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How to check NoneType in Python

Check Prime Number in Python

Last Updated : 10 Apr, 2025

Given a positive integer N, the task is to write a Python program to check if the number is Prime or not in Python. For example, given a number 29, it has no divisors other than 1 and 29 itself. Hence, it is a prime number.

Note: Negative numbers (e.g. -13) are not considered prime number.

Using sympy.isprime() method

In the sympy module, we can test whether a given number n is prime or not using sympy.isprime() function. For n < 2⁶⁴ the answer is definitive; larger n values have a small probability of actually being pseudoprimes. Before using sympy module, we need to install it using this command:

pip install sympy

Python

from sympy import * g1 = isprime(30) g2 = isprime(13) g3 = isprime(2) print(g1) print(g2) print(g3)

Output

False True True

Explanation:

isprime() function from the SymPy library checks if a number is prime or not.
It prints False for 30, True for 13 and True for 2 because 30 is not prime, while 13 and 2 are prime numbers.

Using Math Module

The code implements a basic approach to check if a number is prime or not, by traversing all the numbers from 2 to sqrt(n)+1 and checking if n is divisible by any of those numbers.

Python

import math  n = 11 if n <= 1:     print(False) else:     is_prime = True     for i in range(2, int(math.sqrt(n)) + 1):         if n % i == 0:             is_prime = False             break     print(is_prime)

Output

True

Explanation:

Check if n <= 1, if true, it’s not prime.
Loop from 2 to the square root of n, if n % i == 0, it’s not prime. If no divisors found, n is prime.

Using Flag Variable

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 let’s see the code for the first optimization method

Python

from math import sqrt  n = 17 p_fl = 0  if(n > 1):     for i in range(2, int(sqrt(n)) + 1):         if (n % i == 0):             p_fl = 1             break     if (p_fl == 0):         print("True")     else:         print("False") else:     print("False")

Output

True

Explanation:

If n is greater than 1, it loops from 2 to the square root of n to check if any number divides n evenly.
If any divisor is found, the p_fl is set to 1, indicating n is not prime. If no divisors are found, it prints “True” (meaning n is prime). Otherwise, it prints “False”.
If n <= 1, it directly prints “False” since numbers less than or equal to 1 are not prime.

Using Miller-Rabin Primality Test

We can use the Miller-Rabin Primality Test, a probabilistic method, to check if a number is prime by performing multiple rounds of testing, where each test verifies if a randomly chosen base witnesses the compositeness of the number.

Python

import random  n = 30 k = 5  if n <= 1:     print(False) elif n <= 3:     print(True) elif n % 2 == 0:     print(False) else:     d = n - 1     while d % 2 == 0:         d //= 2      is_prime = True     for _ in range(k):         a = random.randint(2, n - 2)         x = pow(a, d, n)                  if x == 1 or x == n - 1:             continue                  while d != n - 1:             x = (x * x) % n             d *= 2             if x == 1:                 is_prime = False                 break             if x == n - 1:                 break         if not is_prime:             break     print(is_prime)  n = 3 k = 5  if n <= 1:     print(False) elif n <= 3:     print(True) elif n % 2 == 0:     print(False) else:     d = n - 1     while d % 2 == 0:         d //= 2      is_prime = True     for _ in range(k):         a = random.randint(2, n - 2)         x = pow(a, d, n)                  if x == 1 or x == n - 1:             continue                  while d != n - 1:             x = (x * x) % n             d *= 2             if x == 1:                 is_prime = False                 break             if x == n - 1:                 break         if not is_prime:             break     print(is_prime)

Output

False True

Explanation:

Return False if n <= 1 or even, True if n <= 3.
Set d = n-1 and repeatedly divide by 2. Perform k iterations, picking a random base a and checking x = a^d % n.
If any test fails, return False; else, return True.

Using Recursion

We can also find the number prime or not using recursion. We can use the exact logic shown in method 2 but in a recursive way.

Python

from math import sqrt  def Prime(n, i):       if i == 1 or i == 2:           return True     if n % i == 0:           return False     if Prime(n, i - 1) == False:           return False      return True  n = 13 i = int(sqrt(n) + 1)  print(Prime(n, i))

Output

True

Explanation:

Base condition: The recursion ends when the iterator i reaches 1 or 2, returning True because numbers 1 and 2 are prime.
Divisibility check: If n is divisible by i (i.e., n % i == 0), it returns False (meaning n is not prime).
Recursive call: The function calls itself with i – 1, checking for divisibility from i down to 2.
If the function finds no divisors by the time i reaches 2, it returns True, indicating the number is prime.

Recommended Artilce – Analysis of Different Methods to find Prime Number in Python

How to check NoneType in Python

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