# Formulae of the Year $2016$ [closed]

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.

• $$~2^0 = 1^6~$$ Commented Dec 15, 2015 at 19:34
• @G-man I read the text for the tag "recreational-mathematics" and it said "fun". So yes!
– Jens
Commented Dec 15, 2015 at 19:34
• @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.
– user249332
Commented Dec 15, 2015 at 19:37
– Jens
Commented Dec 15, 2015 at 20:09
• Very nice..........+1 Commented Dec 15, 2015 at 20:13

An easy one ;-) $$(2 + 0 + 1)! = 6$$

• If I allow myself inequalities, how easy is $20 > 16$?
– Jens
Commented Dec 15, 2015 at 19:57
• @jens very, but inequalities are not allowed :-P
– Ant
Commented Dec 15, 2015 at 20:56

$$\color{red}{2}\pi i\left(\oint_{|z-\color{red}{0}|=R}\frac{dz}{z}\right)^{-\color{red}{1}}=\lceil \cos(\color{red}{6})\rceil$$

• How? Could you or someone else explain how this is evaluated / equation reached? You have the unit disc (?) Commented Dec 15, 2015 at 20:04
• The integral is taken over $|z|=R$ counterclockwise. It is a circle around $0$ with radius $R$. Commented Dec 15, 2015 at 20:14
• I see. However why does R not have to be a certain number, what is with the 2*pi*i ? Commented Dec 15, 2015 at 20:17
• $R$ is arbitrary, the integral is equal to $2\pi i$, see the Wikipedia Commented Dec 15, 2015 at 20:21

Another easy one) $$\Large(\color{red}{ 2}!)^{\color{blue}{\Large {2}}}+(\Large \color{red}{0}!)^{\color{blue}{\Large{0}}}+(\color{red}{1}!)^{\color{blue}{\Large{1}}}=\color{red }{\Large 6}$$

• Very Nice! +1 :)
– RFZ
Commented Dec 15, 2015 at 20:01
• 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. Commented Dec 16, 2015 at 4:44
• @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 $6$? Commented Dec 16, 2015 at 4:54
• 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). Commented Dec 16, 2015 at 6:19
• That is a point. So it should look like $$\Large(\color{red}{ 2}!)^{\color{blue}{\Large {2}}}+(\Large \color{red}{0}!)^{\color{blue}{\Large{0}}}+(\color{red}{1}!)^{\Large\color{blue}{{1}}}= \Large(\sqrt[\Large \color{red}{6}]{\color{red }{ 6}})^{\Large \color{blue}{6}}$$ Commented Dec 16, 2015 at 6:26

A simple one: $$2\times 0 = \sin\left(1 \times 6 \times \pi \right)$$

• Ok, but the digits are not quite in the same order.
– Jens
Commented Dec 15, 2015 at 19:36
• @Jens, feel free to do edits. Clearer view will be much appreciated.
– user249332
Commented Dec 15, 2015 at 19:45
• $e$ and $\pi$ are both numbers Commented Dec 15, 2015 at 19:47
• @Ataulfo But they are not digits. Creative rule interpretation allowed for fun...
– Jens
Commented Dec 15, 2015 at 19:52
• @Jens : is it better like this? Commented Dec 15, 2015 at 20:01

Arithmetic's fundamental theorem implies $$2016=2^{5}\cdot3^{2}\cdot 7$$

Edit:

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

• $= 21 \cdot 16 \cdot 6$ Commented Dec 15, 2015 at 19:30
• ...which does use some digits not in {2, 0, 1, 6}...
– Jens
Commented Dec 15, 2015 at 19:31
• Which makes of $2016=2^5(2^6-1)$ a pernicious number, but sadly not a perfect number. Commented Dec 15, 2015 at 19:46
• numberempire.com/2016 Commented Dec 15, 2015 at 20:29
• @DanBrumleve, you mean $2^{6-1}\cdot(2^6-1)$ Commented Dec 16, 2015 at 3:23