We Will Be Counting Primes

We can solve it mathematically too. We know that $\Pi (x)=\int _{ 2 }^{ x }{ \frac { dt }{ log(t) } }$ . So, just putting the value of 'x' as 2 x $10^{11}$ we get answer as 8.007 x $10^{9}$

I used Sage for this one:

prime_pi(2*10^11)

But for a true algorithm, see this

The simplest is Sieve of Eratosthenes

Explain me the music symbols in the Question title, please?

Java code:

public class brilliant{
    public static void main(String args[]){
        int res = 0;
        long[] primes = new long[0];
        for(long i=2;i<200000000000L;i++){
            boolean isPrime = true;
            for(int j=0;j<primes.length;j++) if(i%primes[j]==0) isPrime = false;
            if(isPrime){
                primes = add(primes,i);
                res++;
            }
        }
        System.out.println(res);
    }
    private static long[] add(long[]a, long b){
        long[] res = new long[a.length+1];
        for(int i=0;i<a.length;i++) res[i]=a[i];
        res[a.length]=b;
        return res;
    }
}

(Just kidding, I used WolframAlpha for this one)

We already have Sage.

It's rich cousin*, Mathematica:

FromDigits[IntegerDigits[PrimePi[2*10^11]][[;; 3]]]

returns

800

* all respect to Sage, it is wonderful free as it is.

using Eulero's φ(2x10^11)=φ(2x2^11x5^11)=φ(2^12x5^11) we get φ=2^12x5^11x(1-1/2)x(1-1/5)=8x10^10. So answer is 800