Find all positive integers that have as many divisors divisible by six as divisors not divisible by six.
Solution. Let $n$ be a number that satisfies the condition. Write the prime factorization of $n$:

$$
n=2^{x} 3^{y} p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}
$$

where $x, y \geq 1$, otherwise the number of divisors divisible by 6 would be 0, but the number of divisors is never 0. From here, the number of divisors divisible by 6 is:

$$
x \cdot y \cdot\left(\alpha_{1}+1\right) \cdot\left(\alpha_{2}+1\right) \cdot \ldots \cdot\left(\alpha_{k}+1\right)
$$

and the number of all divisors is:

$$
(x+1) \cdot(y+1) \cdot\left(\alpha_{1}+1\right) \cdot \ldots \cdot\left(\alpha_{k}+1\right)
$$

From the condition of the problem,

$$
2 x y=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{k}+1\right)=(x+1)(y+1)\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{k}+1\right)
$$

and from here $2 x y=(x+1)(y+1)$. Performing the indicated operations, we arrive at the following Diophantine equation: $x y=x+y+1$. Solve the equation in the usual way:

$$
\begin{aligned}
& x(y-1)=y+1 \\
& x=\frac{y+1}{y-1}=\frac{y-1+2}{y-1}=1+\frac{2}{y-1}
\end{aligned}
$$

Since $x, y$ are positive integers, $y-1$ must be a divisor of 2. We only need to consider the divisors for which $x \geq 1$, that is, $y-1=1$ or $y-1=2$. The solutions to the equation are the pairs $(3 ; 2)$ or $(2 ; 3)$. The solutions to the problem are given by numbers of the form $72 n$ or $108 n$, where $n$ is not divisible by either 2 or 3.