CODEWITHSUNDEEP
pythonrange
Mahdi D
Mahdi D

Reputation: 3

my isPrime's range is not working, it gives just the first number

#!/usr/local/bin/python3.6

def isPrime(n):


if n == 1:
    print("1 is a special")
    return False


for x in range(2, n):
    if n % x == 0:
        print("{} is equal to {} * {}".format(n, x, n // x))
        return False
    else:
        print(n, " is a prime number")
        return True


for n in range(1, 21):
    isPrime(n)

isPrime(2)
isPrime(21)
isPrime(25)

and the result it gives to me is:

1 is a special
3  is a prime number
4 is equal to 2 * 2
5  is a prime number
6 is equal to 2 * 3
7  is a prime number
8 is equal to 2 * 4
9  is a prime number
10 is equal to 2 * 5
11  is a prime number
12 is equal to 2 * 6
13  is a prime number
14 is equal to 2 * 7
15  is a prime number
16 is equal to 2 * 8
17  is a prime number
18 is equal to 2 * 9
19  is a prime number
20 is equal to 2 * 10

21  is a prime number
25  is a prime number

there is no result of (2), and also this is result of 'even-odd' numbers, not 'isPrime' , because in the code 'for x in range(2, n)' , it has calculated only with number 2 for x

what is wrong in my code ?

Upvotes: 0

Views: 75

Answers (3)

Cezar Cobuz
Cezar Cobuz

Reputation: 1207

for x in range(2, n):
    if n % x == 0:
        print("{} is equal to {} * {}".format(n, x, n // x))
        return False
else:
    print(n, " is a prime number")
    return True

Watch the indentation: else is now is in aligned with for (not with if), and it should solve your problem. Else means that the x got iterated over n, so n%x was never 0 => you got a prime

Upvotes: 0

Henry Heath
Henry Heath

Reputation: 1082

Do this:

def isPrime(n):
    if n == 1: 
        print("1 is a special")
        return False
    for x in range(2, n):
        if n % x == 0:
            print("{} is equal to {} * {}".format(n, x, n // x))
            return False
    print(n, " is a prime number")
    return True

for n in range(1, 21):
    isPrime(n)

Upvotes: 0

user2390182
user2390182

Reputation: 73470

Don't return True form the first iteration. Not being divisible by 2 does not make it prime:

for x in range(2, n):  # it would be enough to loop to sqrt(n)
    if n % x == 0:
        # you know it is NOT prime after first divisor found
        return False 
# you only know it IS prime after you tried all possible divisors
return True

Upvotes: 1

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