6. Find all prime numbers whose representation in base 14 has the form 101010 ... 101 (ones and zeros alternate).
Solution: Let $2n+1$ be the number of digits in the number $A=101010 \ldots 101$. Let $q=$ 14 - the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2 n}=\frac{q^{2 n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.

- $n=2 k \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k+1}-1}{q-1} \cdot \frac{q^{2 k+1}+1}{q+1}$. Thus, the number $A$ is represented as the product of two integer factors (by the theorem of Bezout, the polynomial $q^{2 k+1} \pm 1$ is divisible without remainder by the polynomial $q \pm 1$), each of which is different from 1. Therefore, when $n$ is even, the number $A$ is not prime.
- $n=2 k-1 \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k}-1}{q^{2}-1} \cdot\left(q^{2 k}+1\right)$. For $k>1$, both factors are integers and different from 1; therefore, the number $A$ is composite. It remains to verify that for $k=1$, the number $A=q^{0}+$ $q^{2}=197$ is prime.

Answer: 197.