Which are those positive integers $k$ for which $2 \cdot 3^{k}$ is a perfect number?
I. solution. A number is a perfect number if it is equal to the sum of its proper divisors. The proper divisors of $2 \cdot 3^{k}$ are $1,3, \ldots, 3^{k}, 2,2 \cdot 3, \ldots, 2 \cdot 3^{k-1}$. If it is a perfect number, the following equation can be written:

$$
1+3+\ldots+3^{k}+2+2 \cdot 3+\ldots+2 \cdot 3^{k-1}=2 \cdot 3^{k}
$$

Adding $2 \cdot 3^{k}$ to both sides and using the formula for the sum of a geometric series:

$$
\begin{aligned}
1+3+\ldots+3^{k}+2+2 \cdot 3+\ldots+2 \cdot 3^{k} & =4 \cdot 3^{k} \\
3\left(1+3+\ldots+3^{k}\right) & =4 \cdot 3^{k} \\
3 \cdot \frac{\left(3^{k+1}-1\right)}{2} & =4 \cdot 3^{k} \\
3^{k+2}-3 & =8 \cdot 3^{k} \\
(9-8) 3^{k} & =3 \\
3^{k} & =3
\end{aligned}
$$

All transformations were equivalent.

The function $3^{x}$ is strictly increasing, so the only solution is $k=1$.

II. solution. For $k=1$, $2 \cdot 3^{k}$ is a perfect number, since $2 \cdot 3^{1}=6=1+2+3$.

For $k \geq 2$, $3^{k-2}, 3^{k-1}$, $3^{k}$, and $2 \cdot 3^{k-1}$ are all distinct positive integers and divisors of $2 \cdot 3^{k}$ (since their prime factorizations are different and no prime factor's exponent is greater than the corresponding prime factor's exponent in $2 \cdot 3^{k}$).

Thus, the sum of the divisors of $2 \cdot 3^{k}$ is at least

$$
3^{k-1}+2 \cdot 3^{k-1}+3^{k}+3^{k-2}=3 \cdot 3^{k-1}+3^{k}+3^{k-2}=2 \cdot 3^{k}+3^{k-2}
$$

This is certainly greater than $2 \cdot 3^{k}$.

Therefore, $2 \cdot 3^{k}$ is a perfect number only when $k=1$.

III. solution. The number $2 \cdot 3^{k}$ is certainly even, as it is divisible by two but not by a higher power of two. A known theorem (found in Zsuzsanna Karlik's work "Perfect Numbers" along with the proof) states that an even number $n$ is perfect if and only if $n=2^{p-1} \cdot (2^{p}-1)$, and both $(2^{p}-1)$ and $p$ are prime numbers.

In this case, $2 \cdot 3^{k}=2^{p-1} \cdot (2^{p}-1)$ must hold for some prime number $p$. Since $2^{p}-1$ is odd, $2=2^{p-1}$, which implies $p=2$ due to the strict monotonicity of the $2^{x}$ function. Thus, $2 \cdot 3^{k}=2^{2-1} \cdot (2^{2}-1)=2 \cdot 3$, so $k=1$ is the only solution.

Remark. Some used Euler's theorem and sought the solution in the form $2^{n}(2^{n+1}-1)$. A common mistake in this line of thought was that the solvers did not consider that the existence of odd perfect numbers is an open question, and they used Euler's theorem in general.

IV. solution. If $k=1$, the number is perfect: $6=1 \cdot 2 \cdot 3=1+2+3$.

Let's see what happens when we go from $k$ to $(k+1)$: how will the sum of the divisors and the number itself change.

The proper divisors of $2 \cdot 3^{k}$: $2 \cdot 3^{0}, 2 \cdot 3^{1}, \ldots, 2 \cdot 3^{k-1} ; 1,3^{1}, 3^{2}, \ldots, 3^{k}$.

The proper divisors of $2 \cdot 3^{k+1}$: $2 \cdot 3^{0}, 2 \cdot 3^{1}, \cdots, 2 \cdot 3^{k} ; 1,3^{1}, 3^{2}, \cdots, 3^{k}, 3^{k+1}$.

Thus, the differences are as follows: $2 \cdot 3^{k+1}-2 \cdot 3^{k}=2 \cdot 3 \cdot 3^{k}-2 \cdot 3^{k}=4 \cdot 3^{k}$ is the difference between the two numbers, and $2 \cdot 3^{k}+3^{k+1}=2 \cdot 3^{k}+3 \cdot 3^{k}=5 \cdot 3^{k}$ is the difference between the sums of the divisors. This means that for $k>1$, the sum of the proper divisors will be greater than the number.