random a 512-bit integer N that is not a multiple of 2, 3, or 5

问题

if you are to choose a random a 512-bit integer N that is not a multiple of 2, 3, or 5 What is the probability that N is prime? i don't know the algorithm behind this one... i'm trying to work on a project but this is the starting point.. :)

回答1:

The number of primes less than n=2⁵¹² is approximately n/log(n). The number of numbers you are considering is 4/15*n, so the probability you are looking for is 15/(4*log(n)), which is about 1 %.

回答2:

Probability bounds

You may use the following inequality for the prime pi function:

(Where log is taken in base e)

So:

8.58774*10¹⁵¹ < π(2⁵¹²) < 8.93096*10¹⁵¹

And as you are only leaving alive 4/15 n numbers (because of killing he multiples of 2, 3 and 5), te probability is bounded by:

8.58774*10¹⁵¹/(4/15 2⁵¹²) < P < 8.93096*10¹⁵¹/(4/15 2⁵¹²)

Or:

0.010507 < P < 0.010687

Which is a nice, pretty tight bound.

回答3:

This sounds homeworkish so I suggest you generate some 512bit numbers and use some well known prime tests to get an approximate answer heuristically.

回答4:

Do you want an exact answer or an approximation? For an approximation you can use the prime number theorem or prime counting function.

来源：https://stackoverflow.com/questions/5264605/random-a-512-bit-integer-n-that-is-not-a-multiple-of-2-3-or-5