Make 27 using 1, 1, 1, 1

Question

Make the number 27 using 4 instances of the digit 1. You must use all 4 digits to make it count.

Here are the allowed operators:

1. Addition (+), subtraction (-), multiplication (*), division (/)
2. Square roots (√) and exponents (^)
3. Decimals (.) and repeating decimals (¯)
4. Negative numbers (-n)

That's the allowed operators. Any other operators cannot be used for this problem. That's all, have fun solving the puzzle!

Answer 1

For starters,

I can make $.\bar1 = \frac19 = 3^{-2}$ cheaply, and the target is $3^3$, which suggests the formula $27 = \frac19^{\frac{-3}2}$

I submit

$.\bar1^\frac{-1}{\sqrt{.\bar1}+\sqrt{.\bar1}} = \frac19^{-\frac1{\frac23}} = \frac19^{\frac{-3}2} = 27$

Answer 2

$3/(1/9)=$

$$\frac{1+1+1}{.\bar1}$$

Answer 3

Using repeating decimals, $\ 27 =$

$\Large \frac{1}{.\bar{1}} \times \sqrt{\frac{1}{.\bar{1}}}$

Another solution based on a comment by OP, $\ 27 =$

$\left(\sqrt{.\bar{1}}\right)^{-(1+1+1)}$

Answer 4

Using

$\sqrt{1\over.\bar1} = 3$

the first thing that came to my mind was:

$\left(\sqrt{1\over.\bar1}\right)^\sqrt{1\over.\bar1}$

It may not be the simplest solution, but I like the reuse of the core structure.