For any finite non empty set $$X$$ of integers, let $$\max(X)$$ denote the largest element of $$X$$ and $$|X|$$ denote the number of elements in $$X$$. If $$N$$ is the number of ordered pairs $$(A, B)$$ of finite non-empty sets of positive integers, such that $$\max(A)\times|B|=12$$ and $$|A|\times\max(B)=11$$, and $$N$$ can be written as $$100a+b$$ where $$a, b$$ are positive integers less than $$100$$, find $$a+b$$.

Let $$|A|=m,\; \max(A)=p,\; |B|=n,\; \max(B)=q$$.
The two given relations translate to

$$p\,n = 12 \qquad -(1)$$ $$m\,q = 11 \qquad -(2)$$

Because all quantities are positive integers, we analyse the factorisations of the right-hand sides.

Since $$11$$ is prime, equation $$(2)$$ gives only two possibilities: Case 1: $$m=1,\; q=11$$   or   Case 2: $$m=11,\; q=1$$.

Case 1: $$m=1,\; q=11$$

Here $$A$$ has exactly one element, so $$p=\max(A)$$ equals that lone element. Equation $$(1)$$ becomes $$p\,n = 12$$. All factor pairs $$(p,n)$$ of $$12$$ are

$$(1,12),\; (2,6),\; (3,4),\; (4,3),\; (6,2),\; (12,1).$$

Because $$B$$ is a set (no repetition) whose largest element is $$11$$, its size cannot exceed $$11$$. Thus the pair $$(1,12)$$ is impossible. The remaining five pairs are valid.

For a fixed $$(p,n)$$, set $$A=\{p\}$$ is unique.
Set $$B$$ must contain $$11$$ and exactly $$n-1$$ other distinct numbers chosen from $$\{1,2,\dots,10\}$$. Therefore the number of such $$B$$ is $$\binom{10}{\,n-1\,}$$.

Counting for each valid $$n$$:

$$n=6:\; \binom{10}{5}=252$$
$$n=4:\; \binom{10}{3}=120$$
$$n=3:\; \binom{10}{2}=45$$
$$n=2:\; \binom{10}{1}=10$$
$$n=1:\; \binom{10}{0}=1$$

Total pairs in Case 1: $$252+120+45+10+1 = 428$$.

Case 2: $$m=11,\; q=1$$

Since $$\max(B)=1$$, the only possible set is $$B=\{1\}$$, giving $$n=1$$.
Equation $$(1)$$ now yields $$p\,\cdot 1 = 12 \implies p=12$$.

Set $$A$$ must have size $$11$$ with maximum $$12$$, so it must be $$\{12\}$$ together with any $$10$$ distinct numbers from $$\{1,2,\dots,11\}$$. Number of choices for $$A$$: $$\binom{11}{10}=11$$. With $$B$$ fixed, pairs in Case 2 = $$11$$.

Total ordered pairs: $$N = 428 + 11 = 439$$.

Express $$N$$ as $$100a + b$$ with $$a,b \lt 100$$: $$439 = 100 \times 4 + 39 \implies a = 4,\; b = 39$$.

Hence $$a + b = 4 + 39 = 43$$.

Final answer: 43