Today we will learn the prime number program in c and how to check whether the given number is prime or not. So before learning, you should have a piece of knowledge about a prime number.so first of all,

Table of Contents

### What are Prime Numbers?

Any natural number which can be divisible by itself or the number which has only two factors i.e 1 and the number itself are called prime numbers.

**For example** 2,3,5,7,11 and so on…

**Note**: The number 2 is only even prime number because most of the numbers are divisible by 2.

## Program to check a single number is prime or not in c:

In this program,** we will take** input from the user and check whether the given **input number is prime or not**.

<div class="wp-block-codemirror-blocks-code-block code-block"> <pre>#include<stdio.h> int main(){ int n,i,m=0,flag=0; printf("Enter the number to check prime:"); scanf("%%d",&n); m=n/2; for(i=2;i<=m;i++) { if(n%%i==0) { printf("Number is not prime"); flag=1; break; } } if(flag==0) printf("Number is prime"); return 0; } </pre> </div>

**Output:**

## C Program to check given number is prime or not using for loop:

In this program, we will ask the user to input n number and check whether the given n number is prime or not using for loop.

<div class="wp-block-codemirror-blocks-code-block code-block"> <pre>#include <stdio.h> main() { int n, i, c = 0; printf("Enter any number n to check:"); scanf("%%d", &n); for (i = 1; i <= n; i++) { if (n %% i == 0) { c++; } } if (c == 2) { printf("given n is a Prime number"); } else { printf("given n is not a Prime number"); } return 0; }</pre> </div>

**Output:**

**Explanation:**

let us consider n=5 for loop becomes **for(i=1; i<5; i++)**

**1st Iteration:** for(i=1; i<=5; i++) in 1st iteration i is incremented i.e the value of i for the next iteration is 2.If (n%i==0) then c is incremented i.e (5%1==0) then c is incremented.Here, (5%1=0) so c is incremented i.e c=1.

**2nd Iteration: **for(i=2;i<=5;i++) in 2nd iteration i is incremented i.e the value of i for the next iteration is 3.If (n%i==0) then c is incremented i.e (5%2==0) then c is incremented.here,(5%2!=0) so c is not incremented i.e c=1.

**3rd Iteration: **for(i=3;i<=5;i++) in 3nd iteration i is incremented i.e the value of i for the next iteration is 4.If (n%i==0) then c is incremented i.e (5%3==0) then c is incremented.Here,(5%3!=0) so c is not incremented i.e c=1.

**4th Iteration: **for(i=4;i<=5;i++) in 4th iteration i is incremented i.e the value of i for the next iteration is 5.If (n%i==0) then c is incremented i.e (5%4==0) then c is incremented.Here,(5%4!=0) so c is not incremented i.e c=1.

**5th Iteration: **for(i=5;i<=5;i++) in 5th iteration i is incremented i.e the value of i for the next iteration is 6.If (n%i==0) then c is incremented i.e (5%5==1) then c is incremented.Here,(5%5=0) so c is incremented i.e c=2.

**6th Iteration: **for(i=6;i<=5;i++) in 6th iteration the value of i is 6 which is greater than n i.e i>n(6>5) so the for loop is terminated.Here c=2 so the given number is prime number.

## C Program to check given number is prime or not using while loop:

In this program we will ask user to enter the number which user want to check whether it is prime or not and then check it using while loop.

<div class="wp-block-codemirror-blocks-code-block code-block"> <pre>#include <stdio.h> int main() { int i = 2, Number, count = 0; printf("\n Please Enter number to Check Prime or not \n"); scanf("%%d", &Number); while(i <= Number/2) { if(Number%%i == 0) { count++; break; } i++; } if(count == 0 && Number != 1 ) { printf("\n %%d is a Prime Number", Number); } else { printf("\n %%d is Not a Prime Number", Number); } return 0; }</pre> </div>

**Output:**

In above program we just replace the for loop instead we add while loop.

## C program to check the given number is prime or not using functions:

In this program, we will use the function **find_factors** to check whether the given number is prime or not.

<div class="wp-block-codemirror-blocks-code-block code-block"> <pre>#include <stdio.h> int find_factors(int Number) { int i, Count = 0; for (i = 2; i <= Number/2; i++) { if(Number%%i == 0) { Count++; } } return Count; } int main() { int i, Number, count = 0; printf("\n Please Enter any number to Check Prime or not \n"); scanf("%%d", &Number); count = find_factors(Number); if(count == 0 && Number != 1 ) { printf("\n %%d is a Prime Number", Number); } else { printf("\n %%d is Not a Prime Number", Number); } return 0; }</pre> </div>

**Output:**

## Sieve Of Eratosthenes Algorithm:

In this algorithm, we first declare the numbers from [2:n]. We mark all the multiples of 2 as the number 2 is composite. A proper multiple of number x, which is multiple of number **x** and divisible by **x**.

Then we mark the number which hasn’t be composite.for example 3 which is not composite,It means that 3 is a prime number.

Now we will see the program for Sieve Of Eratosthenes Algorithm.

<div class="wp-block-codemirror-blocks-code-block code-block"> <pre>int count_primes(int n) { const int S = 10000; vector<int> primes; int nsqrt = sqrt(n); vector<char> is_prime(nsqrt + 1, true); for (int i = 2; i <= nsqrt; i++) { if (is_prime[i]) { primes.push_back(i); for (int j = i * i; j <= nsqrt; j += i) is_prime[j] = false; } } int result = 0; vector<char> block(S); for (int k = 0; k * S <= n; k++) { fill(block.begin(), block.end(), true); int start = k * S; for (int p : primes) { int start_idx = (start + p - 1) / p; int j = max(start_idx, p) * p - start; for (; j < S; j += p) block[j] = false; } if (k == 0) block[0] = block[1] = false; for (int i = 0; i < S && start + i <= n; i++) { if (block[i]) result++; } } return result; }</pre> </div>

Here we have a program that counts the no of primes smaller than or equal to n.

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