The product of two numbers is $4107$. If the H.C.F. of these numbers is $37$, then the greater number is

The correct option is B $111$
Let the numbers be $37a$ and $37b$.
Then, $37a\times 37b=4107$
$\Rightarrow ab=3$.
Now, co-primes with product $3$ are $(1,3)$.
So, the required numbers are $(37\times 1,37\times 3)$ i.e., $(37,111)$.
$\therefore$ Greater number $=111$.