Let $S = \{1, 2, 3, \ldots, 40\}$ and let $A$ be a subset of $S$ such that no two elements in $A$ have their sum divisible by $5$. What is the maximum number of elements possible in $A$?

Let $S$ be the set of the first $11$ natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \ge 2 \text{ and the product of all elements of } B \text{ is even}\}$ is . . . . . . .

If $P(A) = \frac{2}{5}$,$P(B) = \frac{1}{4}$ and $P(A \cup B) = \frac{1}{2}$,then $P(A' \cup B') = $

$A$ number is chosen at random from the set $\{1, 2, 3, \ldots, 2000\}$. Let $p$ be the probability that the chosen number is a multiple of $3$ or a multiple of $7$. Then the value of $500p$ is . . . . . .

Let $S = \{4, 6, 9\}$ and $T = \{9, 10, 11, \ldots, 1000\}$. If $A = \{a_{1} + a_{2} + \ldots + a_{k} : k \in N, a_{1}, a_{2}, \ldots, a_{k} \in S\}$,then the sum of all the elements in the set $T - A$ is equal to:

The smallest set $A$ such that $A \cup \{1, 2\} = \{1, 2, 3, 5, 9\}$ is