When decomposing numbers $A$ and $B$ into 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?
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 (we will consider the length of the minimal period of a finite decimal fraction to be one).

Lemma. 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 $A-B$.

Proof. 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. Similarly,

$B=\frac{Y}{10^{m}\left(10^{k}-1\right)}$. Without loss of generality, we can assume that $l \geq m$. 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 proof for $A-B$ is analogous.

Suppose fraction $A$ has a period of length 6, and fraction $B$ has a period of length 12. From the lemma, 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 remaining options are 12 and 4. We will show that both these options are possible:

$A=0,(000001), B=0,(000000000001), A+B=0,(000001000002)$

$A=0,(000001), \quad B=0,(011100110110), A+B=0,(0111)$