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I know the Prime Number Theorem, but 100 digits numbers are too big to be put in a calculator. Is there a way of finding out how many primes numbers as a percentage of the total numbers with 100 decimal digits

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  • $\begingroup$ It is not too big to put in a calculator! The proportion is $\frac{1}{log_e(10^{100})}=\frac{1}{100\ln10}\approx 0.43\%$ $\endgroup$
    – almagest
    Apr 16 '16 at 14:55
  • $\begingroup$ Do you want an "exact" answer, or an approximate one? And if approximate, how precise should the approximation be? $\endgroup$ Apr 16 '16 at 14:59
  • $\begingroup$ I would be to happy to find out an approximate one $\endgroup$
    – chen
    Apr 16 '16 at 15:01
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Big Hint

When $x$ is large$$\pi(x)\sim \frac{x}{\ln(x)}$$ where $\pi(x)$ is the number of prime number in $\{1,...,x\}$.

Added

The biggest number with $100$ digits is $10^{101}-1$. With those kind of number, it's not a problem to say that it's $1O^{101}$. Then, $$\pi(10^{101})\approx\frac{10^{101}}{\ln(10^{101})}=\frac{10^{101}}{101\ln(10)}\approx \frac{10^{101}}{10^2\ln(10)}=\frac{10^{99}}{\ln(10)}.$$ Therefore, the proportion is almost $$\frac{10^{99}}{10^{101}\ln(10)}\approx \frac{1}{230}.$$

Do the same for number with $99$ digits, and you'll get your result. (because you only want the proportion of prime with numbers of $100$ digits).

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  • $\begingroup$ 999 does not have 100 digits! $\endgroup$
    – almagest
    Apr 16 '16 at 14:51
  • $\begingroup$ @almagest: I'm not english, then I had a confusion on the question. But anyway, with number of $100$ digit, the approximation $\pi(x)\sim \frac{x}{\ln(x)}$ is much better than any other approximation (and this approximation work much much better than when I supposed $x=999$) ;-) $\endgroup$
    – Surb
    Apr 16 '16 at 14:52
  • $\begingroup$ i know this theorem, but I can't see a way to apply this theorem with 100 decimal digits numbers such as 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 $\endgroup$
    – chen
    Apr 16 '16 at 14:58
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    $\begingroup$ @chen $\log(10^{100})=100\log10$. $\endgroup$
    – almagest
    Apr 16 '16 at 15:01
  • $\begingroup$ @almagest ah, ok i was stupid $\endgroup$
    – chen
    Apr 16 '16 at 15:02

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