P is a natural number. 2P has 28 divisors and 3P has 30 divisors. How many divisors of 6P will be there? While answering Aptitude questions in a book I faced this question but not able to find the solution So I searched Google and got two answers but didn't get an idea how the answer came. 
Question: 
P is a natural number. 2P has 28 divisors and 3P has 30 divisors. How many divisors of 6P will be there?
Solution 1:
2P is having 28(4*7) divisors but 3P is not having a total divisors which is divisible by 7, so the first part of the number P will be 2^5.
Similarly, 3P is having 30 (3*10) divisors but 2P does not have a total divisors which is divisible by 3. So 2nd part of the number P will be 3^3. So, P = 2^5*3^3 and the solution is 35.
Solution 2:
2P has 28 divisors =4x7,
3P has 30 divisors 
Hence P=2^5 3^3
6p =2^6 3^4
Hence 35 divisors 
I have been trying to understand the steps but not able to get.

First, we want to know how to easily calculate the number of divisors of any number. If we have $n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ where all $p_i$ are distinct, then to construct a divisor we have $e_1+1$ choices for the number of factors $p_1$ in our divisor, $e_2+1$ choices for the number of factors $p_2$, etc., making the number of divisors equal to $(e_1+1)(e_2+1)\cdots (e_k+1)$. So for example, $12=2^2\cdot 3$ so $12$ has $(2+1)(1+1)=6$ divisors.
Let's look at the first hint now, $2P$ having 28 divisors. Let's write $2P=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$. The number of divisors is now $(e_1+1)(e_2+1)\cdots (e_k+1)=28$. So, we can say that one power, say $e_1$, must be either $6,13$ or $27$ (since 28 has a factor 7, and we assumed it was contained in $e_1+1$. Our options are now:
$$2P=p_1^6\cdot p_2\cdot p_3$$
$$2P=p_1^6\cdot p_2^3$$
$$2P=p_1^{13}\cdot p_2$$
$$2P=p_1^{27}$$
The second hint says that 3P has 30 divisors. Since this does not contain a factor 7, we know that the $2$ in $2P$ must be responsible for this (notice that follows $e_1$ is the power of 2 in $2P$). Thus we know that $p_1=2$. Now we can, by our previously stated options, calculate P.
$$P=2^5\cdot p_2\cdot p_3$$
$$P=2^5\cdot p_2^3$$
$$P=2^{12}\cdot p_2$$
$$P=2^{26}$$
In the third option, the number of divisors of $3P$ must be divisible by $12+1=13$, and in the last case, the number of divisors of $3P$ must be divisible by $27$. We conclude the last two are not possible, since 30 is not divisible by either 13 or 27. In the first case, the number of divisors of $3P$ is divisible by $6$ because of the amount of factors 2, and we also know that $3P$ there has (at least) one prime factor that it only contains one time, so we have another factor 2 in the number of divisors of $3P$. Now the number of divisors of $3P$ is divisible by $12$, which is also impossible ($12$ does not divide $30$). We conclude that $P$ must be of the form $p_1^5\cdot p_2^3$. We also know that if $p_2\neq 3$, then the number of divisors of $3P$ is divisible by $6$ because of the factors 2 and by $4$ because of the factors $p_2$. This is again impossible.
Finally, we must have $p_2=3$ so $P=2^5\cdot 3^3$. Now we can easily calculate the number of divisors of $6P$; it must be $(6+1)(4+1)=35$.
Hope this helped!
