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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.

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

4 Answers 4

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Here is one way to do it

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

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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:

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

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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

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

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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!

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

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