Example 24 (1989 50th Putnam Mathematical Competition) 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 $K$?

---

The translation preserves the original text's line breaks and format.
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
$$
1+100+100^{2}+\cdots+100^{n}=\frac{100^{n+1}-1}{100-1}(n \geqslant 2) \text {. }
$$

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.