CareerCup

1. If the number is less than 2 return false.
2. If the numbers is 2 return true.
3. let x = square_root_of (n)
4. for i = 3 to x
5. if ( x % number == 0) return false;
6. return true;

Why Square root? (step 3)
..... because for any non prime number, one of the factors is <= the square root.

Why less than 2? (step1)
......negative numbers and zero are not prime
......1 is not prime

#include<stdio.h>

int main()
{
	int i, n, flag;
	int str[10];
	printf("\nEnter the number\n");
	scanf("%d", &n);
	flag = 1; //----------------Assuming that the number is prime-----//
	for(i = 2; i <= n/2; i++)
	{
		if(n % i == 0)
			flag = 0;
	}
	
	if(0 == n || 1 == n || flag == 1)
		printf("\n%d is a prime number\n", n);
	else
		printf("\n%d is a not prime number\n", n);
}

public static boolean isPrime(int num)
	{
		if(num<=0) 
		{
			System.out.println("not a natural number");
			return false;
		}
		if(num==1)
			{return true;}
		int limit=(int) Math.sqrt((double)num);
		for(int i=2;i<limit;i++)
		{
			if(num%i==0)
			{
				return false;
			}
		}
		return true;

}

public static boolean isPrime(int num)
	{
		if(num<=0) 
		{
			System.out.println("not a natural number");
			return false;
		}
		if(num==1)
			{return true;}
		int limit=(int) Math.sqrt((double)num);
		for(int i=2;i<limit;i++)
		{
			if(num%i==0)
			{
				return false;
			}
		}
		return true;

}

for(int i=2;i<limit;i++) ... Following for loop condition in above code should be modified as for(int i=2;i<=limit;i++)....check for num = 25.

All integers can be expressed as (6k + i) for some integer k and i = -1 0 1 2 3 4.

Integers of the form (6k + 0), (6k + 2) and (6k + 4) can be divided by 2. Integers that can be express as (6k + 3) can be divided by 3. Hence:

public boolean isPrime(int n) {
        if (n <= 1)
            return false;
        if (n <= 3) {
            return true;
        }
        if (n % 2 == 0) {
            return false;
        }
        if (n % 3 == 0) {
            return false;
        }

        int squareRootOfN = (int) Math.sqrt(n);
        int x = 5;

        while (x <= squareRootOfN) {
            if (n % x == 0)
                return false;
            x+=2;
            if (n % x == 0)
                return false;
            x+=4;
        }
        return true;
    }

#include<math.h>

int isPrime( int n)
{
int i;

for( i = 2; i <= sqrt(n); i++)
if( n%i == 0)
return 0;
else
return 1;
}

int main()
{
if(isPrime(16))
printf("Yes it's a prime !!");
else
printf("No it's not a prime !!");
return 0;
}

bool checkPrime(int n)
{
if(n==1) return false;
for(int i=2;i<=sqrt(n);i++)
{
if((n%i)==0) return false;
}
return true;
}

import java.lang.Math.*;
import java.io.*;
public class Prime
{

public static Boolean primeNumber(long x)
{
Boolean flag = true;
if(x<4)
{
if(x==1)
flag=false;
}
else
{
for(int i = 2; i<(int)Math.sqrt(x)+1;i++)
{
if(x%i==0)
{
flag=false;
}
}
}
return flag;
}
public static void main(String args[])
{
Boolean flag;
flag =primeNumber(Long.parseLong(args[0]));
System.out.println(flag);
}
}