Python program to check whether a number is Prime or not

Difficulty Level : Easy
Last Updated : 12 Jul, 2021

Given a positive integer N, The task is to write a Python program to check if the number is prime or not.
Definition: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The first few prime numbers are {2, 3, 5, 7, 11, ….}.

Examples :

Input: n = 11
Output: true
Input: n = 15
Output: false
Input: n = 1
Output: false

The idea to solve this problem is to iterate through all the numbers starting from 2 to (N/2) using a for loop and for every number check if it divides N. If we find any number that divides, we return false. If we did not find any number between 2 and N/2 which divides N then it means that N is prime and we will return True.

Below is the Python program to check if a number is prime:

C

// C program for
// checking if a number is prime
#include <stdio.h>
int main()
{
    int i,v=1;
    int a = 1;
    // iterate through 2 to a/2
    for (i=2;i<=a/2;i++)
      {
          v = a%i;
      // if remainder is zero, the number is not prime
          if(v==0)
            break;
      }
    if (v==0 || a==1)
      printf("%d is not prime number",a);
    else
      printf("%d is prime number",a);
}
// This Code is Contributed by
// Siddharth Verma

Python3

# Python program to check if
# given number is prime or not
  
num = 11
  
# If given number is greater than 1
if num > 1:
  
    # Iterate from 2 to n / 2
    for i in range(2, int(num/2)+1):
  
        # If num is divisible by any number between
        # 2 and n / 2, it is not prime
        if (num % i) == 0:
            print(num, "is not a prime number")
            break
    else:
        print(num, "is a prime number")
  
else:
    print(num, "is not a prime number")

C++

// C++ program for
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
int main() {
    // Given number
    int n=11;
    // checking the given number 
    // whether it is 1 or not
      if(n==1)
      {
          cout<<n<<" is not a prime number";
      }
      else
      {
          int f=0;
          // iterate from 2 to n/2
          for(int i=2;i<=(n/2);i++)
          {
              // If n is divisible by any number between
            // 2 and n/2, it is not prime
              if(n%2==0)
              {
                  f=1;
                  // break out of for loop as 
                  // it is not prime
                  break;
              }
          }
          if(f==1)
          {
              cout<<n<<" is not a prime number";
          }
          else
          {
              cout<<n<<" is a prime number";
          }
      }
    return 0;
}
// This code is contributed by
// Murarishetty Santhosh Charan

Output

11 is a prime number

Optimized Method
We can do the following optimizations:

Instead of checking till n, we can check till √n because a larger factor of n must be a multiple of a smaller factor that has been already checked.
The algorithm can be improved further by observing that all primes are of the form 6k ± 1, with the exception of 2 and 3. This is because all integers can be expressed as (6k + i) for some integer k and for i = -1, 0, 1, 2, 3, or 4; 2 divides (6k + 0), (6k + 2), (6k + 4); and 3 divides (6k + 3). So a more efficient method is to test if n is divisible by 2 or 3, then to check through all the numbers of form 6k ± 1. (Source: wikipedia)

Output:

 true
 false

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