The Triple Number Game (n + n + n) I work at the Science Museum on weekends, and I sometimes present the following puzzle:
$$0\;0 \; 0 = 6$$
$$1 \; 1 \; 1 = 6$$
$$2 \; 2 \; 2 = 6$$
$$3 \;3 \; 3 = 6$$
$$\vdots$$
$$n\;n\;n=6$$
The idea is you can add in any operations you like, as long as they do not require you to write additional numbers. For example:
$$2 + 2 + 2 =6$$
$$\sqrt4+\sqrt4+\sqrt4=6$$
$$ (1+1+1)! = 6$$
 But $\sqrt[3]{8}+\sqrt[3]{8}+\sqrt[3]{8}=6$ would not be allowed, as it requires the additional three to be written.
The operations are thus typically restricted to $+\;-\;\times\;\div\;\sqrt\;\;!$
I have solutions for every integer n so $0\le n\le10$ (available at request), and for every integer $n^{2k}$ where $k\in{\mathbb{N}}$ (this is accomplished by nesting square roots.)
My question is two parts:


*

*Is there any other integers for which there are solutions?

*Are there any other operations that I haven't used that can be used to solve the problems in creative ways?

 A: $\sqrt{35+35/35}=6$  This can be extended to $6^{2n}-1$ with more square roots.  $16/\sqrt{16}+\sqrt{\sqrt{16}}=6\\144/(\sqrt{144}+\sqrt{144})=6$  
Other operators I have seen are concatenation, so $\sqrt {33+3}=6$ (though you probably have $3 \times 3-3=6$), decimals (so you can get $.2$ instead of $2$ and now $94/.94-94=6$), repeats (so you can do $.\overline 2=\frac 29$) and the floor function.  Floor plus factorial is too powerful-I remember seeing a proof that those plus a single $\pi$ could give any natural.  You might want to include base $2$ logs-I have seen $\lg(x)$ but the ISO version seems to be $\operatorname{lb}(x)$.
A: To answer question $2$, if you allow the use of the logarithm, floor or ceiling function, we can practically do anything.
For example:
$$\log\sqrt{a}=\frac12\log(a)$$
Allows you to sort of create numbers indirectly.
Now, I note the problem can be sort of reduced:
$$n/n*f(n)=6$$
So it becomes only a matter of finding a way to get $n$ to become $6$.
And if you really want to get silly, you can mess with people and say $11=3$ because $00=0$, $01=1$, $10=2$, etc. you count in binary.
Or you could use a different number as the base.  Allows for invisible changing of numbers without the use of any symbols.
One could attempt modular arithmetic, but you'd sorta have to cheat this and rewrite the question as follows:
$$6=n\;n\pmod n$$
Maybe too easy, maybe not...
$$6=n\pmod{n/n}$$
See, if you allowed that, we'd have the solution for all $n$, positive or negative...
