TITLE: How many factors of $2^{95}$ are greater than $10^6$
QUESTION [5 upvotes]: The title says it all.
My problem: I can solve this using logarithms but the problem is meant to be solved by combinatorics methods which I can't see how to apply them in this kind of problem.

REPLY [2 votes]: Recall that $10^3=1000 < 1024 =2^{10}$.
So, $2^{20} > 10^6$.
Now, since $10^3$ is so close to $2^{10}$, most probably $2^{19} < 10^6$. (Indeed, $2^{19}=524288$. Or argue that $2^{20} < 2\cdot 10^6$ by squaring $1024=1000+24$.)
So, $2^m > 10^6$ iff $m \ge 20$.
Adding the restriction $m\le 95$ gives $95-20+1=76$ possible factors.