Here is a mathematical riddle:
Find a way to get 20160 with 5 2's and any mathematical operators you want.
Have fun!
Don't forget that I said any. Feel free to look online for crazy whacky math operators :))
(..and please don't close this as too broad. It's just a fun riddle)
Uh...I don't think this'll pass as an answer but how about
$\dfrac{(2+2+2+2)!}{2} = 20160$
Lol lucky guess.
EDIT: Apparently here: A new PSE member with square reputation the OP of this question created a bounty for my already correct answer on that question because he wanted "explanation of answer" so here is my explanation:
The first thing you notice is that 20160 is not some ordinary number. It's dvisible by 8, 7, 6, 5, 4, 3, 2, 1. So first thing you do, consider factorials. Then, you realize that 20160 is exactly $8!/2$. Then, you go like whoaaa how do you calculate $8$ using 4 $2$'s and then you stare at that problem for like 5 minutes and then you facepalm and realize that it's $2+2+2+2$...done.
"Any mathematical operator" gives too much freedom. In fact, here is a way to make any natural number using only 3 twos, credited to Ben Rudiak-Gould:
$$n = -\frac{\log\left[\left(\log\underbrace{\sqrt{\sqrt{\cdots\sqrt2}}}_{n}\right) / \log2\right]}{\log{2}}$$
Where the number of square roots is the number $n$ you want to express. For 20160, this means there would be 20160 nested square roots inside the log.
Any, you say? Well, then, how about
$ \fbox{out-of-the} $ 22222
where $\fbox{out-of-the}$ is of course the out-of-the-box operator, that takes the rules, bends them until they are unrecognizable, and changes any number outside the box boundaries to 20160.
