Here's the question:
Let $N$ be a positive integer, not divisible by $6$. Suppose $N$ has $6$ positive divisors, the number of positive divisors of $9N$ is:
I know how approach these questions by adding $1$ to the powers of the exponent and then multiplying them.
Here the result of the result will be $6*3$ , so the result will be 18 divisors because multiplication of N with $9$ will be $3^2$ which means extra power $2$.
I want to know where I'm wrong in my approach towards this concept.
The question has no options.
If $N$ has $6$ divisors it is of the form $p^2q$ or $p^5$ for $p,q$ prime. The fact that it is not divisible by $6$ means that at least one of $p,q$ is neither $2$ nor $3$, but that doesn't really help with the problem because we only care about factors of $3$. $9N$ could now be $3^4q,p^23^3,3^2p^2q, 3^7,\text { or } 3^2p^5$which have $10, 12,18,8\text { and } 18$ divisors respectively.
Added: $N$ could have $0,1,2, \text {or } 5$ factors of $3$. Adding two more multiplies the number of factors by $3,2,\frac 53, \text { or } \frac 43$ respectively, giving $18,12,10,8$ factors for $9N$