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

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).
