Let $G$ be a group of order $1925$, then G has how many Sylow-$5$-subgroups?
What I know: $o(G)= 5^2\cdot 7\cdot 11$ then using Sylow's third theorem, the number of Sylow-$5$-subgroups are given by $n_5=1+5k \;\text{such that}\,$ $$ 1+5k|o(G)$$
now, as $1+5k \;\text{is coprime to}\; 5$ $$\therefore 1+5k|o(G) \implies 1+5k|77$$ by trial I found that it is possible for $k=0 \;\text{or}\; k=2$ giving $n_5=1 \;\text{or}\; 11$ But here are two possible answer which one is the correct numbers of sylow-$5$-subgroups?
$1$ is always possible for any order $n$ and prime $p\mid n$: Take an abelian group of any given order, e.g. $\mathbb Z/n\mathbb Z$.
In this particular case, $11$ is possible. Let $G=C_7\times C_5 \times (C_5\ltimes C_{11})$ where the second copy of $C_5=\langle a \rangle$ acts on $C_7=\langle b \rangle$ by $a^{-1}ba=b^4$ (as $2$ is a primitive root mod $11$, $x\mapsto x^4$ has order $5$ in $\text{Aut}(C_{11})$), where $C_n$ denotes the cyclic group of order $n$.
You can manually check the above example works. Here is how to come up with it in the first place.
It suffices to find a group $G'$ of order $5^2\cdot 11$ to have $11$ distinct Sylow $5$-subgroups, as we can then cook up $G=G'\times C_7$. Again, the number of Sylow $7$-subgroup is $1$ mod $7$ and also a divisor of $5^2$, hence can only be $1$, so the Sylow $7$-subgroup $N$ must be normal. Let $H$ be a Sylow $5$-subgroup of $G'$, so $H$ acts on $N$ by conjugation, so $G'\simeq H\ltimes N$ is a semi-direct product of $H$ and $N$. And if there is a unique Sylow $5$-subgroup for $G'$, then $H$ must be normal as well. Therefore $G'\simeq H \times N$, where $H$ acts on $N$ trivially by conjugation. This proves that as long as the action of $H$ on $N$ is non-trivial, we have a desired example.
How to get such an example? Well, as the automorphism group $\text{Aut}(N)\simeq \text{Aut}(\mathbb Z/11\mathbb Z)\simeq (\mathbb Z/11\mathbb Z)^*\simeq C_{10}$, we just need to find a group $H$ of order $25$ from which there is a nontrivial homomorphism to $C_{10}$. Such an example is easy: Just let $H=C_5\times C_5=\langle a \rangle \times \langle b \rangle$ where $a, b$ are both of order $5$. By the universal property of direct sum in the category of abelian groups, we just need to find a pair of homomorphisms $\rho_1, \rho_2$ from $C_5$ to $C_{10}$. As long as one of them is non-trivial, we're done. Again this is easy: just send $a$ to any element of order $5$ in $C_{10}$ (and $b$ to the identity).
The same method can be used to handle the problem of classifying groups of order $pq$ (where $p<q$ are distinct primes) as mentioned in the comment.