3. Find the largest natural number $n$, for which the number 999...99 (with 999 nines) is divisible by $9^{n}$.

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Solution. We write the number 999... 99 (with 999 nines) as $10^{999}-1$ and then factor it as follows

$$
\begin{aligned}
10^{999}-1= & \left(10^{666}+10^{333}+1\right)\left(10^{333}-1\right) \\
& =\left(10^{666}+10^{333}+1\right)\left(10^{222}+10^{111}+1\right)\left(10^{111}-1\right) \\
& =\left(10^{666}+10^{333}+1\right)\left(10^{222}+10^{111}+1\right)\left(10^{74}+10^{37}+1\right)\left(10^{37}-1\right) \\
& =\left(10^{666}+10^{333}+1\right)\left(10^{222}+10^{111}+1\right)\left(10^{74}+10^{37}+1\right) \cdot 9 \cdot \underbrace{111 \ldots .11}_{\text {37 ones }}
\end{aligned}
$$

Each of the first three factors is divisible by 3, because the sum of their digits is 3. The fourth factor is $3^{2}$, so it is divisible by $3^{2}$. The fifth factor is not divisible by 3, because the sum of its digits is 37. Therefore, the given number is divisible by $3 \cdot 3 \cdot 3 \cdot 3^{2}=3^{5}$, or $9^{2.5}$. Since $n$ is a natural number, we have $n=[2.5]=2$. Thus, the largest natural number $n$ for which the number 999... 99 (with 999 nines) is divisible by $9^{\text {n }}$ is 2.