# Combinatorial Puzzle

Standard

This is a continuation of previous post:

How many distinct numbers can be formed by using four 2s and the four arithmetic operations $+,-,\times, \div$.

For example:
$1 = \frac{2+2}{2+2}=\frac{2}{2}\times\frac{2}{2}$
$2 = 2+\frac{2-2}{2}=\frac{2}{2}+\frac{2}{2}$
$3 = 2+2 - \frac{2}{2}$
$4 = 2+2+2-2 = (2\times 2) + (2-2)$
(note that some binary operations do not make sense without parenthesis)

I have no idea about how to approach this problem (since I am not very comfortable with combinatorics). So any help will be appreciated.

Edit[29 May 2017]: This problem has been solved in the comments below.

# Arithmetic Puzzle

Standard

Following is a very common arithmetic puzzle that you may have encountered as a child:

Express any whole number $n$ using the number 2 precisely four times and using only well-known mathematical symbols.

This puzzle has been discussed on pp. 172 of Graham Farmelo’s “The Strangest Man“, and how Paul Dirac solved it by using his knowledge of “well-known mathematical symbols”:

$\displaystyle{n = -\log_{2}\left(\log_{2}\left(2^{2^{-n}}\right)\right) = -\log_{2}\left(\log_{2}\left(\underbrace{\sqrt{\sqrt{\ldots\sqrt{2}}}}_\text{n times}\right)\right)}$

This is an example of thinking out of the box, enabling you to write any number using only three/four 2s. Though, using a transcendental function to solve an elementary problem may appear like an overkill.  But, building upon such ideas we can try to tackle the general problem, like the “four fours puzzle“.

This post on Puzzling.SE describes usage of following formula consisting of  trigonometric operation $\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}$ and $\tan(\arcsin(x))=\frac{x}{\sqrt{1-x^2}}$ to obtain the square root of any rational number from 0:

$\displaystyle{\tan\left(\arcsin\left(\cos\left(\arctan\left(\cos\left(\arctan\left(\sqrt{n}\right)\right)\right)\right)\right)\right)=\sqrt{n+1}}$.

Using this we can write $n$ using two 2s:

$\displaystyle{n = (\underbrace{\tan\arcsin\cos\arctan\cos\arctan}_{n-4\text{ times}}\,2)^2}$

or even with only one 2:

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