How many groups of order $2500$ are there? I aborted the GAP-calculation of $Size(ConstructAllGroups(2500))$ after about $3$ hours. $gnu(2500)$ seems to be a very hard case.

Does anyone know $gnu(2500)$ (The number of groups of order $2500$), or at least whether it is smaller than , larger than or equal to $2500$ ?

The number of groups of order $d$ is smaller than $2500$ for every proper divisor $d|2500$, so $gnu(2500)$ could be smaller then $2500$.
 A: There are 227 groups (from a 5 minute calculation in GAP with minor improvements as outlined in previous answers).
A: Applying Sylow's theorem, there is a unique $5$-Sylow which is then normal in $G$ if $G$ is a group of order $2500$. We denote it $N$. 
On the other hand, take $H$ to be one $2$-Sylow of $G$ (it is of order $4$). 
It is clear that :
$$G=N\rtimes H $$
Furthermore (again by Sylow's theory) the isomorphism class of $N$ and $H$ are uniquely defined by $G$, in other words if $N'$ and $H'$ are groups of order $5^4$ and $4$ respectively then :
$$N\rtimes H\text{ is isomorphic to } N'\rtimes H'\implies N\text{ is isomorphic to } N'\text{ and }H\text{ is isomorphic to } H' $$
You have fifteen groups of order $p^4$ and all of them can be written as semi-direct product of cyclic groups (see theorem 1.6.1 and theorem 1.6.14 in https://people.kth.se/~boij/kandexjobbVT11/Material/pgroups.pdf) and two groups of order $4$. 
If you fix $N$ and $H$ then to determine the number of $G$ which are isomorphic to some semi-direct product $N\rtimes H$, it suffices to compute $Hom(H,Aut(N))$. Of course different elements in $Hom(H,Aut(N))$ could lead to isomorphic group. After all, $Aut(N)$ is computable if you use GAP for each group $N$ of cardinal $5^4$. Then computing the cardinality of $Hom(H,Aut(N))$ may not be very difficult... I think you can work this out...
