-5
$\begingroup$

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)

$\endgroup$

closed as too broad by JMP, Rand al'Thor, Ankoganit, ffao, Rubio Jan 10 '18 at 1:43

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Do logic operators count as mathematical operators? $\endgroup$ – NL628 Jan 7 '18 at 7:35
6
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Ummm, the first two 'bored' ones don't work. $\endgroup$ – boboquack Jan 7 '18 at 10:38
  • $\begingroup$ lol sorry @boboquack i'm being dumb $\endgroup$ – NL628 Jan 7 '18 at 22:03
  • $\begingroup$ For bonus points, can you calculate 8 using four 2's differently? $\endgroup$ – Octopus Feb 15 '18 at 0:35
  • $\begingroup$ Yes. 2x2+2x2, 2^2+2^2, (2^2^2)/2 etc. $\endgroup$ – Amit Naidu Feb 15 '18 at 7:24
7
$\begingroup$

"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.

$\endgroup$
  • $\begingroup$ better than my answer :))) $\endgroup$ – NL628 Jan 7 '18 at 7:56
  • $\begingroup$ Should this be edited to clarify we’re talking about the natural log, i.e. $\ln$? Because when I see an unspecified $\log$ I immediately assume $\log_{10}$. $\endgroup$ – DonielF Jan 7 '18 at 23:30
  • $\begingroup$ @DonielF any base should work, this simplifies to n log 2 / log 2 and the logs cancel out in the end. $\endgroup$ – ffao Jan 7 '18 at 23:30
  • $\begingroup$ @ffao Wonder why Wiki specifies $\ln$ then. $\endgroup$ – DonielF Jan 7 '18 at 23:31
7
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ now I realize why the question has -4 votes $\endgroup$ – NL628 Jan 7 '18 at 22:05

Not the answer you're looking for? Browse other questions tagged or ask your own question.