How many prime numbers between 100 and 200 in python

Question

So I'm new to python coming from a java background and it's hard for me to get use to python's syntax and scoping and loops etc. This was a question my instructor asked and I'm lost to what I'm doing wrong

Q. write a function or just use some if-else statement to figure out how many prime numbers are there between 100 and 200

def isPrime(num):
    if num > 1:
        for i in range(2, num):
            if(num % i) == 0:
                return True

def primeNumbers():
    count = 0

    for num in range(100, 200):
        prime = isPrime(num)
        if prime == True:
            count += 1
    
    print(count)

primeNumbers()

What is the *actual* and *specific* question / issue? There is a "Q" and some presented code.. perhaps https://codereview.stackexchange.com/ might be more suitable if just looking for feedback. — user2864740, Oct 21 '21 at 00:40
Question: are you sure it's necessary to use `range(2, num)`? Do you think you could use a smaller range and still get the right answer? — Z4-tier, Oct 21 '21 at 00:41
There are non-Pythonic idioms in the code but this is a logic issue. isPrime is wrong. What happens when i == 2? The function returns True. Is that really what you mean? — Passerby, Oct 21 '21 at 00:43
_I'm lost to what I'm doing wrong_ Tell us what the code is doing that you believe is wrong. Do you get an error? Is a certain number identified as prime when it shouldn't be? — John Gordon, Oct 21 '21 at 00:45
A simple implementation would be to use a sieve of eratosthenes up to 200 and up to 100 and subtract — neuops, Oct 21 '21 at 00:45
Does this answer your question? [How to find prime numbers in python](https://stackoverflow.com/questions/62564607/how-to-find-prime-numbers-in-python) — Z4-tier, Oct 21 '21 at 00:50
The function returns true when the number IS divisible by another number, which is the opposite of what you want. — John Gordon, Oct 21 '21 at 00:50

Selcuk · Accepted Answer · 2021-10-21T01:06:58.357

Basically your isPrime function is wrong. It returns True when the number is divisible (i.e. not prime), and never returns False. A fixed version would be:

def isPrime(num):
    if num > 1:
        for i in range(2, num):
            if (num % i) == 0:
                return False
        return True
    else:
        return False

That being said, a more Pythonic (and PEP-8 compliant) version of your program could be:

def is_prime(num):
    if num > 1:
        for i in range(2, int(num ** 0.5) + 1):
            if num % i == 0:
                return False
        return True
    else:
        return False


def prime_numbers(start, end):
    count = 0
    for num in range(start, end):
        if is_prime(num):
            count += 1
    return count


print("The number of prime numbers between 100 and 200 is", prime_numbers(100, 200))

Matthew Wisdom · Answer 2 · 2021-10-21T01:17:21.710

1

Your program is actually returning the count of numbers that are not prime.
The isPrime function is return True if the number is divisible by by a number between 2 and num. The function should be:

def isPrime(num):
    if num > 1:
        for i in range(2, round(num**0.5)+1): # No need to search up to num
            if(num % i) == 0:
                return False
        return True
    return False

Or print 100 - count in your primeNumbers function.

You could also check out the sieve of Eratosthenes

edited Oct 21 '21 at 01:17

answered Oct 21 '21 at 01:01

Matthew Wisdom

131
3

I've edited the code to account for that. – Matthew Wisdom Oct 21 '21 at 01:17

score 0 · Answer 3 · answered Oct 21 '21 at 01:07

In isPrime(num)

def isPrime(num):
    if num > 1:
        for i in range(2, num):
            if(num % i) == 0: # <-- this condition
                return True # <-- this return value

Notice that if(num % i) == 0: checks that if the remainder of num divided by i is 0, meaning that you want to check that if i is divisible by num. But if that condition is true, then num is not a prime, so you have to return False. And if that condition is False for every i in range(2, num), then you can be assured that num is a prime.

A little thing I want to point out is that if i is a divisor of num then num/i is also a divisor. So in reality you only want to check about from 2 to sqrt(num) or num ** 0.5

Suggested code:

def isPrime(num):
    if num > 1:
        for i in range(2, int(num ** 0.5)):
            if(num % i) == 0:
                return False
        reuturn True
    else:
        return False

p/s: As others have pointed out, you shouldn't really use the normal prime testing in this project, instead, use the Sieve of Eratosthenes

How many prime numbers between 100 and 200 in python

3 Answers3