Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ______.

We need to find the sum of all integral common differences $$d$$ of A.P.'s with first term $$a = 100$$, last term $$l = 199$$, having at least 3 and at most 33 terms.

For an A.P. with $$n$$ terms: $$l = a + (n-1)d$$

$$199 = 100 + (n-1)d \implies (n-1)d = 99$$

We need $$d$$ to be a positive integer (since $$l > a$$) and $$(n-1) = \dfrac{99}{d}$$ must be a positive integer, so $$d$$ must divide 99.

Also, $$3 \leq n \leq 33$$, which means $$2 \leq n-1 \leq 32$$, so $$2 \leq \dfrac{99}{d} \leq 32$$.

$$99 = 9 \times 11 = 3^2 \times 11$$

Divisors of 99: 1, 3, 9, 11, 33, 99.

For each divisor $$d$$, compute $$n - 1 = 99/d$$:

$$d = 1$$: $$n-1 = 99$$, $$n = 100 > 33$$. Not valid.

$$d = 3$$: $$n-1 = 33$$, $$n = 34 > 33$$. Not valid.

$$d = 9$$: $$n-1 = 11$$, $$n = 12$$. Valid ($$3 \leq 12 \leq 33$$).

$$d = 11$$: $$n-1 = 9$$, $$n = 10$$. Valid.

$$d = 33$$: $$n-1 = 3$$, $$n = 4$$. Valid.

$$d = 99$$: $$n-1 = 1$$, $$n = 2 < 3$$. Not valid.

$$9 + 11 + 33 = 53$$

The correct answer is $$\boxed{53}$$.