7.4. Find the largest natural number, all digits of which are different, and the product of these digits is a cube of some natural number.
Answer: 984321. Solution. Obviously, among the digits of the desired number $x$, there is no zero. The product of all digits from 1 to 9 is $2^{7} \cdot 3^{4} \cdot 5^{1} \cdot 7^{1}$. Therefore, $x$ cannot contain the digit 5 or the digit 7 (otherwise, the product of the digits of the number $x$ would have to be divisible by $5^{3}$ or $7^{3}$). This means that $x$ contains no more than seven of the remaining digits, and $2^{7} \cdot 3^{4}$ is divisible by the product of the digits of the number $x$. Since $2^{7} \cdot 3^{4}$ is not a cube of a natural number, $x$ cannot contain all seven of these digits. If we remove the digit 6, the remaining six digits will give a product of $2^{6} \cdot 3^{3}=12^{3}$. Removing any other digit instead of 6 clearly does not yield a product equal to the cube of a natural number. Removing several digits results in a number consisting of five or fewer digits. Therefore, $x$ consists of the digits 1, 2, 3, 4, 8, 9, and the requirement for maximality leads to the result: $x=984321$.