## Task $2 / 83$

We are looking for all natural numbers $n$ for which $3^{n}$, when written in decimal notation, has exactly $0.5 n$ digits.
According to the task, the following should apply:

$$
3^{n}=m \cdot 10^{0.5 n-1}
$$

where $m$ is the rational number with $0.5 n-1$ decimal places, $1 \leq m0$, from which it follows that

$$
n \leq \frac{2}{1-2 \lg 3}=43, \ldots
$$

Since $0.5 n$ must be an integer, all even natural numbers $n=2 i$ with $i \in N, 1 \leq i \leq 21$ satisfy the requirements of the task. Verification with a calculator of sufficient capacity confirms the correctness:

$$
3^{42}=1, \ldots \cdot 10^{20} \quad ; \quad 3^{44}=9, \ldots \cdot 10^{20}
$$

i.e., up to $i=21, n=42$ the condition holds, from $i=22, n=44$ onwards it no longer holds.