11. For which $k$ the number $N=101 \cdots 0101$ with $k$ ones is a prime?
Answer: There is only one such prime: 101 so $k=2$.
To see this, suppose that $N=101 \cdots 0101$ with $k$ ones, for some $k \geq 2$. Then
$$
99 N=9999 \cdots 9999=10^{2 k}-1=\left(10^{k}+1\right)\left(10^{k}-1\right) .
$$

If moreover $N$ is prime, then $N$ divides either $10^{k}+1$ or $10^{k}-1$, and hence one of $\frac{99}{10^{k}-1}=\frac{10^{k}+1}{N}$ and $\frac{99}{10^{k}+1}=\frac{10^{k}-1}{N}$ is an integer. For $k>2$, $10^{k}-1$ and $10^{k}+1$ are both greater than 99 , so we get a contradiction. Therefore $k=2$ and $N=101$ (which is prime).