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

• Difficulty Level : Easy
• Last Updated : 10 Nov, 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:

## 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"``)`

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.

Now lets see the code for the first optimization method ( i.e. checking till √n )

## Python3

 `from` `math ``import` `sqrt``# n is the number to be check whether it is prime or not``n ``=` `1`` ` `# no lets check from 2 to sqrt(n)``# if we found any facto then we can print as not a prime number`` ` `# this flag maintains status whether the n is prime or not``prime_flag ``=` `0`` ` `if``(n > ``1``):``    ``for` `i ``in` `range``(``2``, ``int``(sqrt(n)) ``+` `1``):``        ``if` `(n ``%` `i ``=``=` `0``):``            ``prime_flag ``=` `1``            ``break``    ``if` `(prime_flag ``=``=` `0``):``        ``print``(``"true"``)``    ``else``:``        ``print``(``"false"``)``else``:``    ``print``(``"false"``)`

Output

`false`

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)

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