primes

This is quite an odd problem I know, but I'm trying to get a copy of the current largest prime number in a file. Getting the number in integer form is fairly easy. I just run this. prime = 2**74207281 - 1 It takes about half a second and it works just fine. Operations are fairly quick as well. Dividing it by 10 (without decimals) to shift the digits is quick. However, str(prime) is taking a very long time. I reimplemented str like this, and found it was processing about a hundred or so digits per second. while prime > 0: strprime += str(prime%10) prime //= 10 Is there a way to do this more

#include <iostream> using namespace std; void whosprime(long long x) { bool imPrime = true; for(int i = 1; i <= x; i++) { for(int z = 2; z <= x; z++) { if((i != z) && (i%z == 0)) { imPrime = false; break; } } if(imPrime && x%i == 0) cout << i << endl; imPrime = true; } } int main() { long long r = 600851475143LL; whosprime(r); } I'm trying to find the prime factors of the number 600851475143 specified by Problem 3 on Project Euler (it asks for the highest prime factor, but I want to find all of them). However, when I try to run this program I don't get any results. Does it have to do with how

This is my isPrime method: private static boolean isPrime(int num) { if (num % 2 == 0) return false; for (int i = 3; i * i < num; i += 2) if (num % i == 0) return false; return true; } I put isPrime(9) and it returns true . What is wrong with the method? Tareq Salah Your condition should be i * i <= num private static boolean isPrime(int num) { if (num == 2) return true; if (num < 2 || num % 2 == 0) return false; for (int i = 3; i * i <= num; i += 2) if (num % i == 0) return false; return true; } You didn't take number 9 in your consideration so 9<9 will result false. But you need to check 9.

Hi can anyone tell me how to implement Sieve of Eratosthenes within this code to make it fast? Help will be really appreciated if you can complete it with sieve. I am really having trouble doing this in this particular code. #!/usr/bin/env python import sys T=10 #no of test cases t=open(sys.argv[1],'r').readlines() import math def is_prime(n): if n == 2: return True if n%2 == 0 or n <= 1: return False sqr = int(math.sqrt(n)) + 1 for divisor in range(3, sqr, 2): if n%divisor == 0: return False return True #first line of each test case a=[1,4,7,10,13,16,19,22,25,28] count=0 for i in a: b=t[i]

Possible Duplicate: Fastest algorithm for primality test Would appreciate a reference to sample code for fast primality testing in C#, preferably using BigInteger or other variable size type. This is a Miller Rabin test in c#: bool MillerRabin(ulong n) { ulong[] ar; if (n < 4759123141) ar = new ulong[] { 2, 7, 61 }; else if (n < 341550071728321) ar = new ulong[] { 2, 3, 5, 7, 11, 13, 17 }; else ar = new ulong[] { 2, 3, 5, 7, 11, 13, 17, 19, 23 }; ulong d = n - 1; int s = 0; while ((d & 1) == 0) { d >>= 1; s++; } int i, j; for (i = 0; i < ar.Length; i++) { ulong a = Math.Min(n - 2, ar[i]);

I am doing a project at the moment and I need an efficient method for calculating prime numbers. I have used the sieve of Eratosthenes but, I have been searching around and have found that the sieve of Atkin is a more efficient method. I have found it difficult to find an explanation (that I have been able to understand!) of this method. How does it work? Example code (preferably in C or python) would be brilliant. Edit: thanks for your help, the only thing that I still do not understand is what the x and y variables are referring to in the pseudo code. Could someone please shed some light on

So I've spent hours trying to work out exactly how this code produces prime numbers. lazy val ps: Stream[Int] = 2 #:: Stream.from(3).filter(i => ps.takeWhile{j => j * j <= i}.forall{ k => i % k > 0}); I've used a number of printlns etc, but nothings making it clearer. This is what I think the code does: /** * [2,3] * * takeWhile 2*2 <= 3 * takeWhile 2*2 <= 4 found match * (4 % [2,3] > 1) return false. * takeWhile 2*2 <= 5 found match * (5 % [2,3] > 1) return true * Add 5 to the list * takeWhile 2*2 <= 6 found match * (6 % [2,3,5] > 1) return false * takeWhile 2*2 <= 7 * (7 % [2,3,5] > 1)

I came across this question.A number is called lucky if the sum of its digits, as well as the sum of the squares of its digits is a prime number. How many numbers between A and B are lucky? 1 <= A <= B <= 10 18 . I tried this. First I generated all possible primes between 1 and the number that could be resulted by summing the squares (81 *18 = 1458). I read in A and B find out maximum number that could be generated by summing the digits If B is a 2 digit number ( the max number is 18 generated by 99). For each prime number between 1 an max number. I applied integer partition algorithm. For