1. Find the smallest natural number $n$ for which the number $10 \cdot n$ is a perfect square, and the number $12 \cdot n$ is a perfect cube.
II/1. From the text of the problem, we understand that there exist such natural numbers $a$ and $b$ that $10 n=a^{2}$ and $12 n=b^{3}$. Therefore, $\frac{10}{12}=\frac{a^{2}}{b^{3}}$ or

$$
6 a^{2}=5 b^{3}
$$

Thus $6 \mid b^{3}$, so $6\left|b, 6^{3}\right| b^{3}$ and from this $6^{2} \mid a^{2}$ or $6 \mid a$. Similarly, we conclude that $5 \mid a^{2}$, so $5\left|a, 5^{2}\right| a^{2}$ and from this $5 \mid b^{3}$ and $5 \mid b$. Thus, the number $5 b^{3}$ is divisible by $5^{4}$ and therefore $5^{2} \mid a$.

The number $a$ must be divisible by at least $25 \cdot 6$, and the number $b$ by $5 \cdot 6$. Therefore, $a \geq 150$ and $b \geq 30$. We can verify that $a=150$ and $b=30$ satisfy the conditions of the problem, which gives us $n=\frac{a^{2}}{10}=2250$.
Introduction of $10 n=a^{2}$ and $12 n=b^{3}$ ..... 1 point
Equation $6 a^{2}=5 b^{3}$ ..... 1 point
Conclusion $6 \mid b$ and $6 \mid a$ ..... 1 point each
Conclusion $5 \mid b$ and $5^{2} \mid a$ ..... 1 point each
Minimal solution $n=2250$ 1 point