# Python program to check whether a number is Prime or not

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 = 11Output:true

Input:n = 15Output:false

Input:n = 1Output: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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