ is 101.