Python Prime number checker [duplicate]

You need to stop iterating once you know a number isn't prime. Add a break once you find prime to exit the while loop.

Making only minimal changes to your code to make it work:

a=2num=13while num > a :  if num%a==0 & a!=num:    print('not prime')    break  i += 1else: # loop not exited via break  print('prime')

Your algorithm is equivalent to:

for a in range(a, num):    if a % num == 0:        print('not prime')        breakelse: # loop not exited via break    print('prime')

If you throw it into a function you can dispense with break and for-else:

def is_prime(n):    for i in range(3, n):        if n % i == 0:            return False    return True

Even if you are going to brute-force for prime like this you only need to iterate up to the square root of n. Also, you can skip testing the even numbers after two.

With these suggestions:

import mathdef is_prime(n):    if n % 2 == 0 and n > 2:         return False    for i in range(3, int(math.sqrt(n)) + 1, 2):        if n % i == 0:            return False    return True

Note that this code does not properly handle 0, 1, and negative numbers.

We make this simpler by using all with a generator expression to replace the for-loop.

import mathdef is_prime(n):    if n % 2 == 0 and n > 2:         return False    return all(n % i for i in range(3, int(math.sqrt(n)) + 1, 2))

def isprime(n):    '''check if integer n is a prime'''    # make sure n is a positive integer    n = abs(int(n))    # 0 and 1 are not primes    if n < 2:        return False    # 2 is the only even prime number    if n == 2:         return True        # all other even numbers are not primes    if not n & 1:         return False    # range starts with 3 and only needs to go up     # the square root of n for all odd numbers    for x in range(3, int(n**0.5) + 1, 2):        if n % x == 0:            return False    return True

Taken from:

http://www.daniweb.com/software-development/python/code/216880/check-if-a-number-is-a-prime-number-python

def is_prime(n):    return all(n%j for j in xrange(2, int(n**0.5)+1)) and n>1