# Prime number or Not in C using do-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,

11 is a *Prime Number*, because 11 is not divisible by any number other than 1 and 11.

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

**Related: Prime number or Not in C using While Loop**

## Working:

- First the computer reads the positive integer value from the user.
- Then using do-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 13.
- It assigns the value of flag=0, n=13.
- It assigns the value of i=2 and the loop continues till the condition of the do-while loop is true.

3.1. do

(n!=2) && (n%i==0) ((13!=2) && (13%2==0)) if condition is false

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

i<=sqrt(n) (3<=√13) do-while loop condition is true

3.2. do

(n!=2) && (n%i==0) ((13!=2) && (13%3==0)) if condition is false

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

i<=sqrt(n) (4<=√13) do-while loop condition is false

3.3. It comes out of the do-while loop.

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

So it prints 13 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 9.
- It assigns the value of flag=0, n=9.
- It assigns the value of i=2 and the loop continues till the condition of the do-while loop is true.

3.1. do

(n!=2) && (n%i==0) ((9!=2) && (13%2==0)) if condition is false

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

i<=sqrt(n) (3<=√9) do-while loop condition is true

3.2. do

(n!=2) && (n%i==0) ((9!=2) && (9%3==0)) if condition is true

It assigns flag=1

breaks the loop and comes out of the do-while loop

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

So it goes to else part and prints 9 is not a prime number.

- Thus program execution is completed.

## Output:

### TO DOWNLOAD THE PROGRAM CODE : CLICK HERE