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Make 24 using exactly three 3s

Each number formed with a 3 and the 24 in the equation are all base 10.

You cannot introduce any additional digits or constants.

Plus(+), negation/subtraction (-), division(/ or show it with a fraction bar), multiplication(* or use grouping symbols as appropriate), exponentiation(^ or show it with an exponent). regular factorial(!), square root(sqrt(), or the radical symbol), grouping symbols such as parentheses and/or square brackets, and the regular decimal point(not for the repeating decimal) are allowed. No concatenation of any type is allowed.

No other functions/operators are allowed. This includes floor and ceiling functions not being allowed. (edit)

I am looking for a total of 10 solutions.

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    $\begingroup$ Let's please not make "non-serious" or otherwise very clearly non-pertinent answers here. They do not attempt to answer the posed puzzle, and will be deleted: puzzles with no lateral-thinking tag do not invite such answers. $\endgroup$ – Rubio May 23 at 14:34
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Ten solutions.

(i) $\sqrt{3^{3!}} -3 = 24$
(ii) $\left(\frac{3! + 3!}{3} \right)! = 24$
(iii) $ \left(\frac{3}{.3} - 3! \right)! = 24$
(iv) $\left(\sqrt{\frac{3}{.3} + 3!} \right)! = 24$
(v) $\left(3! - \frac{3!}{3} \right)! = 24$
(vi) $(3\times 3!)+3! = 24$ (Pugmonkey)
(vii) $(3 + \frac{3}{3})! = 24$ (Weather Vane)
(viii) $3^3 - 3 = 24$ (Olive)
(ix) $(\sqrt{3})^{3!} -3 = 24$ (Rupert)
(x) $(3+(3−3)!)! = 24$ (Trenin)

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    $\begingroup$ @JonathanAllan That's what Olive's comment above was referring to. $\endgroup$ – hexomino May 22 at 23:31
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    $\begingroup$ @JonathanAllan the similar variant $(3 + \frac{\sqrt{3}}{\sqrt{3}})! = 24$ was banned too. $\endgroup$ – Weather Vane May 22 at 23:54
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    $\begingroup$ I'd say that i and ix are basically the same since the order of applying the square root and the power are irrelevant in the same way that 3x2 and 2x3 are the same (though that is just my subjective opinion). $\endgroup$ – Chris May 23 at 14:45
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    $\begingroup$ @Chris Yes I think you are right here, not sure if we really can consider them to be different considering what has been disallowed previously. $\endgroup$ – hexomino May 23 at 15:32
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    $\begingroup$ I see, and since it's more of a numerical "fluke" (in that it would only work for 3) that they cancel that stops them being classed as identical I guess. $\endgroup$ – Fifth_H0r5eman May 23 at 15:33
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(edit) I will post one of the 10 solutions as an example:

$3^3 - 3 = 24$

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    $\begingroup$ This solution came to mind after reading just the title of the question. I assumed it was the only one. $\endgroup$ – JollyJoker May 23 at 7:32
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Here is my first:

$(3 + \frac{3}{3})! = 24$

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    $\begingroup$ @ Weather Vane - Let us say that by adding the square root symbols, you have not changed the values of those. They are redundant. It's similar to writing 1! instead of 1 or 2! instead of 2. Although mathematically correct, I would not see them as "primitive solutions." $\endgroup$ – Olive Stemforn May 22 at 21:44
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    $\begingroup$ Ah OK, I was about to add another similar variant - removed. $\endgroup$ – Weather Vane May 22 at 21:45
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Tenth solution from hint

$$\left(3+(3-3)!\right)!$$

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    $\begingroup$ Thank you for this tenth solution. Is it not ironic how relative the lack of complexity this solution has, given the amount of time between its posting and the original problem? $\endgroup$ – Olive Stemforn May 23 at 20:24
  • $\begingroup$ Nice work,can;t believe I missed this. $\endgroup$ – hexomino May 23 at 23:26
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Here is one possible solution:

$3\times3!+3!$

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$(\sqrt{3})^{3!} -3 = 24$

Which is obviously only slightly different to hexomino's:

(i) $\sqrt{3^{3!}} -3 = 24$

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  • $\begingroup$ Ah, yes, good spot! $\endgroup$ – hexomino May 22 at 21:45
  • $\begingroup$ @ Rupert Morrish - Good, your first one listed is a definitely newer type, because, for instance, if the square root symbol were to be dropped, the expression would have a different value. $\endgroup$ – Olive Stemforn May 22 at 21:50
  • $\begingroup$ @OliveStemforn: But the order of powers (which square root basically is) is unimportant (ie $a^{b^c} = a^{c^b}$) so are these two really different? To me it feels like it is the same principle as $3\times3!+3!$ being different from $3!+3\times3!$ - its exactly the same operations, just in a slightly different order... $\endgroup$ – Chris May 23 at 15:15
  • $\begingroup$ @chris The order of powers is definitely important. $2^{3^5} \ne 2^{5^3}$ The first is $2^{3^5} = 2^{243}$. The second is $2^{5^3}=2^{125}$. Perhaps you mean $(a^b)^c = (a^c)^b$. $\endgroup$ – Trenin May 23 at 17:34
  • $\begingroup$ @Chris Because that is more applicable to this case anyways because $\sqrt{3^{3!}}= (3^{3!})^{\frac{1}{2}}=(3^{\frac{1}{2}})^{3!}=(\sqrt{3})^{3!}$. $\endgroup$ – Trenin May 23 at 17:50
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I feel like something should be able to be done with

$\frac{(3!)!}{30} = 24$

with judicious use of

decimal points

but I can't find it.

$\frac{(3!)!}{3^3 + 3} = 24$

uses 4 3s.

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    $\begingroup$ :: : I will have to leave this computer in about five minutes from now (for up to as long as 20 hours later on) before I post for confirmations of remaining solutions. $\endgroup$ – Olive Stemforn May 22 at 21:59

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