Finding prime number between 1 and N - c

Question

In fact my teacher has passed the program that calculates the prime numbers from 1 to N. But I did not understand some things in the code and would like to help.
In line 18 we have the following: for(j=2; j<=i/2; j++), because j was divided by 2? and why j start in 2? should not start i and j on 1?
#include <stdio.h>
int main() {
int i, j, n, isPrime; //isPrime is used as flag variable
/* Reads upper limit to print prime */
printf("Find prime numbers between 1 to : ");
scanf("%d", &n);
printf("\nAll prime numbers between 1 to %d are:\n", n);
/* Finds all Prime numbers between 1 to n */
for(i=2; i<=n; i++) {
/* Assume that the current number is Prime */
isPrime = 1;
/* Check if the current number i is prime or not */
for(j=2; j<=i/2; j++) {
/*
* If i is divisible by any number other than 1 and self
* then it is not prime number
*/
if(i%j==0) {
isPrime = 0;
break;
}
}
/* If the number is prime then print */
if(isPrime==1) {
printf("%d is Prime number\n", i);
}
}
return 0;
}

Answer 1

should not start i ... on 1?
Yes, i should at 1. Consider that the current code will evolve. Later code may look like the below. Notice the comment and code matches your contract "numbers between 1 and N". No implied short-cut starting at 2, but clarity that code is properly testing all numbers [1...N].
/* Finds all Prime numbers between 1 to n */
for(i=1; i<=n; i++) {
if (IsPrime(i)) {
printf("%d is Prime number\n", i);
}
}
The point is that as a programmer, you are given the task "Finding prime number between 1 and N". A common sub-task is to code bool IsPrime(int i). Certainly if you code everything together, i can start at 2 - we know 1 is not a prime. Yet good programming begins with software architecture and that involves divide and conquer. So making a stand-alone helper function that works for all input is a good first step.
should not start j ... on 1?
No. Now code is into its find-a-prime algorithm and starting j=1 would fail.
bool IsPrime(int i) {
if (i <= 1) return false; // Cope with negative, 0, and 1.
for (int j=2; j<=i/2; j++) {
if (i%j==0) {
return false;
}
}
return true;
}
Of course, we can optimize IsPrime() in many ways.

Answer 2

The first prime number is 2 -- which is why you start as 2 -- everything is divisible by 1, and if you start there your algo will fail.
You end with N/2 because testing larger number will not result in anything that you would not have found, simply because to be a non-prime means that you will have to have to have at least 2 factors, and the smallest prime is 2 and 2*(i/2) >= i -- but in reality it is better and safe to stop at square root of N (but maybe that is in the next lesson).
This starting at 2 and increment the loop by one is wasteful -- since all primes except 2 is odd it would be better to start at 3 and increment by 2, and just make a special test for dividing by 2 outside the loop.

Answer 3

The first prime number is known to be 2, so there is no reason to start with 1. Given number i, there is no reason to try to divide it by numbers greater than its half, as the result will always be less than 2 and yield a quotient of 0 and a non-zero remainder (unless we divide it by itself, which is pointless in the prime test).

Answer 4

This is the easiest way:
#include <iostream>
using namespace std;
int main (){
int first,second;
cout<<"Enter starting of limit";
cin>>first; //First value
cout<<"Enter Ending of limit";
cin>>second; //Second value
L:
if(first < second){ //Checking starts here
if(first % 2 != 0 && first % 5 != 0 || first - 5 == 0 && first % 7 != 0){
cout<<first<<endl;
first++ ;
goto L; //Iteration
}
}
}

Answer 5

#include<stdio.h>
int main()
{
int num,i,count,n;
printf("Enter max range:");
scanf("%d",&n);
for(num=1; num<=n; num++)
{
count=0;
for(i=2; i<=num/2; i++)
{
if(num%i==0)
{
count++;
break;
}
}
if(count==0 && num!=1)
printf("%d ",num);
}
return 0;
}