Find an expression equalling 69 using the digits 1 9 9 2 in order A friend gave me a puzzle to find an expression for each of the integers $1$ to $100$ using the digits $1, 9, 9, 2$ such that they appear in that exact order. You're allowed any combination of the four basic operations, brackets, square roots, the factorial, exponentiation and also you can put digits together, e.g. we could have $23=19+(\sqrt{9})!-2$. I've managed every number relatively quickly except $69$ which I've been stuck on for ages and it's infuriating! Any ideas?
Edit: Apologies for my unclear explanation. You're only allowed to put the original digits together in the equation. The equation must make sense as a normal equation so you can't do things like combine $1$ with $\sqrt{9}$ to get $13$ because this can't be written down in a way which makes sense as a normal equation.
Also you aren't allowed to just use powers, unless you're using one of the four digits themselves, e.g. $2=1+(9/9)^2$ is okay because we've used the $2$ as a power but $69=(1-9)^2+\sqrt{9}+2$ isn't because there's now a superfluous $2$.
 A: Since we can put digits together, one solution might be 
$1+ ((\sqrt{9})!)((\sqrt{9})!) +2$
One expression which does not keep the exact order is 
$9*9-12$ or $((\sqrt{9})!)((\sqrt{9}+1) +2$
A: I came up with:
$$(1-9)^2+\sqrt{9}+2.$$
A: I'm not sure if this meets the "makes sense as a normal equation" condition, but if we can apply the square root operator an infinite number of times we have:
$$69 = 71-2 = \sqrt{5041} - 2 = \sqrt{1 + 7!} - 2 = \sqrt{1 +(3!+1)!}-2$$
$$= \sqrt{1 + \Big((\sqrt9)! + 1\Big)!}-2 = \sqrt{1 + \Bigg((\sqrt9)! + \sqrt{\sqrt{\sqrt{ \cdots9}}}\Bigg)!}-2$$
where the dots indicate an infinite sequence of nested square roots.  Note that:
$$\sqrt{\sqrt{\sqrt{ \cdots9}}} = \lim_{n\rightarrow\infty}\Big(9^{(1/2)^n}\Big) = 1$$
A: I can't comment, therefore I am posting this as an answer and delete it later. 
$(\sqrt{9})!9 = 69$, since $(\sqrt{9})!=6$, since you said, it is allowed to put digits together.
Getting rid of the 1 is not a problem $1\cdot n=n$.
But still I need to get rid of the 2. 
edit: 
$1 + \underbrace{(\sqrt{9})! (\sqrt{9})!}_{=66} + 2 = 69$
A: How about $\ \ \ \ $ $$-1+9!!!-92$$
or
$$(1+\sqrt{9})!+9\div.2$$
