If $A = \{x : x$ is a multiple of $4\}$ and $B = \{x : x$ is a multiple of $6\}$ then $A \cap B$ consists of all multiples of

If $A=\{3,6,9,12,15,18,21\}, B=\{4,8,12,16,20\},$ $C=\{2,4,6,8,10,12,14,16\}, D=\{5,10,15,20\} ;$ find

$D-A$

Show that $A \cap B=A \cap C$ need not imply $B = C$

Let $\mathrm{X}=\{\mathrm{n} \in \mathrm{N}: 1 \leq \mathrm{n} \leq 50\} .$ If $A=\{n \in X: n \text { is a multiple of } 2\}$ and $\mathrm{B}=\{\mathrm{n} \in \mathrm{X}: \mathrm{n} \text { is a multiple of } 7\},$ then the number of elements in the smallest subset of $X$ containing both $\mathrm{A}$ and $\mathrm{B}$ is

If $A=\{3,5,7,9,11\}, B=\{7,9,11,13\}, C=\{11,13,15\}$ and $D=\{15,17\} ;$ find

$A \cap C \cap D$

Sets $A$ and $B$ have $3$ and $6$ elements respectively. What can be the minimum number of elements in $A \cup B$