Problem 624

$\triangle$ The desired number $x$ is divisible by 2 or 3. Since it is the smallest of all natural numbers with the given properties, its prime factorization consists only of twos and threes:

$$x=2^{k} \cdot 3^{l} \quad(k \in N, l \in N)$$

After multiplying the number $x$ by 2, the result is a perfect square, and after multiplying by 3, the result is a perfect cube, so

$$2 x=2^{k+1} \cdot 3^{\prime}=a^{2}, \quad 3 x=2^{k} \cdot 3^{\prime+1}=b^{2}$$

where $a$ and $b$ are natural numbers. Then

$$\begin{aligned} & (k+1): 2, \quad l: 2 \\ & k: 3, \quad(l+1): 3 \end{aligned}$$

We need to find the smallest $k$ and $l$ that satisfy these divisibility conditions: $k=3, l=2$. Therefore, $x=2^{3} \cdot 3^{2}=72$.

Answer: 72.