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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.

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I suspect Stirling's approximation will help. – David Mitra Apr 25 '12 at 15:43
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The simplest way is just to compute $20!$. $20$ is small enough that this shouldn't take too much time (or paper). – Chris Eagle Apr 25 '12 at 15:45
See "D. FACTORIALS OF LARGE NUMBERS" in groups.google.com/group/sci.math/msg/d12962e3af2c74b7 – Dave L. Renfro Apr 25 '12 at 15:45

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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

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(Exponents of more than one character require {} to render properly) – The Chaz 2.0 Apr 25 '12 at 16:00
Apparently I fail basic math. If you actually count 2 for all of the 2-digit numbers, you come up with 28, which is quite a way out. sigh – MathematicalOrchid Apr 25 '12 at 20:34
@TheChaz Thanks for that... – MathematicalOrchid Apr 25 '12 at 20:35

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