11.3. Positive rational numbers $a$ and $b$ are written as decimal fractions, each of which has a minimal period consisting of 30 digits. The decimal representation of the number $a-b$ has a minimal period length of 15. For what smallest natural $k$ can the minimal period length of the decimal representation of the number $a+k b$ also be 15? (A.S. Golyanov)
# Answer. $k=6$.

Solution. By multiplying, if necessary, the numbers $a$ and $b$ by a suitable power of ten, we can assume that the decimal representations of the numbers $a, b, a-b$, and $a+k b$ are purely periodic (i.e., the periods start immediately after the decimal point).

We will use the following known fact: the decimal representation of a rational number $r$ is purely periodic with (not necessarily minimal) period length $T$ if and only if $r$ can be written as $\frac{m}{10^{T}-1}$ for some integer $m$.

Applying this to the problem, this means that $a=\frac{m}{10^{30}-1}$ and $b=\frac{n}{10^{30}-1}$. We also know that the numbers $a-b=\frac{m-n}{10^{30}-1}$ and $a+k b=\frac{m+k n}{10^{30}-1}$ can be written as decimal fractions with a period length of 15, meaning they can be written as common fractions with the denominator $10^{15}-1$. Therefore, their difference $(k+1) b=\frac{(k+1) n}{10^{30}-1}$ can also be written with the denominator $10^{15}-1$. Thus, the number $(k+1) n$ is divisible by $10^{15}+1$, while the number $n$ is not (otherwise, $b$ would be written as a fraction with a period length of 15). Therefore, the number $k+1$ is divisible by some prime divisor of the number $10^{15}+1$. The smallest such divisor is 7. Indeed, the number $10^{15}+1$ gives a remainder of 1 when divided by 2 and 5, and a remainder of 2 when divided by 3. On the other hand, it is divisible by $10^{3}+1=7 \cdot 143$. Thus, $k+1 \geqslant 7$ and $k \geqslant 6$.

In a certain sense, the minimal example of numbers satisfying the condition for $k=6$ is obtained by setting $a-b=\frac{1}{10^{15}-1}$ and $a+6 b=\frac{2}{10^{15}-1}$. Then $a=\frac{8}{7\left(10^{15}-1\right)}$ and $b=\frac{1}{7\left(10^{15}-1\right)}$. It is clear that the lengths of the minimal periods of the numbers $a-b$ and $a+6 b$ are 15. Furthermore, the lengths of the minimal periods of the numbers $a$ and $b$ are greater than 15 and are divisible by 15 (since $10^{T}-1$ must divide $10^{15}-1$). On the other hand, since $10^{30}-1 \vdots 7\left(10^{15}-1\right)$, the numbers $a$ and $b$ are periodic with a period length of 30. Therefore, the lengths of their minimal periods are 30.