Using only 1s, make 29 with the minimum number of digits

Question

Use the minimum number of ones to make 29.

Here is the list of operations permitted:

• "Standard" operations, such as: $x+y$, $x-y$, $x\times y$, $x\div y$
• Negation: $-x$
• Exponentiation of two numbers: $x^y$
• The square root of a number: $\sqrt{x}$
• The factorial: $x!$
• Concatenation of the original digits: $x|x$, $x|x|x$, $x|x|x$, etc. This means that you cannot concatenate like this: $(1+1)|1 = 2|1 = 21$

You can only use ones, and the result must be exactly $29$, not "about" 29.

You may not use any other operations. That includes $\log{x}$, $\left \lfloor{x}\right \rfloor$, $\left \lceil{x}\right \rceil$ and any others. You must only use base 10, and you not use any decimal points (i.e. no .1, .11, etc.).

The record to beat is 7.

Answer 1

Here's a 7-digit solution:

$(11-1)\times(1+1+1)-1$

Searching exhaustively is somewhat complicated. Let's say, broadly, impossible. One is able to take indefinite factorials and raise to ludicrous powers and take arbitrary numbers of square roots. These endeavors would almost certainly lead nowhere but computers don't necessarily know that.

Not only so, but a search that doesn't use symbolic algorithms to handle sqrts and fractions might find something that looks like a solution but is only 29.0000000.

Nonetheless, one can arbitrarily put "reasonable" restrictions on such things and say that we are not going to do, for example, $a^b$ unless $0<a<1000$ and $-20<b<20$.

Doing this, I find that there are 28,948 different numbers that can be reached with six 1s. None of them is 29. My code just tracked one solution for each number that it found and spat out the above solution which suggests to me that it's the only one with 7 digits. I haven't tried to do list them all, but if someone else wants to wade through my code it is below

def factorial(n):
    if n<=1: return 1
    return n*factorial(n-1)

sols={}
sols[1] = {}
sols[1][1] = "1"
vals = set([1])

digits=2
while digits<=7:
    sols[digits] = {}

    #concat
    concat = "1"*digits
    if eval(concat) not in vals:
        sols[digits][eval(concat)] = concat
        vals.add(eval(concat))

    #simple sum
    if digits not in vals:
        ssum = "1+"*digits
        sols[digits][digits] = ssum[:-1]
        vals.add(digits)

    #partitions
    for part1 in range(1,digits):
        part2 = digits-part1
        if part1>part2: break
        for val1 in sols[part1]:
            for val2 in sols[part2]:
                #multiply
                if val1*val2 not in vals:
                    sols[digits][val1*val2] = "("+sols[part1][val1]+")*("+sols[part2][val2]+")"
                    vals.add(val1*val2)
                #add
                if val1+val2 not in vals:
                    sols[digits][val1+val2] = "("+sols[part1][val1]+")+("+sols[part2][val2]+")"
                    vals.add(val1+val2)
                #divide
                if val2 !=0:
                    if val1/val2 not in vals:
                        sols[digits][val1/val2] = "("+sols[part1][val1]+")/("+sols[part2][val2]+")"
                        vals.add(val1/val2)
                if val1 !=0:
                    if val2/val1 not in vals:
                        sols[digits][val2/val1] = "("+sols[part2][val2]+")/("+sols[part1][val1]+")"
                        vals.add(val2/val1)
                #subtract
                if val1-val2 not in vals:
                    sols[digits][val1-val2] = "("+sols[part1][val1]+")-("+sols[part2][val2]+")"
                    vals.add(val1-val2)
                if val2-val1 not in vals:
                    sols[digits][val2-val1] = "("+sols[part2][val2]+")-("+sols[part1][val1]+")"
                    vals.add(val2-val1)
                #exponent
                if val1 > 0 and val1 < 1000 and abs(val2)<20:
                    if val1**val2 not in vals:
                        sols[digits][val1**val2] = "("+sols[part1][val1]+")^("+sols[part2][val2]+")"
                        vals.add(val1**val2)
                if val2 > 0 and val2 < 1000 and abs(val1)<20:
                    if val2**val1 not in vals:
                        sols[digits][val2**val1] = "("+sols[part2][val2]+")^("+sols[part1][val1]+")"
                        vals.add(val2**val1)

    #adjustments
    k = list(sols[digits].keys())[:]
    for val in k:
        #Sqrt
        if val>0:
            if val**0.5 not in vals:
                sols[digits][val**0.5] = "sqrt("+sols[digits][val]+")"
                vals.add(val**0.5)
            if val==int(val) and val<=20:
                if factorial(val) not in vals:
                    sols[digits][factorial(val)] = "("+sols[digits][val]+")!"
                    vals.add(factorial(val))

    if 29 in vals:
        break
    #next number
    print(digits,len(sols[digits]))
    digits+=1

Answer 2

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!
$\sqrt{((1+1+1)!)!+11^{\frac{1}{1/2}}}$

Answer 3

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.

Answer 4

I managed three 8's:

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

and a very dodgy 6:

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

and 4:

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

Answer 5

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.

Answer 6

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

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

Answer 7

Lowest I managed so far is 9 digits:

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

• 11*(1 + 1 + 1) - (1 + 1 + 1 + 1)

Some other ways I came up with:

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

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

• (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)

Answer 8

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:

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

Answer 9

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

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

Answer 10

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