1. Nov 21, 2013

### ananthu017

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

2. Nov 21, 2013

### Pythagorean

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

3. Nov 21, 2013

### Staff: Mentor

4. Nov 21, 2013

### SteamKing

Staff Emeritus
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5. Nov 22, 2013

### 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! = en(ln(n) - 1)) (approximately)

6. Nov 29, 2013

### lurflurf

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