How many Prime Nos. r there which are less than N(any given number)?

Question

Is there a general method? If no, what if N=multiple of 100, say 2300?

-Thanks.

Answer 1

It is approximately pi(N), where pi is the famous number theory function, pi(N)= # of primes <= N. The famous "prime number theorem" says pi(N) is approximately N/ln(N). So for 2300, 2300/ln(2300) is about 297.132 whereas there are actually 342 primes <2300. There is an error term in the theorem. See http://mathworld.wolfram.com/PrimeNumberTheorem.html

Answer 2

The simplest formula has already been given.

The next simplest formula is (no. of primes < N) = N / (log(N) - 1) where log means natural logarithm. When N is 2300, this gives 341 primes instead of the actual 342 . . . hey, that's pretty good!

For N = 4000, 6000, 9000 this formula gives 548, 779, 1110 as the expected number of primes, compared with the actual numbers of 550, 783, 1117. Okay, it's getting a bit worse, but it's still a lot better than N / log(N).

There are more complicated formulae which are more accurate.

Answer 3

I'd guess there would be no general method

For instance, between 61 and 69, N could be any one of 62, 63, 64, 65, 66, 67 and 68 with the same number of primes less than N.

Answer 4

According to the prime number theorem, the number of primes less than N is approximately N/logN

Answer 5

I just know that, if for any prime 1< x <= sqrt(N) N is not divisible by x, then N is a prime number.