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I tried this way, I only need to know if this is correct or if there are better ways to solve this:

$2^{1000}$ does not have a factor of $5$ obviously therefore we can assume

$$ 10^{m} < 2^{1000} < 10^{m+1}$$ for some $m$

Assume $ k = 2^{1000}$, then take log on both sides $\log k = 1000 \log 2 \approx 301.02999 > 301$

Therefore $2^{1000}$ has $302$ digits.

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302 is correct. –  user2468 Mar 14 '12 at 2:57
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That's essentially what I would have done. If you want to double check, feel free to count the number of digits in 10715086071862673209484250490600018105614048117055336074437503883703510511249361‌​224931983788156958581275946729175531468251871452856923140435984577574698574803934‌​567774824230985421074605062371141877954182153046474983581941267398767559165543946‌​077062914571196477686542167660429831652624386837205668069376. –  Nate Eldredge Mar 14 '12 at 3:17
    
@Nate: Awesome! –  JavaMan Mar 14 '12 at 4:03
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For a quick back-of-the-envelope computation, you can note that $2^{10}$ is only a little larger than $10^3$, so $2^{1000} = (2^{10})^{100}$ is larger than $10^{300}$, though not by much; so $2^{1000}$ should have close to, but perhaps a few more, than 300 digits. –  Arturo Magidin Mar 14 '12 at 4:34
    
Offtopic: on Unix-like machine, echo "2^1000" | bc | tr -d '\n\r\\' | wc -m can do such count a la Nate's comment above :) –  user2468 Mar 14 '12 at 4:46
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1 Answer

up vote 5 down vote accepted

Recall that $10^{d-1}$ has $d$ digits. So for any number $n,$ the number of digits of $n$ is given by solving $ 10^{d-1} = n,$ or $$d = 1 + \lfloor \log_{10}(n) \rfloor$$

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