Is there any neat way to solve how many digits the number $20!$ have? I'm looking a solution which does not use computers, calculators nor log tables, just pen and paper.
Factorial – Calculating the Number of Digits in a Factorial
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Is there any neat way to solve how many digits the number $20!$ have? I'm looking a solution which does not use computers, calculators nor log tables, just pen and paper.
Best Answer
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.