2. Find the smallest natural number $n$, divisible by 20, for which $n^{2}$ is a perfect cube, and $n^{3}$ is a perfect square.
I/2. Since the number $n$ is divisible by 20, it is of the form $n=2^{2+a} \cdot 5^{1+b} \cdot k$, where $k$ is a natural number that is not divisible by either 2 or 5, and $a$ and $b$ are non-negative integers. If $n^{2}=2^{2(2+a)} \cdot 5^{2(1+b)} \cdot k^{2}$ is a perfect cube, then $3 \mid 2(2+a)$ and $3 \mid 2(1+b)$. If $n^{3}=2^{3(2+a)} \cdot 5^{3(1+b)} \cdot k^{3}$ is a perfect square, then $2 \mid 3(2+a)$ and $2 \mid 3(1+b)$. From this, it follows that $6 \mid 2+a$ and $6 \mid 1+b$, so $a$ is at least 4 and $b$ is at least 5. The smallest possible natural number $k$ is $k=1$. With these values, we get $n=2^{6} \cdot 5^{6}=1000000$. Since $n^{2}$ is a perfect cube and $n^{3}$ is a perfect square, this is the smallest such natural number.

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