1. Does there exist a three-digit number such that in its representation all digits are distinct and arranged in ascending order, in the representation of its square all digits are distinct and arranged in ascending order, and in the representation of its cube all digits are distinct and arranged in ascending order? (20 points)
Solution: Suppose such a number exists. Then, in the decimal representation of its square, there are no fewer than five digits, and in the representation of its cube, there are no fewer than seven digits. This means that the last digit of its square is no less than 5, and the last digit of its cube is no less than 7. The last digit of a square cannot be 5, because then the cube of the number would also end in 5, and it also cannot be 7 or 8, because such squares do not exist. If the last digit of the square is 6, then the three-digit number ends in either 4 (in which case its cube ends in 4) or 6 (in which case its cube ends in 6) - neither of these options work. If the last digit of the square is 9, then the three-digit number ends in either 7 (in which case its cube ends in 3, which is impossible) or 3. The latter is only possible if the three-digit number is 123. But \(123^2 = 15129\), which does not meet the condition. Therefore, such a number does not exist.

## Answer: does not exist