12
$\begingroup$

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.

  • 4
    302 is correct.2012-03-14
  • 11
    That's essentially what I would have done. If you want to double check, feel free to count the number of digits in 10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376.2012-03-14
  • 0
    @Nate: Awesome!2012-03-14
  • 3
    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.2012-03-14
  • 1
    *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 :)2012-03-14
  • 1
    $2^{1000} = 1024^{100} = 10^{300}x1.024^{100}$. $1.024^{100}$ = $1 + 100*0.024 + 100*99/2*0.024^2 + 100*99*98/6*0.024^3...}$, about 1 + 2.4 + 2.8512 + 2.2353 + 1.3019 + 0.5994 > 10 but definitely less than 100.2014-06-02
  • 0
    5 years too late but to make this *perfectly* clear: $2^m = (10^{\log_{10}2})^m=10^{m\log_{10} 2}$. $10^k$ will have $k + 1$ digits (1 and k zeros) so $10^x$ with $k < x < k+1$ will have $x$ rounded up to the next higher integer ($k+1$). So $2^{m}$ will have $\lceil m\log_{10} 2 \rceil= \lceil m*0.30102999566398119521373889472449.... \rceil$ .2017-01-07

2 Answers 2

11

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 $$d = 1 + \lfloor \log_{10}(n) \rfloor$$

-7

you can simply right it in the form of log as
Log2x=1000
Log x/log2=1000
Taking value of log 2 which is 0.301
Log x=301
Converting in exponents form
10^301=x
It means x has 301 digits

  • 1
    There are most definitely 302 digits. Count them: $$107150860718626732094842504906000181056140481170553360744375038837035105112\\ 4936122493198378815695858127594672917553146825187145285692314043598457757469\\ 85748039345677748242309854210746050623711418779541821530464749835819412673\\ 98767559165543946077062914571196477686542167660429831652624386837205668069376.$$2015-06-03