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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!

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  • $\begingroup$ Welcome to PSE (Puzzling Stack Exchange)! $\endgroup$ Jan 23 at 4:49
  • 3
    $\begingroup$ 11^11 (but the numbers need to be interpreted as binary numbers, so probably a bit cheating to use a different base than base 10) $\endgroup$
    – Alderath
    Jan 23 at 14:44

4 Answers 4

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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

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

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  • $\begingroup$ Great job, i had a different solution in mind but this is still correct. $\endgroup$ Jan 21 at 3:48
  • $\begingroup$ @AlejandroGarcia $9\sqrt 9$, perhaps? $\endgroup$ Jan 21 at 4:45
  • $\begingroup$ No, it's (1/9)^-3. $\endgroup$ Jan 21 at 6:15
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+100
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$3/(1/9)=$

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

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    $\begingroup$ I like this answer the best. To emphasise how elegant it is, I switched your solution to use the other notation for division, which gets rid of all the parentheses too. (If you prefer the original formatting, please do feel free to revert my edit.) $\endgroup$
    – Bass
    Jan 23 at 7:55
  • $\begingroup$ @Bass Lovely edit! $\endgroup$ Jan 24 at 9:23
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Using repeating decimals, $\ 27 =$

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

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

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

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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.

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  • $\begingroup$ +1 Very elegant! $\endgroup$ Jan 23 at 4:12
  • $\begingroup$ you puzzled frac with overline $\endgroup$
    – Timo
    Jan 24 at 9:10

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