# What is the percentage of prime numbers among all numbers with 100 decimal digits?

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

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

Big Hint

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

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).

• 999 does not have 100 digits! Apr 16 '16 at 14:51
• @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 $x=999$) ;-)
– Surb
Apr 16 '16 at 14:52
• i know this theorem, but I can't see a way to apply this theorem with 100 decimal digits numbers such as 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
– chen
Apr 16 '16 at 14:58
• @chen $\log(10^{100})=100\log10$. Apr 16 '16 at 15:01
• @almagest ah, ok i was stupid
– chen
Apr 16 '16 at 15:02