def 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 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 True
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 smaller
a=int(input('print number:'))
for i in range(2,a):
if a%i !=0:
continue
else:
print("Its not a prime number")
break # here break is exicuted then it means else would not be exicuted.
else:
print("Its a prime number")
def 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 all
#prime number verification program
a=int(input('print number:'))
for i in range(2,a):
if a%i !=0:
continue
else:
print("Its not a prime number")
break # here break is exicuted then it means else would not be exicuted.
else:
print("Its a prime number")#this is out of the for loop suite.
# Prime number:
n = int(input("Please enter your input number: "))
if n>1:
for i in range(2,n):
if n%i == 0:
print("%d is a Not Prime."%n)
break
else:
print("%d is a Prime."%n)
else:
print("%d is a Not Prime."%n)
# Use to Definition Function:
'''
def Prime_number_chcek(n):
if n>1:
for i in range(2,n):
if n%i == 0:
return ("%d is a Not Prime."%n)
return ("%d is a Prime."%n)
return ("%d is a Not Prime."%n)
# Main Driver:
if __name__=="__main__":
n = int(input("Enter your input number: "))
print(Prime_number_chcek(n))
'''
# This shows how we can use for + else using a break in between
for x in range(1,101):
# if you want to find whether a user input is a prime number
# use the following insted of the first for loop
# x = int(input("Type a number: "))
for i in range(2, x):
if x % i == 0:
print(x, "is not a prime number.")
break
else:
print(x, "is a prime number.")
# This will print all the numbers from 1-100,
# in the same line will print whether it is a prime or not
# if you use the user input method
# when you type 9, the output will be:
# 9 is not a prime number.
# when you type 7, the output will be:
# 7 is a prime number.
def prime(n):
if n>1:
if n==2 or n==3:
print("it is a prime number")
for i in range(2,int(n/2)+1):
if n%i==0:
print("it is not a prime number")
break
else:
print("it's a prime number")
break
else:
print("it is not a prime number")
start_num , end_num = input("enter 2 number sepreted by ,:").split(",")
start_num , end_num = int(start_num) , int(end_num)
for number in range(start_num , end_num+1):
is_prime = True
for counter in range(2,number):
value = number % counter
if value == 0:
is_prime = False
break
if is_prime == True:
print(number)
from math import sqrt, floor;
def is_prime(num):
if num < 2: return False;
if num == 2: return True;
if num % 2 == 0: return False;
for i in range(3,floor(sqrt(num))+1,2):
if num % i == 0: return False;
return True;
'''Write a Python script that prints prime numbers less than 20'''
print("Prime numbers between 1 and 20 are:")
ulmt=20;
for num in range(ulmt):
# prime numbers are greater than 1
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
print(num)
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
prime=int(input("Enter a number:"))
buffer=0
for i in range(2,prime):
if prime%i==0:
print(prime," is not a prime number")
buffer=1
break
if buffer==0:
print(prime," is a prime number")