Problem 11.3. A natural number $n$ is called interesting if $2 n$ is a perfect square, and $15 n$ is a perfect cube. Find the smallest interesting number.
Answer: 1800.

Solution. Factorize the number $n$ into prime factors. For a number to be a square, it is necessary that all prime numbers in this factorization appear in even powers, and for a number to be a cube, it is necessary that all prime numbers appear in powers divisible by 3.

Let's look at the power of two that divides $n$. First, this power is odd, since $2 n$ is a perfect square. Second, this power is divisible by 3, since $15 n$ is a perfect cube. Therefore, the minimum power of two is 3.

Now let's look at the power of three that divides $n$. First, this power is even, since $2 n$ is a perfect square. Second, this power gives a remainder of 2 when divided by 3, since $15 n$ is a perfect cube. Therefore, the minimum power of three is 2.

Similarly, for five, we get that its minimum power is 2.

Therefore, $n$ is divisible by $2^{3} 3^{2} 5^{2}=1800$, i.e., $n \geqslant 1800$. It is not difficult to check that $n=$ 1800 satisfies all the conditions of the problem.