By using for loop how i printed prime numbers?

Answer 1

for(i=0;i<n;i++)
{ count=0;
for(j=1;j<=i;j++)
{
if(i%j==0)
{
count++;
}
}
if(count==1&&count==2)
{
// print i
}
}
// this will print all the prime number between limit
i.e.,0 to n numbers

Answer 2

The iteration limit to find prime numbers can be square root of the tested number, instead its half:

x = sqrt( x ) * sqrt( x ), so, all x values greater than sqrt( x ) were already tested, by testing divisors lighter than x ^^

Answer 3

int numDiv = 0;
int max = 2947;
//max serves as a limit for how many primes, highest number you want to test.
for(int x = 0; x < max; x++){
numDiv = 0;
for(int y = 2; y <= x/2; y++){
//tests divisors of x
//prime numbers have no divisiors other than 1 and themselves
if(x%y == 0){
numDiv++;
}
}
if(numDiv == 0){
System.out.println(x + " is prime");
}
}

Answer 4

Abdul,
having your for loop to x < n, is a waste of computational time. the lowest number that a non prime can be divisible by is 2, therefore the highest divisor would be n/2. testing any case over the half way point is a wast of time and processing power.
what if you were trying to determine if a Large number was prime. you wouldn't want to look at twice the numbers you dont need to. think about that.