If a perfect square $N$ has nine factors less than its square root, how many factors does $N^5$ have? A perfect square number $N$ has nine factors below its square root. Find the number of factors of $N^5$.
I want to know: if a square root has $9$ factors below the square root then how come it can have $9$ factors above the square root?
 A: If $N$ has $9$ factors less than its square root, it must have $9$ factors above its square root, too. Since it is a perfect square, its square root is also a factor, so $N$ has $9+1+9=19$ total factors.
If the prime factorization of $N$ is given by $$\Pi_i p_i^{\alpha_i}$$ then we've shown that $$\Pi_i (\alpha_i+1)=19$$ But $19$ is prime. So we conclude $\alpha_i=18$ for some $i$, and all other $\alpha_j$ are $0$.
Therefore $N$ is the $18$th power of some prime number. Hence $N^5$ is the $90$th power of some prime number and thus has $91$ factors.

EDIT
Note that the factors of a positive integer $N$ come in pairs, arranged symmetrically around the square root of $N$. Take $N=60$, for example. The factors of $60$, listed in increasing order, are
$$\color{red}{1, 2, 3, 4, 5, 6}\,\,\,\, \color{blue}{10, 12, 15, 20, 30, 60}$$
The red factors are below $\sqrt{60}\approx 7.7$, and the blue factors are above $\sqrt{60}$. Indeed, each factor below the square root pairs up with one above the square root.
This happens in general, unless $N$ happens to be a perfect square, in which case $\sqrt{N}$ is itself a factor. Take $N=36$, for example. The factors are
$$\color{red}{1, 2, 3, 4}\,\,\,\,\color{green}{6}\,\,\,\,\color{blue}{9, 12, 18, 36}$$
Again each factor below the square root pairs up with one above the square root,  but this time the square root, $6$, is itself a factor.
A: Let $N$ have $a$ as a factor.  Then $b = \frac Na$ is also a factor because $a*b = a\frac Na = N$.
So factors come in pairs.
If $a < \sqrt{N}$ the $N = a*b < \sqrt{N}*b$.
So $b > \frac N{\sqrt{N}} = \sqrt{N}$.
So every fact less than $\sqrt{N}$ is "matched up" with a factor greater then $\sqrt{N}$.  
So if $N$ has $9$ factors less than $\sqrt N$ then there will be $9$ matching factors that are greater than $\sqrt N$.  And as $N$ is a perfect square, $\sqrt N$ is also a factor of $N$.  So there are $9 + 9 + 1 = 19$ factors.
That answers your question.
Now to finish:
So how many factors does $N^5$ have?
Well, that depends one the prime factorization of $N$.  If for example; $N=p^3$ for some prime, which has $4$ factors $1,p,p^2,p^3$ then $N^5 = p^{15} $ will have $16$ factors.  (All the $1,p, p^2,....,p^{15}$.
But if $N = pq$ for $p$ and $q$ then $N$ also has $4$ factors, $1,p,q,pq$ and $N^{5} = p^5*q^5$ which has $36$ factors.  
It has $36$ factors because all the factors are of the form $p^iq^j; j\le 5;i\le 5$.  As there are $6$ possible $i$s ($i = 0....5$) and $6$ possible $j$s ($j = 0....5$, there are $6*6=36$ possible pairs of $i$ and $j$.
So what we know about the prime factorization of $N$?  Well, $N$ has $19$ factors and $19$ is prime.  
Why is that significant?
Because if $N = \prod p_i^{k_i}$ is the prime factorization of $N$ then the factors will be in the form $\prod p_i^{a_i}; ai \le k_i$. For each $a_i$, $a_i = 0....,k_i$ so there will be $k_i+1$ choices.  So the total number of factors will be $\prod (k_i + 1)$.
So we have $19 = \prod(k_i+1)$.  But $19$ is prime so there can only be one term; $k_1 + 1 = 19$.
So $N = p^{18}$ for some prime $p$.  Those are the only numbers that have $19$ factors.  (Note:  $N = p^{18} = (p^{9})^2$ so $N$ is a perfect square.)
So $N^5 = p^{90}$ and the factors are: $1,p, p^2,...., p^{90}$ so there are  $91$ factors.
