1. When the numbers $A$ and $B$ are expressed as infinite decimal fractions, the lengths of the minimal periods of these fractions are 6 and 12, respectively. What can the length of the minimal period of the number $A+B$ be?
1. As is known, the length of the minimal period of a fraction is a divisor of the length of any other period of the fraction, see fact 4 (the length of the minimal period of a finite decimal fraction will be considered equal to one).

Let's prove the following statement: if $k$ is the length of one of the periods (not necessarily minimal) of each of the fractions $A$ and $B$, then $k$ will be the length of some period of the fractions $A+B$ and $A-B$.

We will prove the statement for $A+B$ (the proof for $A-B$ is analogous). A periodic fraction $A$ with period length $k$ can be represented as

$$
A=\frac{X}{10^{l}\left(10^{k}-1\right)}
$$

where $X$ is an integer (see fact 13). Similarly, we can write $B=\frac{Y}{10^{m}\left(10^{k}-1\right)}$. Without loss of generality, we can assume that $l \geqslant m$. Then

$$
A+B=\frac{X+10^{l-m} Y}{10^{l}\left(10^{k}-1\right)}
$$

This number is of the same form, so the corresponding fraction has a period of length $k$. The statement is proved.

Let the fraction $A$ have a period of length 6, and the fraction $B$ have a period of length 12. From the proved statement, it follows that 12 is the length of some period of the fraction $A+B$. Therefore, the length of the minimal period of the fraction $A+B$ is a divisor of the number 12.

On the other hand, the period length of the fraction $A+B$ cannot be 6 - otherwise, the fraction $B=(A+B)-A$ would have a period of length 6. Therefore, the length of the minimal period of the fraction $A+B$ cannot be 6, 3, 2, or 1.

The only two options remain: 12 and 4. We will show that both these options are possible:

$$
\begin{gathered}
A=0,(000001), \quad B=0,(000000000001) \\
A+B=0,(000001000002) \\
A=0,(000001), \quad B=0,(011100110110), \quad A+B=0,(0111)
\end{gathered}
$$

Comments. $1^{\circ}$. Let's show how to come up with an example of fractions with minimal period lengths of 6 and 12, the sum of which has a minimal period length of 4. We need to subtract a fraction with a minimal period length of 6 from any fraction with a minimal period length of 4, and we will get a fraction with a minimal period length of 12.

$2^{\circ}$. Let's describe all possible minimal period lengths of the fraction $A+B$ in the general case. Let $m, n$, and $k$ be the lengths of the minimal periods of the fractions $A, B$, and $A+B$. Then $k$ is a divisor of the least common multiple of the numbers $m$ and $n$. Similarly, $m$ is a divisor of $\operatorname{LCM}(n, k)$, and $n$ is a divisor of $\operatorname{LCM}(m, k)$. Let $p_{1}, \ldots, p_{s}$ be all the prime divisors of $\operatorname{LCM}(m, n)$, and

$$
m=p_{1}^{\alpha_{1}} \ldots p_{s}^{\alpha_{s}}, \quad n=p_{1}^{\beta_{1}} \ldots p_{s}^{\beta_{s}}
$$

where $\alpha_{i}, \beta_{j}$ can be zero (see fact 10). Then it follows from the above that $k=p_{1}^{\gamma_{1}} \ldots p_{s}^{\gamma_{s}}$, where $\gamma_{i}=\max \left(\alpha_{i}, \beta_{i}\right)$, when $\alpha_{i} \neq \beta_{i}$, and $\gamma_{i}$ is any number from 0 to $\alpha_{i}$, when $\alpha_{i}=\beta_{i}$. It can be shown that the reverse is also true: any such $k$ is the length of the minimal period of some fraction $A+B$, where the length of the minimal period of the fraction $A$ is $m$, and the length of the minimal period of the fraction $B$ is $n$.