A formula with only one $0$ that evaluates to a given integer Just found this math puzzle online:


  Original puzzle: Using one zero (and no other numbers) and
  mathematical (including trigonometric and advanced) operations, how
  can you achieve a final result of 6?
  
  Advanced one: Using only trigonometric functions and a single instance
  of the number zero, derive a forumula to calculate any positive
  integer n.


I am sure there is an answer somewhere online but I can't find one - very interesting.

And the answer I can think of for the original one is


  $$\lim_{x\to0}\;(a^x+ a^x+ a^x+ a^x+ a^x+ a^x)$$
This may not be the most efficient solution for 6, but it works in general.  First, note that $cos(0) = 1$, so we can start with $1$.  The important formulas are then

(a) $\tan(\cot^{-1}(x)) = \frac{1}{x}$
(b) $\sin(\cot^{-1}(x)) = \frac{1}{\sqrt{1 + x^2}}$

These can be checked by drawing a triangle with the appropriate sides lengths.  Then, starting with $\sqrt{n}$, applying (b) and then (a) gives $\sqrt{n+1}$.  Thus, starting with $\sqrt{n} = n = 1$, repeated applications of (a) and (b) can give the square root of any natural number.  Since every natural number is the square root of another natural number, we can get any natural number this way.
Edit: After looking at the link Eugene gave below, this same process can be done more simply by iterating

$\sec( \tan^{-1}(x)) = \sqrt{x^2+1}$
(0!+0!+0!)!=6

As 0!=1 then it becomes (1+1+1)! => 3! = 6.