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Factorial of avogadro number

  1. Nov 21, 2013 #1
    can anyone explain to me how to find the factorial of avogadro number ? what is its value ?
     
  2. jcsd
  3. Nov 21, 2013 #2

    Pythagorean

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    It's such a large number that you might be able to justify using Stirling's approximation.
     
  4. Nov 21, 2013 #3

    Borek

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    Google for Stirling's approximation.
     
  5. Nov 21, 2013 #4

    SteamKing

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    It's Huuuuuuuuuuuuuuuuuuuuge!
     
  6. Nov 22, 2013 #5
    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! = en(ln(n) - 1)) (approximately)
     
  7. Nov 29, 2013 #6

    lurflurf

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

    assumes usual metric value of
    6.022141×10^23 mol^(-1) (reciprocal moles)
     
    Last edited: Nov 29, 2013
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