Find the intersection of the the sets given If $aN= \{ ax:x\in N \}$ then $3N\cap 7N$ =
Options are 
a)3N
b)7N
c)N
d)21N
I think the answer should be 3N as 7N would contain 3N and we have to find their intersection. But the answer is 21N, and I am not able to figure out the reason for that answer. 

Think about it this way:
$3N=\{n:n\in N\land 3\mid n\}$
$7N=\{n:n\in N\land 7\mid n\}$
For an two sets $A$ and $B$, such that $A=\{a:\varphi(a)\}$ and $B=\{b:\psi(b)\}$, the intersection is $A\cap B= \{c:\varphi(c)\land\psi(c)\}$
So...
$$3N\cap7N=\{n:n\in N\land (3\mid n)\land (7\mid n)\}$$
If $3$ divides $n$ and $7$ divides $n$, then $n$ must be a multiple of $21$ (since $3$ and $7$ are prime and $21$ is their least common multiple).
Hence, $3N\cap7N=21N$.
