import math
def isPrimeNumber(n):
if (n < 2):
return False;
sq = int(math.sqrt(n))
for i in range(2, sq + 1):
if (n % i == 0):
return False
return Trueuntil = 20
[n for n in range(2, until) if all(n % m != 0 for m in range(2, n-1))]from math import sqrt
for i in range(2, int(sqrt(num)) + 1):
if num % i == 0:
print("Not Prime")
break
print("Prime")
# Note: Use this if your num is big (ex. 10000 or bigger) for efficiency
# The result is still the same if the num is smallerdef is_prime(n):
if n in (2, 3):
return True
if n <= 1 or not (n%6==1 or n%6==5):
return False
a, b= 5, 2
while a <= n**0.5:
if not n%a:
return False
a, b = a+b, 6-b
return True
# this method is much faster than checking every number because it uses the fact
# that every prime is either 1 above or 1 below a multiple of 6
# and that if a number has no prime factors, it has no factors at alldef prime_number(n):
c = 0
for x in range(2, n):
if n % x == 0:
c = c + 1
return c
n = int(input("Enter a number = "))
if prime_number(n) == 0:
print("Prime number.")
else:
print("Not prime number.")import math
def main():
count = 3
while True:
isprime = True
for x in range(2, int(math.sqrt(count) + 1)):
if count % x == 0:
isprime = False
break
if isprime:
print count
count += 1