How to Calculate the Factorial of Avogadro Number?

[ananthu017]

can anyone explain to me how to find the factorial of avogadro number ? what is its value ?

 

[Pythagorean]

It's such a large number that you might be able to justify using Stirling's approximation.

 

[Borek]

Google for Stirling's approximation.

 

[SteamKing]

It's Huuuuuuuuuuuuuuuuuuuuge!

 

[cp255]

I can tell you that the number of base 10 digits is > 6.02x10^23. You can't calculate this. Recently I have been coding my own big number c++ library. I often test it on 1,000,000!. It takes 5 minutes, and produces somewhere around 5 million digits. I doubt you are asked to calculate the exact value of 6.02E23!. If you want an approximation use Stirling approx. I use it in my e-calculator to find how many terms I need to sum to get the the desired accuracy.

ln(n!) = n(ln(n) - 1) Note this is only a good approximation.
This therefore means that n! = e^{n(ln(n) - 1))} (approximately)

 

[lurflurf]

http://www.wolframalpha.com/ gives
10^(10^(10^1.400640864781007))

assumes usual metric value of
6.022141×10^23 mol^(-1) (reciprocal moles)