# How to make 2012 by using 2, 0, 1, 2?

How to make $$2012$$ by using $$2, 0, 1, 2$$?

Allowed Operations:

Addition, Subtraction, Multiplication, Division, $$!$$ (factorial), subfactorial ( $$!n$$), primorial (product of the first $$n$$ primes), square root($$\sqrt{}$$), exponentiation ($$a^{b}$$), double factorial ($$n!!$$), triple factorial ($$n!!!$$), radical and ceiling function, decimal point, tetration (iterated exponentiation), infinite roots.

Brackets and parenthesis () are also allowed.

• 2012 is 2,0,1,2 concatenated (did I spell it correctly?) Feb 7 at 4:33
• Yea, if concatenation is allowed, then 2012 = 2.0.1.2 where . is concatenation operation. Feb 7 at 7:45

## 4 Answers

Here is one way to do it

$$-1 -\left \lceil-\sqrt{ \left(\left(\left\lceil \sqrt{((2 + 0!)!)!!!} \right\rceil!!\right)!!\right) \times 2 } \right \rceil = -1 -\left \lceil-\sqrt{ \left(\left(5!!\right)!!\right) \times 2 } \right \rceil = 2012$$

A shortened version based on @hexomino's nice solution:

$$-0! - \left \lceil - \sqrt{ \left(\left( \left( \frac{1}{.2} \right) !!\right)!!\right) \times 2 } \right \rceil = 2012$$

I don't understand why the ceiling function is allowed, but why the floor function should not be allowed. If it is allowed you could shorten it further and arrange it to:

$$\left \lfloor \sqrt{ \left(\left( \left( \frac{1}{.2} \right) !!\right)!!\right) \times 2 } \right \rfloor - 0! = 2012$$

Here is another possibility

We have
$$6!!!=18, \lfloor\sqrt{(6!!!)}\rfloor=4, 4!!=8, 8!!=384, \lceil\sqrt{(8!!)}\rceil=20, 20!!!=4188800, \lfloor \sqrt{(20!!!)}\rfloor=2046$$
and $$6!!!=18, \lceil\sqrt{(6!!!)}\rceil=5, 5!=120, \lfloor\sqrt{(5!)}\rfloor=10, !10=1334961, \lceil\sqrt{\sqrt{(!10)}}\rceil=34$$

Hence

$$\lfloor\sqrt{\left(\lceil \sqrt{\left(\left(\lfloor\sqrt{\left(\left(0!+2\right)!\right)!!!}\rfloor\right)!!\right)!!}\rceil\right)!!!}\rfloor - \lceil\sqrt{\sqrt{!\left(\lfloor\sqrt{\left(\lceil\sqrt{\left(\left(1+2\right)!\right)!!!}\rceil\right)!}\rfloor\right)}}\rceil = 2046-34 = 2012$$

My approach relies primarily on the Primorial and Multi-factorial operations. To simplify and make it easier to read, I am using the notation P(n) to represent the primorial of n.

$$P\left(\left \lceil \sqrt{P(2)!!!} \right \rceil \right) = 2310$$

$$P\left(\left \lceil \sqrt{\left \lceil \sqrt{P\left(P(1)\right)!!!}\right \rceil !!!} \right \rceil \right) = 210$$

$$\left \lceil \sqrt{\left \lceil \sqrt{P(2)!!}\right \rceil!!}\right \rceil = 11$$

$$\left \lceil \sqrt{\left \lceil \sqrt{P\left(P(0!)\right)!!!}\right \rceil!!}\right \rceil!! = 8$$

$$P\left(\left \lceil \sqrt{P(2)!!!} \right \rceil \right) - P\left(\left \lceil \sqrt{\left \lceil \sqrt{P\left(P(1)\right)!!!}\right \rceil !!!} \right \rceil \right) - \left(\left \lceil \sqrt{\left \lceil \sqrt{P(2)!!}\right \rceil!!}\right \rceil \times \left \lceil \sqrt{\left \lceil \sqrt{P\left(P(0!)\right)!!!}\right \rceil!!}\right \rceil!!\right) = 2012$$

Feel free to let me know if there are any mistakes!

• I'm not clear about your P(n). I think P(2)=2,hence P(2)!!!=2 and finally $$P\left(\lceil \sqrt{P(2)!!!} \rceil \right) = 2$$ Feb 28 at 22:00
• But the Primorial of n is the product of the first n primes, so P(2) = 2*3 = 6 Feb 29 at 3:30
• oh, I see, I misunderstood the definition, +1 for your solution Feb 29 at 18:25