Is there a general method of finding the number of Sylow p-subgroups, given only the order of the group? Given that one is only told the order of the group, one will often reach conclusions as "its either $1$ or $x$ Sylow $p$-subgroups." Is there a general method to find out if one can rule one of the alternatives out, given only the order of the group? (That is, I am aware of the fact that for some orders we have to content with "$1$ or $x$, but I have also seen--seen, not understood--cases where one can rule one of the alternatives out.")

To answer you question quickly, yes. There are more parts to Sylow's Theorem than more people are usually aware. Two of them are
Thm: If $G$ is a finite group, $p$ a prime and $p^n \mid o(G)$ but $p^{n+1} \nmid o(G)$, then any two subgroups of $G$ of order $p^n$ are conjugate. 
Corollary: The number of $p$-Sylow subgroups in $G$ equal $o(G) / o(N(P))$, where $P$ is any $p$-Sylow subgroup of $G$.
Does this help?
