What is the percentage of 2-groups out of all the groups of order less than $10,000$? For a my class in Abstract Algebra, our teacher asked us to answer the question above, noting it's a surprisingly large number.
I have found this question More than 99% of groups of order less than 2000 are of order 1024?. It seems like the answer should be even greater when looking at groups of order less than 10,000 but I would not know how to prove it or state it.
Now, I think the number 10,000 was arbitrary, used as an example, what my teacher is actually looking for is an answer like "99%" or "almost all". Still, my problem persists - I have no idea how to prove this. A hunch tells me that it should boil down to the fact that 2 is a prime factor of so many numbers (all the evens).
Any help is appreciated.
 A: The maximal $2-$group in this situation is of order $2^{13}.$ Now, the question comes down to deciding how many non-isomorphic $2-$groups exist with order less than or equal to $2^{13},$ and comparing that to the totality of groups of order less than $10,000.$ I can't speak to what these numbers should be . . . more funding is needed.
Just looking at the non-isomorphic groups of order less than or equal to $100$ is rather taxing. There are 1048, see:
https://www.mimuw.edu.pl/~zbimar/small_groups.pdf
Now, in terms of $2-$groups, there are relatively few. The total number of non-isomorphic $2-$groups of order $64$ is $267$, of order $32$ is $51$, of order $16$ is 14, or order $8$ is $5$, of order $4$ is $2,$ and of order $2$ is $1.$ In total 
$$\text{total number of $2-$groups}= 267 + 51 +16 +5+2+1=342.$$
(http://oeis.org/A000679)
So we get $342/1048\approx 32\%.$ Rober Bell's comment, gives an idea of why the number of 2-groups of a given order is essential to the number of groups less or equal to a given order.
This can be quite difficult to compute, as the number of non-isomorphic groups with order $1024$ is $49,487,365,422$ . . . 
(https://mathoverflow.net/questions/46855/how-to-generate-all-finite-groups-of-order-n)
Suppose we know the number of $2-$groups less than a given order $N.$ And let $n$ be the maximal value so that $2^n\leq N.$ We can give $N$ a prime factorization $$N=2^n(p_1^{e_1}\cdots p_m^{e_m}).$$
Suppose we know the total number of $2-$groups of orders $2,2^2,\ldots,2^n.$ Let us denote these as $m_1,\ldots,m_n$ respectively.
Then for each $p_i$  we have at least
$m_kp_i^\ell$ groups of order $2^kp_i^\ell,$ for $1\leq k\leq n,$ $1\leq \ell\leq e_i.$ Hopefully this gives the idea of how one might compute the number of groups, up to isomorphism, less than a given order. 
