What is the largest value of n for which n! has fewer that 100 decimal digits?

Question

undeadknight26 · Answer

I believe the answer is 70, not 100% sure.

sillybilly123 · Answer

reckon it's 69, using the Stirling approximation, calculator and trial and error - all of which can then be checked on Wolfram(:-/).  is that how you are supposed to be doing this?!

i initially thought it was a mod arithmetic thing but can't seem to get that started.

sillybilly123 · Answer

$10^1$ has 2 digits, $10^2$ has 3, etc ... so we are looking for n where $\log n! = 99.xxxxxx$ to get a 100 digit factorial

Take Stirling:

$n! \approx  \left(\dfrac{n}{e} \right)^n\sqrt{2 \pi n}$

re-write it in terms of $\log$  **base 10**, as:

$\log n! \approx n  \log \dfrac{n}{ e}  + \log \sqrt{2 \pi n}$

then **trial and error** in calculator gives:

- $\log 69! \approx 98.2$, 99 digits

-  $\log 70! \approx 100.1$, ie 101 digits

No 100 digit factorial....

celticcat · Answer

Thanks