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In this question, I am asking you to get to 24 in the most complicated way using the numbers below.

enter image description here

One way to solve this is (9 x 8) / ( 6 / 2 ) = 24. But this is an easy way.

I am looking for the most complicated way of getting to 24. The person scoring the highest amount of complicated points wins.

Generic rules for complicated

there are no strict rules that apply for 'complicated' answers, as the community will vote for the most complicated answer. However the following is generally encouraged:

  • getting to higher numbers within your equation wins you more 'complicated points' (e.g I got to a maximum of 72 in the above equation)

  • use of non-common operators (factorials, square roots, absolute signs) is deemed more complicated than common operators

Other Rules

  • you are allowed to rearrange the numbers on the left hand side to get to 24

  • you are allowed to combine numbers if you deem it more complicated

  • you are only allowed to use each number once

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  • $\begingroup$ Should the numbers be in order? $\endgroup$
    – user41805
    Jun 22, 2015 at 19:42
  • $\begingroup$ @KritixiLithos - "you are allowed to rearrange the numbers on the left hand side to get to 24" $\endgroup$
    – Walter
    Jun 22, 2015 at 19:44
  • $\begingroup$ @randal'thor Sorry, my bad... $\endgroup$
    – user41805
    Jun 22, 2015 at 19:45
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    $\begingroup$ This question seems to be too broad in that the answer is chosen instead of found, so I have voted to close it. $\endgroup$
    – Mark N
    Jun 22, 2015 at 19:50
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    $\begingroup$ @MarkN I was going to ask why you dislike your own question, but then I saw the N! $\endgroup$
    – Rand al'Thor
    Jun 22, 2015 at 20:04

15 Answers 15

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25
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One that is mathematically precise (rounding answers are such cop-outs! (p.s. I upvoted them in any case))

Using the Riemann Zeta function $\zeta$ , the sign function $\text{sgn}$, the Gamma function $\Gamma$ , the sigma function $\sigma_0$ and the negation function

$$\begin{align} 24 & =\frac{\frac{\text{sgn}(2)}{-\zeta(-9)}}{\Gamma(\sigma_{0}(8))+\zeta(\zeta(-6))}\\ & =\frac{\frac{1}{-1/-132}}{\Gamma(4)+\zeta(0)}\\ & =\frac{132}{6-0.5}\\ & =\frac{132}{5.5}\\ \end{align}$$

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  • $\begingroup$ This is quite complicated. $\endgroup$
    – Ian MacDonald
    Jun 23, 2015 at 0:48
  • $\begingroup$ This is crazy complicated! I voted for this being the most complicated answer $\endgroup$
    – Walter
    Jun 23, 2015 at 5:02
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How about:

$\sum_{8}^{9} (6 * 2) = 24$

A little bit of notation mutilation, but hey...

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  • 4
    $\begingroup$ Wow, that's brilliant! $\endgroup$
    – Rand al'Thor
    Jun 22, 2015 at 19:45
  • $\begingroup$ What function does the sigma have? I know it means "sum of" but sum of what numbers? $\endgroup$
    – user41805
    Jun 22, 2015 at 19:47
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    $\begingroup$ Gotta love the genius in this equation $\endgroup$
    – Walter
    Jun 22, 2015 at 19:49
  • $\begingroup$ @KritixiLithos Generally you would insert numbers into the formula, but since there is no variable to place them into, you would simply run add up the numbers for each value in the range of the sumation. $\endgroup$
    – Aggie Kidd
    Jun 22, 2015 at 19:49
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    $\begingroup$ @AggieKidd With your avatar, I imagine you as someone who does nothing but answer various StackExchange questions with clever uses of sum expressions. $\endgroup$
    – nitro2k01
    Jun 22, 2015 at 20:21
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How about this?

$\Big\lceil\frac{8!}{9!}\Big\rceil\times(6-2)!=24$

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  • $\begingroup$ I am just curious, what are the big bracket-like symbols which are around 8!/9! called and what function do they do? $\endgroup$
    – user41805
    Jun 22, 2015 at 19:39
  • $\begingroup$ @KritixiLithos They're called ceiling symbols. $\lceil x \rceil$ is the smallest integer greater than or equal to $x$ (so the ceiling of 3.7 is 4 and the ceiling of 4 is also 4). $\endgroup$
    – Rand al'Thor
    Jun 22, 2015 at 19:40
  • $\begingroup$ Aah, I see. Now I get the answer, thank you. $\endgroup$
    – user41805
    Jun 22, 2015 at 19:42
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    $\begingroup$ Wow - took me a while to actually to just understand this complicated answer. Learnt something new today. $\endgroup$
    – Walter
    Jun 22, 2015 at 19:49
  • $\begingroup$ @Rand isn't it the smallest integer function? $\endgroup$
    – Sharad Gautam
    Jun 22, 2015 at 20:37
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Here's one with a number that goes really big:

$$\lceil{\ln(9^{(8 + 2)} \times 6)}\rceil = 24$$

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  • $\begingroup$ It's the ceiling function, since obviously the natural log can't be met precisely. $\endgroup$
    – user88
    Jun 22, 2015 at 20:24
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I'm fairly certain I'm going to terribly mess up the markup on this, but here goes:

$\left\lceil\sqrt{\sqrt{\dfrac{(8+2)!}{\sqrt[\sqrt{\sqrt{6!}}]{9!}}}}\right\rceil=24$

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(λn.λf.λx.n (λg.λh.h (g f)) (λu.x) (λu.u)) π2$(6*89)$

Explanation:

π2 is the Prime Counting Function nested twice, and the rest is apparently the simplest way to express the predecessor function in lambda calculus. So we multiply 6 by 89 (534), count the number of primes less than or equal to it (99), count the number of primes less than or equal to that (25), then subtract 1 (24).

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6
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Maybe a little bit less complicated than the others

$\sqrt[\sqrt{9}]{8} * 6 * 2 = \sqrt[3]{8} * 6 * 2 = 2 *6 * 2 = 24$

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Edit: Added what will surely win as the "highest number" prize and dropped the spoiler tags as they don't seem necessary with this question.

How about an infinite score using an infinite number of square roots and, just for fun, factorials? Do you count complication based on the number of different terms used or just the number used in total?

$$\left[\left(\sqrt{\sqrt{\sqrt{\cdots\sqrt{\sqrt{\sqrt{9}}}}}}\right)!!!\ldots!!!\right]\times8\times6\div2=1\times8\times6\div2=24$$

Or, just to claim the "highest number" prize using Knuth's up-arrow notation:

$$\left\lfloor{ \sqrt{\sqrt{\sqrt{\ldots m \ldots\sqrt{\sqrt{\sqrt{ 9 \uparrow\uparrow\uparrow\ldots n \ldots\uparrow\uparrow\uparrow 8 }}}}}} }\right\rfloor \times\frac{6!!}{2}=0$$

where $n$ is as large as needed to have the highest number and $m$ is as large as needed so that the floor function evaluates to $1$. I am not mathematical enough to prove that up-arrow notation will grow faster than the double factorial method proposed by Rodolvertice but I am fairly sure that it does so quite easily.

(Note that the $6!!$ is the double factorial and $6!!=48$)

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  • $\begingroup$ @grgarside Thanks for the additional MathJax. I keep forgetting about the formatting that helps make it prettier. $\endgroup$
    – Engineer Toast
    Jun 23, 2015 at 19:46
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Claims both the infinite operators score and the largest number score.

$n!!$ is the double factorial. There is no factorial used here, only double factorials.

$$\cfrac{6!!}{2!!!!\ldots\ldots!!!!}-{\left\lfloor{\cfrac{8!!!!... ...!!!!}{9!!!!... ...!!!!}}\right\rfloor}=24$$

The number $2$ can have an infinite number of factorials (double or single) applied to it and it will still be $2$.

$$\cfrac{6!!}{2!!}=24$$

this is because $6!!=48$, then dividing that by $2$ results in $24$, and $2=2!=2!!=2!!!\dots!!!$

$$\left\lfloor{\frac{8!!!!\dots n\dots!!!!}{9!!!!\dots n\dots!!!!}}\right\rfloor=0$$

Subtracting $0$ does nothing. n (the number of double factorials) can be very VERY large but not infinite, and it will still evaluate to $0$. Just enough so that it is more than the largest number here.

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  • 3
    $\begingroup$ Where could I learn to use MathJax? My answers never look pretty... $\endgroup$
    – vero
    Jun 22, 2015 at 20:41
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    $\begingroup$ I picked up after a few weeks here. $\endgroup$
    – Conor O'Brien
    Jun 22, 2015 at 21:40
  • $\begingroup$ Thanks! I assume I can learn by pressing the edit button on posts to see the source? Or is there an easier way that is less error prone? $\endgroup$
    – vero
    Jun 22, 2015 at 21:50
  • $\begingroup$ That's about the only way to do it! You can also make equations at codecogs.com/latex/eqneditor.php (LaTeX friendly) $\endgroup$
    – Conor O'Brien
    Jun 22, 2015 at 22:04
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    $\begingroup$ @Rodolvertice For the floor bracket problem: use \left\lfloor and \right\rfloor to make them scale. $\endgroup$
    – AlexR
    Jun 22, 2015 at 23:51
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Here's one

$(6 << 2) + \Big\lfloor\frac{8}{9}\Big\rfloor=24$

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  • $\begingroup$ Bit-shifting, eh? Why don't you just multiply by $9 - 8$ at the end, rather than using the floor expression to just reduce things? $\endgroup$
    – user88
    Jun 22, 2015 at 20:25
  • $\begingroup$ I don't understand this one. What does the << operation mean? $\endgroup$
    – Rand al'Thor
    Jun 22, 2015 at 20:28
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    $\begingroup$ @JoeZ Because I wanted to complicate things more than necessary, per the request of this question. $\endgroup$
    – tfitzger
    Jun 22, 2015 at 20:47
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    $\begingroup$ @randal'thor It's a bit shift operator. Basically, the bits (binary representation) for 6 are $0000 0110$. This shifts the bits two to the left, $0001 1000$, which is 24 in decimal. $\endgroup$
    – tfitzger
    Jun 22, 2015 at 20:53
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    $\begingroup$ In other words $n<<k = n \cdot 2^k$. $\endgroup$
    – AlexR
    Jun 22, 2015 at 23:54
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I have a few contributions. $\Gamma$ is the Gamma function; $x!!$ is the double factorial; $\text{Res}$ is the residue of a function (I found the one for $\Gamma$ on the first link); $\phi$ is the Totient Function.

1

Equation

$$\tag{1}\frac{\frac{9!}{\Gamma(8)}}{\frac62}$$ or $$\tag{1b}(9!/\Gamma(8))/(6/2)$$

Explanation

$$\Gamma(n)=(n-1)!$$ $$\begin{align}(9!/\Gamma(8)/(6/2))&=24\\ (9!/7!)/(6/2)&=\\ (72)/(3)&=\\ 24&= \end{align}$$ BONUS: $(1)$ maintains order.

2

Equation

$$\tag{2}\Gamma(9+6-(8+2!))$$

Explanation

$$\begin{align}\Gamma(9+6-(8+2!))&=24\\\Gamma(15-(8+2))&=\\\Gamma(5)&=\\4!&=\\24&=\end{align}$$

3

If you allow $n^{-1}$ as an operation and not a number:

Equation

$$\tag{3}-\text{Res}(\Gamma,-8/2)^{-1}+\left\lfloor\text{Res}(\Gamma,-9/6!)^{-1}\right\rfloor$$

Explanation

$$\text{Res}(\Gamma,-n)=\frac{(-1)^n}{n!}$$ $$-\text{Res}(\Gamma,-8/2)^{-1}=-\text{Res}(\Gamma,-4)^{-1}=-\frac{-1}{4!}^{-1}=-\frac{-1}{24}^{-1}=--24=24$$ Of course, $0\lt\Gamma(9/6!)^-1\lt1$, so it becomes $0$ in the floor bit. Thus, the answer is $24$.

4

Equation

$$\tag{4}\phi(9+8)+6+2$$

Explanation

$$\phi(9+8)=\phi(17)=16$$ check here $$16+6+2=22+2=24$$

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$$ \frac{d^3}{dx^3}(\frac {8}{2} x^\sqrt9+ 6!x) = 24 $$

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  • $\begingroup$ Arguably the $^3$ are numbers though. $\endgroup$
    – March Ho
    Jun 22, 2015 at 21:11
  • $\begingroup$ How about that? the $$ \frac{d^3}{dx^3} $$ is notation so i dont think it should count $\endgroup$
    – Tim
    Jun 23, 2015 at 13:06
  • $\begingroup$ @Tim All math writing is notation. The 3 indicates 3rd derivative just as 3 by itself indicates the third successor of 0. $\endgroup$
    – Deusovi ♦
    Jun 23, 2015 at 21:18
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If you want something that looks complicated...but really isn't:

$\phi$ - The Golden ratio
$\displaystyle\int\limits_{-\infty}^{+\infty}(-(\sqrt{9-8+e^{\pi\cdot i}} \times \sinh(f'(\phi)))^3) \ \mathrm dx + (6\cdot2!)$

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  • $\begingroup$ What are $f$ and $\delta$? $\endgroup$
    – Rand al'Thor
    Jun 22, 2015 at 20:04
  • $\begingroup$ Sheez that looks complicated $\endgroup$
    – Walter
    Jun 22, 2015 at 20:04
  • $\begingroup$ @randal'thor, this was modified from my troll bridge answer [they don't really matter].... ;) $\endgroup$
    – Mark N
    Jun 22, 2015 at 20:06
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    $\begingroup$ This answer's invalid since it uses $e$ and $\pi$ too many times. $\endgroup$
    – user88
    Jun 22, 2015 at 20:18
  • $\begingroup$ @JoeZ. You can never have too much $\pi e$ (pi-e ;)). $\endgroup$
    – Conor O'Brien
    Jun 23, 2015 at 0:17
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Does this involve big enough numbers for you?

I define the operation $*$ on pairs of integers in such a way that $9*8$ is a googolplex and $6*2$ is 24 times a googolplex. Then $\frac{6*2}{9*8}=24$.

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    $\begingroup$ points for getting the highest number. $\endgroup$
    – Walter
    Jun 22, 2015 at 19:56
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    $\begingroup$ -1 for being too arbitrary in the "creativity" involved. $\endgroup$
    – user88
    Jun 22, 2015 at 20:20
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$$\lfloor\tan(\cos(\sin\lceil(\sqrt{8})^{[\frac{d}{dx}(-9x)]}\rceil))\rfloor+\lfloor\sqrt{6!}-2\rfloor$$

Where $x$ is an arbitrary number.

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    $\begingroup$ Is using a variable "x" allowed? $\endgroup$
    – Rand al'Thor
    Jun 22, 2015 at 21:08

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