Question

If $a.N=\{ax:x\in N\}$ then $3N\cap 7N=$

• A. $21N$
• B. $10N$
• C. $4N$
• D. $none$

Answer 1

The correct option is A $21N$

Clearly the set $aN$ will include all multiples of $a$.

Hence $3N$ and $7N$ contain multiples of 3 and 7 respectively.

Their intersection must have the multiples of their L.C.M,i.e,21

Hence $3N\cap 7N=21N$