Factorial number of digits

I come from a background in computers, so here's my two cents. Taking the logarithm to the base 10 of n!. If the log comes out to be x, it is not hard to see that the number of digits must be the lowest integer greater than or equal to x, i.e, $floor(x)+1$. Now the question comes down to approximating the $log(n!)$ It is possible to prove by induction that n! lies between $(\frac{n}{2})^n$ and $(\frac{n}{3})^n$. Thus the log(n!) lies between $nlog(\frac{n}{2})$ and $nlog(\frac{n}{3})$. We can get a pretty tight bound if we used log tables for log(20/3), but as you have disallowed that, using the upper limit(which becomes log(10) = 1) will do quite nicely too. The answer comes to 20*log(10) = 20, which should tell you that the expected number of digit is about 19 or 20.(Since 20 is only the upper bound). And 19 happens to be the answer.

There is unclean way of solving it.

Number of digits in a number 'N' is CEILING(log(N)).

So number of digit in N! is CEILING(log(N!)) = CEILING(log(N*(N-1)*(N-2)) ... 2*1)

Thus,

CEILING(log(N!)) = CEILING(log(N) + log(N-1) +log(N-2)+ ... log(2)*log(1))

Which is pretty easy to compute.

Source: http://pitcher.digitalfreehold.ca/code/computeSize

As a rough approximation, multiplying an $n$-digit number by an $m$-digit number yields a result with about $n+m$ digits. So the numbers from 2 to 9 are all 1-digit numbers. From 10 to 20 are all 2-digit numbers. That suggests we should have about 18 digits or so.

Wolfram|Alpha claims that $20! = 2.4 \times 10^{18}$. Not far off! :-D