2A. The numbers $5, 52, 525, 5252, 52525, \ldots$ are given. Determine the smallest among them that is divisible by 99.
Solution. Let $S_{1}$ be the sum of the digits in the odd positions, and $S_{2}$ be the sum of the digits in the even positions of an arbitrary number $a_{N}$ from the given sequence. A number is divisible by 11 if $11 \mid S_{1}-S_{2}$. A number is divisible by 9 if $9 \mid S_{1}+S_{2}$.

1) Let the number of digits $N$ be even. Then the sum of the digits of the number $a_{N}$ is equal to $\frac{7 N}{2}$, and $S_{1}-S_{2}=\frac{3 N}{2}$. For the number $a_{N}$ to be divisible by 99, it must be that $9 \mid S_{1}+S_{2}$ and $11 \mid S_{1}-S_{2}$. This condition is satisfied by all numbers $N$ for which $99 \mid N$, i.e., $N=99 k$. The smallest $k$ for which this is true is $k=2$. The smallest such $a_{N}$ will have $2 \cdot 99=198$ digits. ( $N=198$ ).
2) Let $N$ be odd. Then

$$
\begin{aligned}
& S_{1}+S_{2}=\frac{7(N-1)}{2}+5=\frac{7 N+3}{2} \\
& S_{1}-S_{2}=\frac{3(N-1)}{2}+5=\frac{3 N+7}{2}
\end{aligned}
$$

Furthermore,

$$
\left.9\left|S_{1}+S_{2} \Leftrightarrow 9\right| \frac{7 N+3}{2} \Leftrightarrow 9 \right\rvert\,(7 N+3) \Leftrightarrow N \equiv 5(\bmod 9)
$$

In addition,

$$
\begin{aligned}
& \left.11\left|S_{1}-S_{2} 11\right| \frac{3 N+7}{2} \Leftrightarrow 11 \right\rvert\,(3 N+7) \Leftrightarrow \\
& 11|4(3 N+7) \Leftrightarrow 11|(N+6) \Leftrightarrow N \equiv 5(\bmod 11)
\end{aligned}
$$

The smallest odd number $N$ such that $N \equiv 6(\bmod 9)$ and $N \equiv 5(\bmod 11)$ is 159.