Number primes digits How do I find out the number of prime numbers that have exactly 100 digits?
I know the Prime Number Theorem, but 100 digits numbers are too big to be put in a calculator.
 A: The prime number theorem will give an approximate answer. You want the number of primes between $10^{99}$ and $10^{100}$  For the simplest approximation, this is just 
$$\frac{10^{100}}{\log 10^{100}}-\frac{10^{99}}{\log 10^{99}}=\frac{10^{100}}{100\log 10}-\frac{10^{99}}{99\log 10}=10^{99}\left(\frac {10}{\log 100}-\frac 1{\log 99}\right)$$ 
If this is too big for your calculator, it is.  Wolfram alpha will give the answer as approximately $3.9\cdot 10^{97}$.  You can use the more accurate formula for the prime number in terms of the logarithmic integral if you want, but the issues are the same.
A: To determine the number of primes with $100$ digits exactly, you would need to calculate, at least effectively, the value of $\pi\left(10^{100}\right) - \pi\left(10^{99}\right)$, where $\pi(x)$ is the count of # of primes up to $x$. However, regarding determining $\pi\left(10^{n}\right)$ for larger values of $n$, according to Prime-counting function,

The value for $10^{27}$ was published in $2015$ by David Baugh and Kim Walisch.

Since the number of primes up to $10^{27}$ was only calculated about $4$ years ago, I highly doubt there would be anything for either $10^{99}$ or $10^{100}$ yet.
Nonetheless, there are various estimates for the # of primes in a larger interval which give considerably more accurate values than you would get by just using the Prime Number Theorem. For example, the Inequalities section of Wikipedia's "Prime-counting function" article gives several good ones you can use instead. Also, the upper & lower bounds used in the inequalities will allow you to determine a maximum error of the estimate you get when you use one of them.
Update: This answer gives an approximation for the # of primes using the Prime Number Theorem formula. Although Wolfram Alpha can calculate it directly, you can also actually fairly easily factor out $10^{99}$, with this allowing any calculator or program which supports natural logarithms to do the calculations of the remaining parts. In particular, you get
$$\begin{equation}\begin{aligned}
\pi\left(10^{100}\right) - \pi\left(10^{99}\right) & \approx \frac{10^{100}}{100\ln 10} - \frac{10^{99}}{99\ln 10} \\
& = 10^{99}\left(\frac{10}{100\ln 10} - \frac{1}{99\ln 10}\right) \\
& = 10^{99}\left(\frac{1}{\ln 10}\right)\left(\frac{1}{10} - \frac{1}{99}\right) \\
& \approx 10^{99} \times 0.0390426 \\
& = 3.90426 \times 10^{97}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note I used the Windows $7$ calculator program to do the calculations to get the final answer shown above.
