How many factors does 6N have? Given a number $2N$ having 28 factors another number $3N$ having 30 factors, then find out the number of factors of $6N$.  

If you multiply a number $n$ by any prime number, then its number of divisors is multiplied by $\frac{m+1}m$ where $p^{m-1}$ is the highest power of $p$ dividing$~n$. This is because the multiplicity of $p$ in divisors of $n$ can be any one of the $m$ values $0,1,\ldots,m-1$, but that multiplicity can be chosen among the $m+1$ values $0,1,\ldots,m$ for divisors of $pn$ (for prime numbers other than $p$, the same set of multiplicities is available in both cases).
So in the problem here let $k$ be the number of divisors of $n$ without factors $2$ or $3$, and $a-1,b-1$ the multiplicities of the primes $2$ and $3$ in$~n$, respectively. So $k,a,b$ are positive integers and $n$ has $kab$ divisors in all. We are given that $28=k(a+1)b$ and $30=ka(b+1)$, which implies that $\frac{28}{30}=\frac{14}{15}=\frac{(a+1)b}{a(b+1)}$. The factor $7$ in the numerator cannot come from $b$ (given that $b+1$ divides $~30$ and that $\frac{a+1}a\neq1$), so it must come from the factor $a+1$. Since $a$ divides $30$, the only possibility is $a=6$, which indeed gives a solution namely $(k,a,b)=(1,6,4)$. So we find $n=k2^{a-1}3^{b-1}=2^53^3$, and $6n=2^63^4$ has $k(a+1)(b+1)=35$ divisors.
