Kdtyjb

I thought I had it with this:

$sqrt{((1+1+1)!)!+11^{1+1}}$.

Until I remembered that

there's no function to square a number without using any digits, even though there is a function to take a square root without using any extra digits. For the past hour I've been trying to figure out how to square a number just using the square root.

Since square root is a number raised to the power of $frac{1}{2}$, I'm wondering if I can just take a reciprocal of that for a total of six 1s. My guess is that this isn't allowed because there isn't a way to actually write this down!
$sqrt{((1+1+1)!)!+11^{frac{1}{1/2}}}$

As stated by luchonacho, exhaustion will show that it's not possible in less than 7 digits with those operations.

Not showing negative values, as they are symmetric to positive ones, here are the easy attainable integer values:

1 digit:  1

2 digits: 0, 2, 11, 11!, ...

3 digits: 3, 6, 10, 12, 111, 720, 10!, ...

4 digits: 4, 5, 7, 9, 13, 22, 24, 110, 112, 120, 121, 719, 721, 1111, ...

5 digits: 8, 14, 17, 20, 21, 23, 25, 33, 36, 64, 66, 100, 109, 113, 119, 122, 132, 144, 222, 360, ...

And now we know we're stuck, because to achieve 29 (a prime number) in 6 digits, we would actually need any of those values in 5 digits: 28, 30, 784, 900, ... and we don't have them.

As such, any solution for 29 in 7 digits is optimal:

29 = (11−1)×(1+1+1)−1 (found by Dr Xorile)

Note that there may be room for a non-trivial sum of factorial numbers to reach a square power of 29 (or of 29±1), as found by Amorydai, but it's unlikely to beat 7 digits.

For illustration, while all values below 27 are attainable trivially with 6 digits or less, here are some solutions in 7 digits for numbers greater or equal to 28:

28 in 7 digits = (1+1+1)^(1+1+1)+1

28 in 7 digits = (1+1)×(11+1+1+1)

29 in 7 digits = (11−1)×(1+1+1)−1

30 in 6 digits = (11−1)×(1+1+1)

31 in 7 digits = (11−1)×(1+1)+11

31 in 7 digits = (11−1)×(1+1+1)+1

31 in 7 digits = (1+1+1)×11-1-1

32 in 6 digits = (1+1+1)×11-1

33 in 5 digits = (1+1+1)×11

34 in 6 digits = (1+1+1)×11+1

35 in 6 digits = (1+1+1)!^(1+1)-1

36 in 5 digits = (1+1+1)!^(1+1)

37 in 6 digits = (1+1+1)!^(1+1)+1

38 in 7 digits = (1+1+1)!^(1+1)+1+1

39 in 7 digits = (11+1+1)×(1+1+1)

40 in 7 digits = (11-1)×(1+1+1+1)

40 in 7 digits = (11-1)×(1+1)×(1+1)

15 is the lowest positive value requiring 6 digits with this reasoning.
28 is the lowest positive value requiring 7 digits with this reasoning.
41 is the lowest positive value requiring 8 digits with this reasoning.

I managed three 8's:

$(1+1+1+1)!+(1+1+1)!-1=24+6-1$
$(1+1+1)!times((1+1+1)!-1)-1=6times(6-1)-1$
$(11+1+1+1)times(1+1)+1=14times2-1$

and a very dodgy 6:

$11!!!!!!!!-(1+1+1+1)=11times3-4$

and 4:

$(11-1)!!!!!!!-1=10times3-1$

According to my analysis, there is no solution using less than 7 digits. The proof would require an exhaustion of every possible operation from 29, reducing the dimensionality of the problem one digit at a time, like a tree. I haven't followed all the leaves of the tree, but it seems clear no one leads to a solution.

To see how my logic works, consider an example:

$29 + 1 = 30$

Here we have a sub problem, which is to find a way to get to 30 using 5 or less digits. You can then follow the leaves of the tree, reducing the dimensionality one at a time, until you see there is no solution. For example, 30 + 1 = 31, and finding 31 from 4 digits, etc. Or 30 + 11 = 41, and so on.

Importantly, many of the branches of the original problem are identical. For example, $29 + 1 = 30$ is the same as $29+1! = 30$; and the subbranch above $30 + 1 = 31$ is exactly the same as $29+(1+1) = 31$ and $29+(1+1)! = 31$.

Same logic applies to, for instance, $29-1=28$. You need to get to 28 using 5 digits. Following the branches seem to lead to nowhere.

Some routes are evidently discarded. For instance, 11! = 39916800, too high. Would need to be scaled with a relatively high number, like 111. But that uses 5 of 6 digits. No way to use the last one meaningfully. Same logic with 111! or other like (11-1)! Other routes, like (1+1+1)!=6, requires you to produce 23 (so 23+6=29) with only 3 digits. Impossible.

No idea why you would ever consider square root as a relevant operation. Temptation here is that somehow a squared root times a natural number ($sqrt{a}b$) will give a natural number. However, this is only true if $ab^2$ is itself the square of a natural number. So you should start by searching for the latter. For example, take 36. It's the square of 6. Can you find natural numbers a and b such that $ab^2=36$? Yes, a=4 and b=3, or a=9 and b=2. But these use too many digits. Follow down the tree and you are done very quickly. What about a=11? It can be seen that it is not possible to find a natural number b such that $sqrt{11b^2}$ is a natural number.

The above applies the same with division, e.g. $frac{sqrt{a}}{b}$ or $frac{b}{sqrt{a}}$ routes.

I haven't given you any proof of my claim and I might be totally wrong. But my methodology (reducing the dimensionality of the problem one at a time) led me to nowhere.

I can get an answer accurate to within 0.12% with six 1s:

$$sqrt{sqrt{sqrt{11^{11}}}}+1+1 approx 29.0343$$

Lowest I managed so far is 9 digits:

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  (1 + 1 + 1 + 1)! + 1 + 1 + 1 + 1 + 1;

  
  




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  11*(1 + 1 + 1) - (1 + 1 + 1 + 1)

  
  

Some other ways I came up with:

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  (1 + 1)^(1 + 1 + 1 + 1 + 1) - 1 - 1 - 1 (10 digits)

  
  




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  (1 + 1 + 1 + 1 + 1)!/(1 + 1 + 1 + 1) - 1 (10 digits)

  
  




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  (1 + 1 + 1 + 1 + 1)^(1 + 1) + 1 + 1 + 1 + 1 (11 digits)

  
  




• 
  

  11*(1 + 1) + 1 + 1 + 1 + 1 + 1 + 1 + 1 (11 digits)

  
  

For the sake of completeness, here's an 8 digit solution:

(1+1+1)^(1+1+1) + 1 + 1

This was posted before the question was edited to demand an integer, and requires that the rules accept rounding, but it does produce the digits 29. Six is achievable in this way:

$sqrt{( ( 1 + 1 + 1 )! + 1 )} * 11 approx 29.1033$

11+11+11-1-1-1-1=29 This is my best guess.