# 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:

1. Let us assume that a user enters the positive integer value as 17.
2. It assigns the value of flag=0, n=17.
3. 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.

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

So it prints 17 is a prime number.

1. Thus program execution is completed.

For Not a Prime Number:

1. Let us assume that a user enters the positive integer value as 21.
2. It assigns the value of flag=0, n=21.
3. 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

1. flag==0    (1==0)    if condition is false

So it goes to else part and prints 21 is not a prime number.

1. Thus program execution is completed.

## Output:  ### 2 Responses

1. Khushboo jain says:

What if you enter 2,how the loop works?

• devang says:

1. If you enter 2.

2. It assigns the value of flag=0, n=2.

3. 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<=√2) while loop condition is false.

4. flag==0 (0==0) if condition is true
So it prints 2 is a prime number.