Let's determine those positive integers which are one and a half times as large as the product of their digits.
Solution. For the $n$-digit integer, we write the condition as:

$$
\overline{a_{n} a_{n-1} a_{n-2} \ldots a_{2} a_{1}}=\frac{3}{2} \cdot a_{n} \cdot a_{n-1} \cdot \ldots \cdot a_{2} a_{1}
$$

where $a_{i} \neq 0$ and $1 \leq a_{i} \leq 9(i=1,2, \ldots, n)$.

Since no digit can be 0 (otherwise, the right side would be 0), the following two inequalities are true:

$$
\begin{gathered}
\overline{a_{n} a_{n-1} a_{n-2} \ldots a_{2} a_{1}}>\overline{a_{n} \underbrace{00 \ldots 0}_{n-1}}=a_{n} \cdot 10^{n-1}, \quad \text { and } \\
\frac{3}{2} \cdot a_{n} \cdot 9^{n-1} \geq \frac{3}{2} \cdot a_{n} \cdot a_{n-1} \cdot \ldots \cdot a_{2} \cdot a_{1} .
\end{gathered}
$$

By comparing these two inequalities with equation (1), we get:

$$
\frac{3}{2} \cdot a_{n} \cdot 9^{n-1}>a_{n} \cdot 10^{n-1}
$$

Dividing by $a_{n} \neq 0$, we get $\frac{3}{2}>\left(\frac{10}{9}\right)^{n-1}$. Since the exponential function with a base greater than 1 is strictly increasing, we get:

$$
(n-1) \cdot \lg \frac{10}{9} < \lg \frac{3}{2}
$$

Since $\lg \frac{10}{9} > 0$, we have $n-1 < \frac{\lg \frac{3}{2}}{\lg \frac{10}{9}} \approx 2.71$, so $n \leq 3$.

For a two-digit number:

$$
\overline{a_{2} a_{1}}=\frac{3}{2} \cdot a_{2} \cdot a_{1}, \quad \text { hence } \quad a_{2}=\frac{10 a_{1}}{3 a_{1}-20}
$$

Since $a_{2}>0$, we have $3 a_{1}>20$. The possible values are: $a_{1}=7,8$ or 9. Substituting into (2): $a_{2}=\frac{14}{1}$, not a single digit, $a_{2}=\frac{16}{24-20}=4$ and $a_{2}=\frac{18}{7}$ not an integer. Only when $a_{1}=8$, we get a solution, and then $a_{2}=4$, and indeed: $48=\frac{3}{2} \cdot 4 \cdot 8$.

For three-digit numbers:

$$
\overline{a_{3} a_{2} a_{1}}=\frac{3}{2} \cdot a_{3} \cdot a_{2} \cdot a_{1}, \quad \text { hence } \quad a_{3}=\frac{20 a_{2}+2 a_{1}}{3 \cdot a_{2} \cdot a_{1}-200}
$$

Since $a_{3}>0$, from the condition $a_{2} a_{1} \geq 67$, the following possible values are obtained: $a_{1}=8, a_{2}=9 ; a_{1}=9, a_{2}=8 ; a_{1}=9$ and $a_{2}=9$. Substitution shows that in none of these cases do we get a single-digit integer value for $a_{3}$.

Finally, for four-digit numbers, similarly:

$$
a_{4}=\frac{200 a_{3}+20 a_{2}+2 a_{1}}{3 \cdot a_{3} \cdot a_{2} \cdot a_{1}-2000}
$$

From here, $a_{3} a_{2} a_{1} \geq 667$, which is only true for $a_{3}=a_{2}=a_{1}=9$, but this does not give an integer solution for $a_{4}$.

The only solution to the problem is 48.