If $A$ and $B$ are subsets of universal set $X$ such that $n(X)=200, n(A)=90, n(B)=80$ and $n(A' \cap B')=40$,then $n(A \cap B')=$

Let $A = \{n \in N \mid n^{2} \leq n + 10,000\}$,$B = \{3k + 1 \mid k \in N\}$,and $C = \{2k \mid k \in N\}$. Then the sum of all the elements of the set $A \cap (B - C)$ is equal to $.....$

Let $A_1, A_2, \ldots, A_m$ be non-empty subsets of $\{1, 2, 3, \ldots, 100\}$ satisfying the following conditions:
$1.$ The numbers $|A_1|, |A_2|, \ldots, |A_m|$ are distinct.
$2.$ $A_1, A_2, \ldots, A_m$ are pairwise disjoint.
(Here $|A|$ denotes the number of elements in the set $A$).
Then,the maximum possible value of $m$ is:

The number of elements in the set $\{n \in \{1, 2, 3, \ldots, 100\} \mid (11)^{n} > (10)^{n} + (9)^{n}\}$ is $.....$

If $n(X)=700, n(A)=200, n(B)=300,$ and $n(A \cap B)=100$,where $X$ is the universal set and $A$ and $B$ are subsets of $X$,then $n(A' \cap B')=$

In a school,there are $21$ students in the cricket team,$26$ in the hockey team,and $29$ in the football team. Among these,$14$ play both hockey and cricket,$15$ play both hockey and football,and $12$ play both football and cricket. If $8$ students play all three games,find the total number of students in the three athletic teams.