What is the maximum number of prime numbers by which N can be divided? 
MyApproach
$$(a+1)(b+1)(c+1)\cdots=45$$
To see maximum number of prime numbers by which $N$ can be divided.
$a,b,c$ are the powers of prime numbers.
I took factors of $45$. I got $3^2 \cdot 5$. From this, I think maximum $3$ factors could be there that will make up $45$.
Is my approach right? Please correct me if I am wrong?
 A: Suppose $p,q,r$ are prime numbers, and consider $\{ p^i q^j r^k : 0\le i\le 2,\  0\le j\le 2,\ 0\le k\le 4 \}$.  This is a set of $45$ numbers, all factors of $N = p^2 q^2 r^4$.
The highest power of $p$ that divides a factor of $N$ is either $1$ or $p$ or $p^2$.
The highest power of $q$ that divides a factor of $N$ is either $1$ or $q$ or $q^2$.
The highest power of $r$ that divides a factor of $N$ is either $1$ or $r$ or $r^2$ or $r^3$ or $r^4$.
Three choices for the first, three for the second, and five for the third.
Since $3\times3\times5$ is the prime factorization of $45$, there is no way to write it as a product of more than three factors, so we cannot do what we did above with more prime numbers than $p,q,r$ and still get exactly $45$ factors.
Some numbers with exactly $45$ factors have exactly one prime factor $p$, and the $45$ factors are $p^i$, $i=0,1,2,\ldots,44$.
Some have two, $p$ and $q$, and the the $45$ factors are either
$$
p^i q^j, \quad i=0,1,2, \quad j=0,1,2,\ldots,14
$$
or
$$
p^i q^j, \quad i=0,1,2,3,4, \quad j=0,1,2,\ldots,8.
$$
