10.4. Find the largest natural number, all digits of which are different, and the product of these digits is a square of a natural number.
Answer: 986431. Solution: Obviously, among these digits there is no zero. Further, we have $1 \cdot 2 \cdot \ldots \cdot 9=2^{7} \cdot 3^{4} \cdot 5^{1} \cdot 7^{1}$. Therefore, we need to remove the digits 5 and 7, and also need to make the odd power of two even. This means we need to remove the digit 2: obviously, we should not remove the digits 6 and 8, as we need the maximum result. Clearly, the remaining digits should be arranged in descending order: 986431. As a result, the product of these digits will be $5184=(72)^{2}$.