3. Find a three-digit number that, when multiplied by 7, gives a cube of a natural number.
Solution. Let the three-digit number be $a$. From the condition of the problem, it follows that $7 a=n^{3}, n \in \mathbb{N}$. Therefore, 7 is a divisor of $n^{3}$. Since 7 is a prime number, it follows that 7 is a divisor of $n$. Let $n=7 k$. Then $a=7^{2} k^{3}=49 k^{3}$. If $k \geq 3$, then $a$ has more than 3 digits. If $k=1$, then $a$ has two digits. Therefore, $k=2$ and $a=392$.