7
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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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closed as too broad by Mark N, xnor, singletee, mmking, Aza Jun 23 '15 at 1:47

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Should the numbers be in order? $\endgroup$ – Cows quack Jun 22 '15 at 19:42
  • $\begingroup$ @KritixiLithos - "you are allowed to rearrange the numbers on the left hand side to get to 24" $\endgroup$ – Mark H Jun 22 '15 at 19:44
  • $\begingroup$ @randal'thor Sorry, my bad... $\endgroup$ – Cows quack Jun 22 '15 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 '15 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 '15 at 20:04

15 Answers 15

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 '15 at 0:48
  • $\begingroup$ This is crazy complicated! I voted for this being the most complicated answer $\endgroup$ – Mark H Jun 23 '15 at 5:02
18
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How about:

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

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

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    $\begingroup$ Wow, that's brilliant! $\endgroup$ – Rand al'Thor Jun 22 '15 at 19:45
  • $\begingroup$ What function does the sigma have? I know it means "sum of" but sum of what numbers? $\endgroup$ – Cows quack Jun 22 '15 at 19:47
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    $\begingroup$ Gotta love the genius in this equation $\endgroup$ – Mark H Jun 22 '15 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 '15 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 '15 at 20:21
14
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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? $\endgroup$ – Cows quack Jun 22 '15 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 '15 at 19:40
  • $\begingroup$ Aah, I see. Now I get the answer, thank you. $\endgroup$ – Cows quack Jun 22 '15 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. $\endgroup$ – Mark H Jun 22 '15 at 19:49
  • $\begingroup$ @Rand isn't it the smallest integer function? $\endgroup$ – Sharad Gautam Jun 22 '15 at 20:37
13
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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. $\endgroup$ – Joe Z. Jun 22 '15 at 20:24
9
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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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8
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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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6
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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 '15 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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    $\begingroup$ Where could I learn to use MathJax? My answers never look pretty... $\endgroup$ – rodolphito Jun 22 '15 at 20:41
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    $\begingroup$ I picked up after a few weeks here. $\endgroup$ – Conor O'Brien Jun 22 '15 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? $\endgroup$ – rodolphito Jun 22 '15 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 '15 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 '15 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? $\endgroup$ – Joe Z. Jun 22 '15 at 20:25
  • $\begingroup$ I don't understand this one. What does the << operation mean? $\endgroup$ – Rand al'Thor Jun 22 '15 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 '15 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 '15 at 20:53
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    $\begingroup$ In other words $n<<k = n \cdot 2^k$. $\endgroup$ – AlexR Jun 22 '15 at 23:54
2
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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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2
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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 '15 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 '15 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 '15 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 '15 at 20:04
  • $\begingroup$ Sheez that looks complicated $\endgroup$ – Mark H Jun 22 '15 at 20:04
  • $\begingroup$ @randal'thor, this was modified from my troll bridge answer [they don't really matter].... ;) $\endgroup$ – Mark N Jun 22 '15 at 20:06
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    $\begingroup$ This answer's invalid since it uses $e$ and $\pi$ too many times. $\endgroup$ – Joe Z. Jun 22 '15 at 20:18
  • $\begingroup$ @JoeZ. You can never have too much $\pi e$ (pi-e ;)). $\endgroup$ – Conor O'Brien Jun 23 '15 at 0:17
1
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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. $\endgroup$ – Mark H Jun 22 '15 at 19:56
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    $\begingroup$ -1 for being too arbitrary in the "creativity" involved. $\endgroup$ – Joe Z. Jun 22 '15 at 20:20
0
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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 '15 at 21:08

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