C++ Program to find Prime Number or Not using While Loop

Before going to the program for Prime Number or Not first let us understand what is a Prime Number?

Prime Number:

A Prime Number is a number greater than 1 and which is only divisible by 1 and the number itself.

For example,

17 is a Prime Number, because 17 is not divisible by any number other than 1 and 17.

To find whether a Number is Prime Number or Not it is enough to check whether ‘n’ is divisible by any number between 2 and √n. If it is divisible then ‘n’ is not a Prime Number otherwise it is a Prime Number.

Related: C Program to display Prime Numbers between Two Intervals

Program code for Prime Number or Not in C++:

/* Prime Number or Not */
#include<iostream.h>
#include<conio.h>
#include<math.h> void main()
{ int n, i=2, flag=0; clrscr(); cout<<"Enter a positive integer value: "; cin>>n; /* Loop to check whether 'n' is divisible by any number between 2 and sqrt(n) */ while(i<=sqrt(n)) { if(n%i==0) { flag=1; break; } i++; } /* if else condition to print Prime Number or Not */ if (flag==0) cout<<n<<" is a prime number."; else cout<<n<<" is not a prime number."; getch();
}

Related: Prime number or Not in C using For Loop

Working:

First the computer reads the positive integer value from the user.
Then using while loop it checks whether ‘n’ is divisible by any number between 2 and √n.
Finally if else condition is used to print the number is prime number or not .

Step by Step working of the above Program Code:

For Prime Number:

Let us assume that a user enters the positive integer value as 17.
It assigns the value of flag=0, n=17.
It assigns the value of i=2 and the loop continues till the condition of the while loop is true.

3.1. i<=sqrt(n) (2<=√17) while loop condition is true

n%i==0 (17%2==0) if condition is false

i++ (i=i+1) So i=3

3.2. i<=sqrt(n) (3<=√17) while loop condition is true

n%i==0 (17%3==0) if condition is false

i++ (i=i+1) So i=4

3.3. i<=sqrt(n) (4<=√17) while loop condition is true

n%i==0 (17%4==0) if condition is false

i++ (i=i+1) So i=5

3.4. i<=sqrt(n) (5<=√17) while loop condition is false

It comes out of the while loop.

flag==0 (0==0) if condition is true

So it prints 17 is a prime number.

Thus program execution is completed.

For Not a Prime Number:

Let us assume that a user enters the positive integer value as 21.
It assigns the value of flag=0, n=21.
It assigns the value of i=2 and the loop continues till the condition of the while loop is true.

3.1. i<=sqrt(n) (2<=√21) while loop condition is true

n%i==0 (21%2==0) if condition is false

i++ (i=i+1) So i=3

3.2. i<=sqrt(n) (3<=√21) while loop condition is true

n%i==0 (21%3==0) if condition is true

It assigns flag=1

breaks the loop and comes out of the while loop