A number of ten different digits, divisible by 8, 9, 10, and 11

Question

Each of the digits 0 through 9 is used exactly once to create a ten-digit integer. Find the greatest ten-digit number which uses each digit once and is divisible by 8, 9, 10, and 11.

Answer 1

For a multiple of 10

The number must end with a 0. So put it at the end.

For a multiple of 9

The sum of digits must be a multiple of 9. This is the case when using all digits 0 to 9. So you don't need to worry about that.

For a multiple of 8

The last 3 digits must form a multiple of 8. Since the last digit is 0, the 2 digits before the 0 must be a multilpe of 4.

For a multiple of 11

The sum of digits at odd positions minus the sum of digits at even positions must be a multiple of 11. This requires some calculations.
Starting from 9876543210, the odd digits are 9,7,5,3,1 with sum 25. The even digits are 8,6,4,2,0, summing to 20.
To get a multiple of 11, you can swap some digits to transfer 3 units to the odd digits. This gives 25+3 = 28 for the odds, 20-3 = 17 for the evens, and 28-17 = 11.
The simplest swap (affecting the smallest digits) seems to be to swap the 1 and the 4.

This results in the number 9876513240. Luckily, 240 is a multiple of 8. (or 24 is a multiple of 4).

So the number is

9876513240.

One can verify easily that is is a multiple of 8, 9, 10 and 11.
It is even a multiple of all integers from 3 to 12, except 7.

Note

Note that swapping 1 and 4 is only an educated guess. To prove that this choice is optimal note that any permutation that affects larger digits than 4 results in a smaller number. So you only need to try permutations of digits 1,2,3,4 to see there is no better way to fix the multiple of 11.
Being a multiple of 8 means the number must terminate with 120, 320 or 240. This reduces the number candidates from 24 to 6.