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Can you use only two instances of the digit $2$, along with the mathematical operations below, to create an expression that evaluates to $2015$?

Allowed operations:

  • arithmetic operations: addition ($+$), subtraction ($-$), multiplication ($\times$), division ($\div$, or $\frac{x}{y}$), exponentiation (like $2^2$);

  • factorial ($!$), absolute value ($|...|$);

  • extraction of the root of any degree in a form $\sqrt[a]{b}$ or the square root in a form $\sqrt{b}$;

  • trigonometric functions: sine, cosine, tangent, cotangent, secant, cosecant

  • inverse trigonometric functions: arcsine, arccosine, arctangent, arccotangent, arcsecant, arccosecant

  • natural logarithm ($\ln b$), or logarithm with any base ($\log_a b$)

Parentheses $()$ are also allowed.

What is not allowed:

  • digits other than $2$, or more than two instances of the digit $2$

  • named constants such as $\pi, e$, etc...

  • defining and using your own functions

  • other variables

Example with three $2$s:

A couple of years ago, I managed to solve the same kind of problem, with three $2$s:

We can express any natural number $A$ using three $2$s and the above operations, like this: $$-\log_2\log_2\underbrace{\sqrt{\sqrt{...\sqrt{2}}}}_{A\,\mathrm{square\,roots}}$$

But I couldn't crack the problem using only two $2$s...

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  • $\begingroup$ You could use the incrementing method described in this answer to get from 2 to sqrt(2015) with a lot of trigonometric operations and then use the other 2 to square it. $\endgroup$
    – f''
    Commented Aug 7, 2015 at 11:55
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    $\begingroup$ sorry but i can't understand "2 digits 2"? did you mean we can use only digit 2 for max 2times? $\endgroup$
    – Saurabh Prajapati
    Commented Aug 7, 2015 at 12:41
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    $\begingroup$ @SaurabhPrajapati, yes, you got it right. We can use only digit 2 but no more than 2 times. $\endgroup$
    – Glinka
    Commented Aug 7, 2015 at 12:45
  • $\begingroup$ @f'', I think this is what I was looking for! Could you post it as an answer? $\endgroup$
    – Glinka
    Commented Aug 7, 2015 at 12:46
  • $\begingroup$ I think $-log_2^2\underbrace{\sqrt{\sqrt{...\sqrt2}}}_{A\,\mathrm{square\,roots}}$ is simpler and shorter. $\endgroup$
    – EKons
    Commented Jun 24, 2016 at 15:38

3 Answers 3

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This answer describes a method using trigonometric operations to obtain the square root of any rational number from 0. In this answer, Daniil Agashiyev notes that $\tan\arcsin\cos\arctan\cos\arctan\sqrt{n}=\sqrt{n+1}$.

Using this, we can write 2015 as:

$$(\underbrace{\tan\arcsin\cos\arctan\cos\arctan}_{2011\text{ times}}\,2)^2$$

or even with only one 2:

$$\underbrace{\tan\arcsin\cos\arctan\cos\arctan}_{2015^2-4\text{ times}}\,2$$

Unlike the linked problem, we're allowed to use the secant and cosecant here, so we can use $\sec\arctan$ or $\newcommand{\arccot}{\operatorname{arccot}}\csc\arccot$ for the same effect.

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  • $\begingroup$ In the first case why 2011 ans not 2013? And I'm not sure the second case is right... $\endgroup$
    – Glinka
    Commented Aug 7, 2015 at 13:36
  • $\begingroup$ @Dennis 2011 because 2 is sqrt(4), so applying the sequence 2011 times gives sqrt(2015). $\endgroup$
    – f''
    Commented Aug 7, 2015 at 13:52
  • $\begingroup$ "using the digit 2 only twice" I can see three 2s in both your answers, counting the '2' in '2015' $\endgroup$
    – Colonel Panic
    Commented Mar 3, 2016 at 15:45
  • $\begingroup$ @ColonelPanic The number under the brace is not actually written. It represents how many times the thing above it is to be written. It is being used as a shorthand because the full form would be too long to fit in this answer. $\endgroup$
    – f''
    Commented Mar 3, 2016 at 16:51
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    $\begingroup$ @OliveStemforn The formulas written in this answer are not the full form of the solution, but rather, an abbreviated form. The full solution has the trigonometric functions written out 2011 times, and was not included because it would be too long to fit in the post. $\endgroup$
    – f''
    Commented Aug 4, 2016 at 0:22
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It looks a bit like cheating but we are not defining any new function, we are just naming one that is allowed to use by the rules:

Name: m: x -> m(x) to be: x -> log2(x) (log in base 2)

Then we apply the square root trick:

- m(m( sqrt(sqrt( .... sqrt(2) ... )) )) = 2015
   (taking square roots 2015 times)
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  • $\begingroup$ That is exactly why defining new functions are forbidden :) It would generate the whole series of easy to find solutions. But still thumbs up :) I would have voted, but I don't have enough points $\endgroup$
    – Glinka
    Commented Aug 7, 2015 at 15:04
  • $\begingroup$ I will delete it if you think it doesn't match your specs. But your requirements is a bit arbitrary. You allow a whole range of trigonometric functions but only ln from the logarithmic/exponential function and for loga(b), one has to use 2 parameters. No exp() or any of the other related functions. For square root one parameter is required but for other roots two. $\endgroup$
    – ypercubeᵀᴹ
    Commented Aug 7, 2015 at 15:10
  • $\begingroup$ I realize my answer looks a bit silly when one thinks that we could define m = 2/2 and then do m+m+...+m (2015 times). $\endgroup$
    – ypercubeᵀᴹ
    Commented Aug 7, 2015 at 16:24
  • $\begingroup$ Your answer has the right to be, but the best one yet is given by f''. As I said, I would have voted up, but i haven't got this privilege. $\endgroup$
    – Glinka
    Commented Aug 7, 2015 at 16:36
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We can express 2015 using 0 digit like this:

$$\dfrac {\underbrace{e+e+e+...+e}_{2015 ~e~s} }{e}$$

Next Answer

15

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  • $\begingroup$ Sadly, we cannot use named constants... I'm sorry I did't mention it in the question, now it's there. $\endgroup$
    – Glinka
    Commented Aug 7, 2015 at 12:27
  • $\begingroup$ ok bro ;). We will think another solution $\endgroup$
    – apm
    Commented Aug 7, 2015 at 12:31
  • $\begingroup$ is it correct ? $\endgroup$
    – apm
    Commented Aug 7, 2015 at 12:35
  • $\begingroup$ What do you mean by that? It's just a 15 :) $\endgroup$
    – Glinka
    Commented Aug 7, 2015 at 12:39
  • $\begingroup$ We wrote 7th aug 2015 as 7th aug 15. Like that we can wrote 2015 as 15 :p $\endgroup$
    – apm
    Commented Aug 7, 2015 at 12:48

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