Sylow theorem; Sylow $III$ Just curious, I have't seen any statement that answers this so far. And I'm not at the stage of being able to or given a proof of the theorem so I am not sure.

Sylow $III$; If $G$ has order $n=p^kr$ for some prime $p$ such that it does not divide $r$, then let $n_p$ be the number of distinct sylow $p$-subgroups. Then $n_p|r$ and $n_p \equiv 1$ mod $p$.

Simply, what if there are multiple possible $n_p$s? Say a situation like when $G$ has order $30=2.3.5$ and therefore, I can try say thinking about $n_2$, but 
$n_2=3,5,15$ say satisfies the conditions given in the theorem. Then what? Is the theorem inconclusive? 
 A: In the case $n=30$ and $p=2$, then the theorem says that
$$n_2|15,\qquad n_2\equiv1\text{ mod }2,$$
thus $n_2$ is either $3$, $5$ or $15$. You can't get more information that that directly from the theorem. Sometimes it is anyway possible to discover more from purely group-theoretic arguments and/or knowing more about the group.
A: The statement is true for any prime $p$ dividing the order of $G$. So if $G$ has order $30$, which has prime factors $2$, $3$ and $5$, it tells you no more and no less than the following:
For $p=2$, the number $n_2$ of Sylow $2$-subgroups divides $15$ and equals $1$ mod $2$ (so it is $1$ or $3$ or $5$ or $15$); for $p=3$, the number $n_3$ of Sylow $3$-subgroups divides $10$ and equals $1$ mod $3$ (so it is $1$ or $10$); for $p=5$, the number $n_5$ of Sylow $5$-subgroups divides $6$ and equals $1$ mod $5$ (so it is $1$ or $6$).
In particular, it does not tell you which of the possible values $n_p$ is equal to, and in fact this depends on the underlying group.
A: In that case, what you wrote is all the information that you get from the theorem, so there are multiple candidates for the number of p-sylow subgroups. 
There is a refined version of the third sylow theorem, which says that $n_p$ is the index of the normalizer of any p-sylow group in $G$, but the normalizer is typically hard to compute. So typically you will need to use knowledge specific to the problem to eliminate the candidates.
