The problem is, first of all, translated incorrectly. The length of the minimal period of $a-b$ and $a+6b$ is exactly 15, and it asks for the smallest value of $k$ such that it is possible that there may exist two $a$ and $b$ with this property. That said, the problem is still very nice.
Multiply $a$ and $b$ by arbitrary powers of $10$ to remove any factors of $2$ and $5$ in the denominators. Then, we can write
$a = \frac{m}{10^{30}-1}$ and $b = \frac{n}{10^{30}-1}$. The condition rapidly rearranges to $10^{15}+1 \mid m - n$, and we wish to find some value of $k$ such that $10^{15}+1 \mid m + kn$. By subtracting, we get the relation $10^{15} + 1 \mid (k+1)n$. But note that $10^{15} + 1 \nmid n$, because the minimal period of $n$ is $30$. Hence, $k+1$ must be at least the smallest prime factor of $10^{15} + 1$, which can be checked to be $7$, so $k$ is at least $6$. On the other hand, it is easy to check that $6$ works by setting $m = 8n$, and then letting $n = (10^{15} + 1)/7$. It is straightforward to check that the resulting fractions have minimal period $30$, so we are done.