CODEWITHSUNDEEP
python
bvdabjorn
bvdabjorn

Reputation: 69

Prime Numbers python

I'm a beginning programmer in python and have a question about my code I'm writing:

number = int(input("Enter a random number: "))

for num in range(1, number + 1) :
    for i in range(2, num) :
         if (num % i) == 0 :
            break
         else :
             print(num)
             break

When I run this program I also get 9, 15 21 as output. But these are not prime numbers. What is wrong with my code?

Thanks!

Upvotes: 5

Views: 1849

Answers (5)

user8616404
user8616404

Reputation:

This is my method to generate first ten prime numbers with no imports, just simple code with components we are learning on college. I save those numbers into list.

def main():
    listaPrim = []
    broj = 0
    while len(listaPrim) != 10:
         broj+= 1
         brojac = 0
         for i in range(1,100):
             if broj % i == 0:
                 brojac += 1
         if brojac == 2:
             listaPrim.append(broj)

    print(listaPrim)

if __name__=='__main__':
    main()

Upvotes: 0

Kaboom
Kaboom

Reputation: 1

I am assuming the random number is the range you want the numbers to be within.

I found that the variable i is always equal to 2 in your code.This destroys the purpose of having a second for loop

Prime numbers are numbers that cannot be divisible by 2, 3 or 7, excluding 2, 3 or 7!

With this knowledge I adapted your code to show the true prime numbers.

solution

number = int(input("Enter a random number: "))

for num in range(2,number +1) :
    printnum = True
    if num == 2:
        printnum = True
    elif num == 3:
        printnum = True
    elif num == 7:
        printnum = True
    elif num % 2 == 0:
        printnum = False
    elif num % 3 == 0:
        printnum = False
    elif num % 7 == 0:
        printnum = False

    if printnum == True:
        print(num)

Upvotes: 0

cafebabe1991
cafebabe1991

Reputation: 5176

Your problem

The else statement that you put inside the for loop means that if that number(num) is divisible by this current no represented by i, return true considering it as a prime which IS NOT CORRECT.

Suggestion

Your outer loop starts with 1 which should change to 2, as 1 is not a prime number

Solution 

def fun(number):
 #Not counting 1 in the sequence as its not prime
 for num in range(2, number + 1) :
    isPrime = True
    for i in range(2, num) :
         if (num % i) == 0 :
        isPrime = False  
            break
    if isPrime:
        print num
fun(10)

Output

1

2

3

5

7

Upvotes: 0

arx5
arx5

Reputation: 336

With if (num % i) == 0: you go to the else block for every num, which is not a multiply of 2, as you start i with 2, thus printing num. I got all odd numbers printed with your code.

You can use something like this:

number = int(input("Enter a random number: "))

for num in range(1, number + 1):
    prime = True
    for i in range(2, num):
        if (num % i) == 0:
            prime = False
            break
    if prime:
        print(num)

It sets prime to False when it encounters a divisor without rest.

Upvotes: 2

bmbigbang
bmbigbang

Reputation: 1378

the trouble is in your else: statement which has to run outside of the first loop. I recommend the following:

def isprime(num):
    for i in range(2, num):
        if (num % i) == 0:
            return False
    return True


for num in range(1, number + 1) :
    if isprime(num):
        print num

Upvotes: 0

Related Questions