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4!!!!


Cyanide

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Well, this isn't a homework problem or anything, but we're incorporating factorials into our Calc II now..and I hate them to be honest :)

 

I was just wondering what would be 4!!!! or if it's even possible to calculate. I did 3!!! which made my head hurt :P

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4! is 24, but 4!!!! isn't! 4!! is 24!, for example. And 4!!!! is 24!!!. In other words, we're talking about an obscenely, obnoxiously big number. I'm not going to try to calculate it.

 

A good point, serves me right for not reading it correctly. My excuse is it's 12:30am I've been awake since 6:30am

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4!!!! = 24!!! > (24!/9!)!! > 10^10 !! > (10^10 ! / 10^9 !) ! > 10^90 ! > 10^90 ! / 10^89 ! > 10^(90*89) = 10^8010.

 

In other words: 4!!!! has more than eight thousand digits (probably a lot more). You´d better be prepared to need a large pice of paper if you want to calculate it.

 

EDIT: To emphasize that above was only a very rough lower limit: 24! = 6.2 *10^23 already, which already is 13 orders of magnitude greater than my approximation. And as a side-effect, we now know how to write the Avogadro constant in a way that no one will understand :D.

EDIT2: You (abskebabs) are right. I somehow though it was 6.22*10^23 when I wrote it; only remembered the ...22 .. ^23 part and forgot about the zero, somehow.

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Sorry if I'm being hopelessly ignorant here, but; couldn't the gamma function be used and adapted to solve this problem? The answer would still be ridiculous tho....

EDIT: Atheist, I'm sure you know this but 24! is not avagadro's constant, sry if thats pedantic...

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Just a really quite, pedantic note. You should be careful when using multiple factorial symbols, as 4!! is generally regarded as the double factorial:

 

[math]n!! = \begin{cases} n(n-2) \cdots 5 \cdot 3 \cdot 1, & n \text{ odd} \\ n(n-2) \cdots 4 \cdot 2, & n \text{ even} \end{cases}[/math]

 

Indeed, multiple factorial symbols are generally regarded as multifactorials. Just so you know :)

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