1. Determine the smallest natural number $n$ for which the numbers $\sqrt{2 n}, \sqrt[3]{3 n}, \sqrt[5]{5 n}$ are natural.
SOLUTION. We will explain why the prime factorization of the sought number must contain only suitable powers of the primes 2, 3, and 5. Any additional prime number would have to appear in the factorization of the number $n$ in a power whose exponent is divisible by two, three, and five simultaneously (see introductory problem 1). After striking out such a prime number, the number $n$ would decrease, and the examined roots would remain integer.

Let us therefore set $n=2^{a} 3^{b} 5^{c}$, where $a, b, c$ are natural numbers. The numbers $\sqrt[3]{3 n}$ and $\sqrt[5]{5 n}$ are integers, so the exponent $a$ must be a multiple of three and five. Also, $\sqrt{2 n}$ is an integer, so the number $a$ must be odd. Therefore, it is an odd multiple of fifteen: $a \in\{15,45,75, \ldots\}$. Analogously, the exponent $b$ is such a multiple of ten that when divided by three, it gives a remainder of 2: $b \in\{20,50,80, \ldots\}$. The number $c$ is then a multiple of six that when divided by five gives a remainder of 4: $c \in\{24,54,84, \ldots\}$. From the condition that $n$ is the smallest, it finally follows that $n=2^{15} 3^{20} 5^{24}$.

Let us also verify that the given roots are natural numbers:

$$
\sqrt{2 n}=2^{8} 3^{10} 5^{12}, \quad \sqrt[3]{3 n}=2^{5} 3^{7} 5^{8}, \quad \sqrt[5]{5 n}=2^{3} 3^{2} 5^{5} .
$$

Conclusion: $n=2^{15} 3^{20} 5^{24}$.

EXERCISES FOR PRACTICE: