Question

If $A=\{x:xisamultipleof4\}andB=\{x:xisamultipleof6\},$ then $A\cap B$ consists of all multiples of

Answer 1

Finding the multiples of $\mathit{A}\mathbf{\cap }\mathit{B}$:

Given $A=\{x:xisamultipleof4\}andB=\{x:xisamultipleof6\},$

$$\begin{gathered} SoA=\{4,8,12,16,20\dots \} \\ B=\{6,12,18,24\dots ..\} \\ A\cap B=\{12,24,36\dots .\} \end{gathered}$$

$A\cap B$ has multiples of $12$

$A\cap B$ has multiples of $4$

Hence, the set $\mathit{A}\mathbf{\cap }\mathit{B}$ consists of all multiples of $\mathbf{4}and\mathbf{12}$.