Question

On the set of all natural numbers $N,$ set $F_n$ and set $M_n$ gives all factors and all multiples of $n$ respectively for all $n\in N.$ Which of the following statements is/are true?
$1.F_{132}\cap F_{96}=F_{12}2.M_{12}\cup M_{18}=M_{36}3.M_6\cap M_9=M_{18}4.F_{72}\cap F_{60}=F_{12}$

• A. $1,2$ and $3$ only
• B. $1$ and $3$ only
• C. $1,3$ and $4$ only
• D. All statements are true

Answer 1

The correct option is C $1,3$ and $4$ only

$1.F_{132}\cap F_{96}=F_{12}$
$F_{132}\cap F_{96}$ is set of all common factors of $96$ and $132.$
$\text{HCF}(96,132)=12\Rightarrow F_{132}\cap F_{96}=F_{12}$
So, statement $1$ is true.

$2.M_{12}\cup M_{18}=M_{36}$
$M_{12}=\{12,24,36,48,...\}$
$M_{18}=\{18,36,54,72,...\}$
But numbers like $\{12,18,24,...\}\notin M_{36}$
So, statement $2$ is not true.

$3.M_6\cap M_9=M_{18}$
$M_6\cap M_9$ is the set of all common multiples of $6$ and $9.$
$\text{LCM}(6,9)=18\Rightarrow M_6\cap M_9=M_{18}$
So, statement $3$ is true.

$4.F_{72}\cap F_{60}=F_{12}$
$F_{72}\cap F_{60}$ is set of all common factors of $60$ and $72.$
$\text{HCF}(60,72)=12\Rightarrow F_{72}\cap F_{60}=F_{12}$
So, statement $4$ is true.
Hence, statements $1,3$ and $4$ are true.