We are looking for 5 digit multiples of 9 that don't have any 0s or 9s in them. Brute force counting looks awkward.

Multiples of 9 have the property that the sum of their digits will also be a multiple of 9.

Let the 5-digit number be abcde. Pick a, b, and c from {1, 2, ..., 8}. From here, it seems like it's a matter of organization to find combinations de (with no 0s or 9s) that can complete the sum and force the multiple of 9 to happen.

There are only 64 choices for the last two digits. We will sort them by their digit sums mod 9:

So for any choice of a, b, c, we will be in one of the two situations:
* If a+b+c is a multiple of 9, we will get 8 numbers.
* Otherwise, we will get 7 numbers.

Let's count the number of ways to get a multiple of 9. We can pick the first two digits in any combination that doesn't add up to 9, which forces the last digit. We have 8 choices for the first digit and 7 choices for the second digit (since there's only one digit will make the digit sum equal to 9).

So there are 56 ways to create a 3 digit number whose digits add up to a multiple of 9. We get 8 numbers for each of these, giving 56*8 = 448 numbers.

There are a total of 8^3 = 512 choices for the first three digits, so after removing the 56 from above, we have 456 numbers to go. Each o these contributes 7 numbers, giving 456*7 = 3192.

So in total, there are 448+3192 = 3640 such numbers.

This is very easy compared to the last one unless I'm misunderstanding something. S(n+1) equals the sum of the remaining digits of S(n) after removing the trailing nines, plus one. 2018 mod 9 = 2, so the smallest S(n+1) = 3.

f(4) = 4a-1
f'(x) = 2x+a
f'(4) = a+8
a+8 = 4a-1
a = 3