Solution:

Given, total number of factors of a natural number N is 45.

We know that the number of factors of a number can be calculated using its prime factorization.

Let, the prime factorization of N be $p_1^{a_1}p_2^{a_2}p_3^{a_3}...p_k^{a_k}$

Then, the total number of factors of N is given by $(a_1+1)(a_2+1)(a_3+1)...(a_k+1) = 45$

We need to find the maximum number of prime numbers by which N is exactly divisible.

Let's consider the following cases:

Case 1: N is divisible by only one prime number

If N is divisible by only one prime number, then its prime factorization will be of the form $p^a$, where p is a prime number and a is a positive integer.

In this case, the total number of factors of N will be (a+1).

Given, (a+1) = 45, which gives a = 44.

So, N will be of the form $p^{44}$.

Therefore, in this case, N is divisible by only one prime number.

Case 2: N is divisible by two prime numbers

If N is divisible by two prime numbers, then its prime factorization will be of the form $p_1^{a_1}p_2^{a_2}$, where $p_1$ and $p_2$ are two different prime numbers and a1, a2 are positive integers.

In this case, the total number of factors of N will be (a1+1)(a2+1).

Given, (a1+1)(a2+1) = 45.

Possible pairs of (a1,a2) are (2,14), (4,8) and (6,6).

So, N will be of the form $p_1^2p_2^{14}$ or $p_1^4p_2^8$ or $p_1^6p_2^6$.

Therefore, in this case, N is divisible by two prime numbers.

Case 3: N is divisible by three prime numbers

If N is divisible by three prime numbers, then its prime factorization will be of the form $p_1^{a_1}p_2^{a_2}p_3^{a_3}$, where $p_1$, $p_2$ and $p_3$ are three different prime numbers and a1, a2, a3 are positive integers.

In this case, the total number of factors of N will be (a1+1)(a2+1)(a3+1).

Given, (a1+1)(a2+1)(a3+1) = 45.

Possible combinations of (a1,a2,a3) are (2,2,6), (2,3,4) and (3,3,3).

So, N will be of the form $p_1^2p_2^2p_3^6$ or $p_1^2p_2^3p_3^4$