prime number betn 1 to 100

Question

Can someone please explain to me what happens in the second loop when i=6 elaborately?
//problem when i=6 and j=3
int main()
{
int num1,num2;
int fnd=0 //1st for loop prime number counter
, ctr=0;//2nd for loop counter;

cout<<" Input number for starting range: ";
cin>> num1;//1
cout << " Input number for ending range: ";
cin>> num2;//100

cout << "\n The prime numbers between "<<num1<<" and "<<num2<<" are:"<<endl;

for( int i= num1; i<=num2; i++) // 1 2 3 4 5
{
for(int j=2; j<=sqrt(i); j++) // 1 2 3 not enter coz 2> 1.414...4 enter coz 2<=2 so j now becomes 3>root 5 so 5 not enter
{
if(i%j==0) //4%2==0
{
ctr++; //1 always as ctr=0 at last
}
}
if(ctr==0 && i!=1) //so ctr 1 !=0 not enter this condition //2 enter coz ctr == 0 2!=1 // 4 not enter coz ctr==1
{
fnd++; //1 2 3
cout<< i << " "; // 2 3 5

}
ctr=0; //again ctr =0
}

cout<<endl<<"total prime numbers: " <<fnd;
}

Answer 1

j can't be equal 3, when i=6, cuz max j value is j=sqrt(6)
So if you cast sqrt(6)(=2,449) to int, it becomes lower integer, in this case it's 2

So it is not possible to have i=6 and j=3 at the same time

Answer 2

thank you