So this 28-digit number needs to be divisible by 9, 4 and 1 1 at the same time.

Numbers divisible by 4, ten digits, each digit must be divisible by 4. Now the last two digits are 76, which can be divisible by 4, so this 28-digit number can be divisible by 4.

A number divisible by 9. The sum of all numbers should be a multiple of 9.

Now this 28-digit number, after all 0 to 9 are filled in, the sum of all the numbers is 135, which is a multiple of 9.

So this 28-digit number must also be a multiple of 9.

So the question is simplified as: What is the possibility that this 28-digit number is a multiple of 1 1?

A number divisible by 1 1 The difference between the sum of all odd numbers and all even numbers is an integer multiple of 1 1.

Then the sum of odd digits is:

5 + 3 + 3 + 8 + 2 + 9 + 6 + 5 + 8 + 2 + 3 + 9 + 3 + 7 = 73

The sum of even numbers is:

8 + 3 + 0 + 6 + (0+ 1+2+3+4+5+6+7+8+9) = 62

73-62= 1 1

So this 28-digit number is also an integer multiple of 1 1.

To sum up, this 28-digit number must be an integer multiple of 396, no matter how 0 to 9 are filled in.

I hope it works.