For 3072-bit RSA, how many decimal digits must each prime $p$ and $q$ have? And roughly how many 400-digit primes exist?

Question

Accepted Answer

$2^{3072} \approx 10^{925}$, so $N$ has about 925 digits and each prime about 460 digits. There are roughly $10^{397}$ primes with exactly 400 digits.
Digits of the primes: use $2^{10} \approx 10^{3}$:
$$2^{3072} = (2^{10})^{307.2} \approx (10^{3})^{307.2} = 10^{921.6} \approx 10^{460} \cdot 10^{460}$$
So $p$ and $q$ are each about 460 digits long. The exact value is $3072 \cdot \log_{10} 2 \approx 924.8$ digits for $N$, i.e. about 462 per prime; the estimate is close enough.
Counting 400-digit primes with the prime number theorem $\pi(x) \approx \dfrac{x}{\ln x}$:
$$\pi(10^{400}) - \pi(10^{399}) \approx \frac{10^{400}}{400 \ln 10} - \frac{10^{399}}{399 \ln 10} = \frac{10^{399}}{\ln 10}\left(\frac{10}{400} - \frac{1}{399}\right) \approx 10^{399} \cdot 10^{-2} = 10^{397}$$
Why it matters: there are vastly more large primes than atoms in the universe (about $10^{80}$), so nobody can precompute a list of all RSA primes, and two key generators practically never pick the same prime. About one in every $400 \ln 10 \approx 920$ numbers of this size is prime, so a random search finds one quickly.