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Soon it's the year $2016$. Time to ponder how we can arrange the digits in 2016 to form a valid equation. Use any symbols you like (please explain the less obvious ones). Keep digits in the same order (should this be relaxed?).

Examples:

$$\lfloor e^2\rfloor + 0 - 1! = 6$$ $$\left\lfloor\sqrt{\sqrt{201}}\right\rfloor = \lceil\sqrt{6}\rceil$$

where $\lfloor x\rfloor$ denotes the floor function and $\lceil x\rceil$ the ceiling.

Don't overuse constants (i.e. avoid adding up several $\pi$ and $e$ just to get to some arbitrary value).

EDIT: clarification: use each of the digits $2$, $0$, $1$, $6$ in this order only once. Combine digits giving $20$, $201$, $16$, etc as you like (I won't argue whether in a fraction the numerator or denominator comes first :-). Please don't criticize answers that violate this rule, as this clarification came late.

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closed as too broad by mrf, user642796 Dec 16 '15 at 9:57

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.

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    $\begingroup$ $$~2^0 = 1^6~$$ $\endgroup$ – Kibble Dec 15 '15 at 19:34
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    $\begingroup$ @G-man I read the text for the tag "recreational-mathematics" and it said "fun". So yes! $\endgroup$ – Jens Dec 15 '15 at 19:34
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    $\begingroup$ @G-man, if you think this is off-topic, then the tag (soft-question) should be outlawed in this site! I don't see anything wrong in asking such question under these tags sometimes. By the way, winter-bash is going on. So, enjoy and stop accusing such posts for some days. $\endgroup$ – user249332 Dec 15 '15 at 19:37
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    $\begingroup$ 7 downvotes already? Would you please read the text for the "recreational-mathematics" tag? I refuse to use reddit; its too addictive a time sink. $\endgroup$ – Jens Dec 15 '15 at 20:09
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    $\begingroup$ Very nice..........+1 $\endgroup$ – Bhaskara-III Dec 15 '15 at 20:13
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An easy one ;-) $$(2 + 0 + 1)! = 6$$

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    $\begingroup$ If I allow myself inequalities, how easy is $20 > 16$? $\endgroup$ – Jens Dec 15 '15 at 19:57
  • $\begingroup$ @jens very, but inequalities are not allowed :-P $\endgroup$ – Ant Dec 15 '15 at 20:56
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$$\color{red}{2}\pi i\left(\oint_{|z-\color{red}{0}|=R}\frac{dz}{z}\right)^{-\color{red}{1}}=\lceil \cos(\color{red}{6})\rceil$$

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    $\begingroup$ How? Could you or someone else explain how this is evaluated / equation reached? You have the unit disc (?) $\endgroup$ – Imago Dec 15 '15 at 20:04
  • $\begingroup$ The integral is taken over $|z|=R$ counterclockwise. It is a circle around $0$ with radius $R$. $\endgroup$ – Kamil Jarosz Dec 15 '15 at 20:14
  • $\begingroup$ I see. However why does R not have to be a certain number, what is with the 2*pi*i ? $\endgroup$ – Imago Dec 15 '15 at 20:17
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    $\begingroup$ $R$ is arbitrary, the integral is equal to $2\pi i$, see the Wikipedia $\endgroup$ – Kamil Jarosz Dec 15 '15 at 20:21
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Another easy one) $$\Large(\color{red}{ 2}!)^{\color{blue}{\Large {2}}}+(\Large \color{red}{0}!)^{\color{blue}{\Large{0}}}+(\color{red}{1}!)^{\color{blue}{\Large{1}}}=\color{red }{\Large 6}$$

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    $\begingroup$ Very Nice! +1 :) $\endgroup$ – ZFR Dec 15 '15 at 20:01
  • $\begingroup$ This is perhaps the most elegant. May be you can write RHS as sixth root of six raised to six to put it in the same form as LHS. $\endgroup$ – Shailesh Dec 16 '15 at 4:44
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    $\begingroup$ @Sailesh: thanks for suggestion but OP asked to arrange the digits $2, 0, 1, 6$ in the same order as in $\Large 2016$ so if it is $(6^{\large 1/6})^{\Large 6}$ then isn't it complex more than writing simply $6$? $\endgroup$ – Harish Chandra Rajpoot Dec 16 '15 at 4:54
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    $\begingroup$ What I meant is $(\sqrt[6]{6})^6$ Now you are preserving order and you can probably render this in the same colorful way so that the equation looks balanced. Every number gets a hat (power). $\endgroup$ – Shailesh Dec 16 '15 at 6:19
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    $\begingroup$ That is a point. So it should look like $$\Large(\color{red}{ 2}!)^{\color{blue}{\Large {2}}}+(\Large \color{red}{0}!)^{\color{blue}{\Large{0}}}+(\color{red}{1}!)^{\Large\color{blue}{{1}}}= \Large(\sqrt[\Large \color{red}{6}]{\color{red }{ 6}})^{\Large \color{blue}{6}}$$ $\endgroup$ – Harish Chandra Rajpoot Dec 16 '15 at 6:26
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A simple one: $$2\times 0 = \sin\left(1 \times 6 \times \pi \right) $$

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  • $\begingroup$ Ok, but the digits are not quite in the same order. $\endgroup$ – Jens Dec 15 '15 at 19:36
  • $\begingroup$ @Jens, feel free to do edits. Clearer view will be much appreciated. $\endgroup$ – user249332 Dec 15 '15 at 19:45
  • $\begingroup$ $e$ and $\pi$ are both numbers $\endgroup$ – Piquito Dec 15 '15 at 19:47
  • $\begingroup$ @Ataulfo But they are not digits. Creative rule interpretation allowed for fun... $\endgroup$ – Jens Dec 15 '15 at 19:52
  • $\begingroup$ @Jens : is it better like this? $\endgroup$ – Tryss Dec 15 '15 at 20:01
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Arithmetic's fundamental theorem implies $$2016=2^{5}\cdot3^{2}\cdot 7$$

Edit:

According to Dan Brumleve $$2016=2^{0-1+6}\cdot\left(2^{0+1\cdot6}-2^{0}\cdot1^{6}\right)$$

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    $\begingroup$ $= 21 \cdot 16 \cdot 6$ $\endgroup$ – Dan Brumleve Dec 15 '15 at 19:30
  • $\begingroup$ ...which does use some digits not in {2, 0, 1, 6}... $\endgroup$ – Jens Dec 15 '15 at 19:31
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    $\begingroup$ Which makes of $2016=2^5(2^6-1)$ a pernicious number, but sadly not a perfect number. $\endgroup$ – Raskolnikov Dec 15 '15 at 19:46
  • $\begingroup$ numberempire.com/2016 $\endgroup$ – janmarqz Dec 15 '15 at 20:29
  • $\begingroup$ @DanBrumleve, you mean $2^{6-1}\cdot(2^6-1)$ $\endgroup$ – janmarqz Dec 16 '15 at 3:23

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