It can be seen from the question that 1, 3, 7 and 9 are all just finished, 9 is a multiple of 3, and 7 and 9 are coprime, so the least common multiple of 7 and 9 is 63, and the result must be a multiple of 63! 2, 4 and 8 all take 1, which means odd numbers, and 5 takes multiples of 4 and 5, so it can be judged that the unit is 9, so the product of 63 and the number with 3 can get the unit of 9. Try to calculate.

63x3= 189, if not satisfied, take 8 and leave1; 63x 13=8 19, which does not meet the requirements, and 8 is obtained, leaving1; 63x23= 1449, enough to take 8, leaving1;

There is no need to judge the remainder of 6, because a number minus 3 of 9 is a multiple of 2, and the characteristic of a multiple of 9 is that the sum of all numbers is a multiple of 9, so the sum of all numbers of 9 without a number is still a multiple of 9, and 6 minus 3 is also a multiple of 3, so it can be judged that the numbers that meet the above conditions must be divisible by 6 plus 3.

So the answer should be 1449!