Perfect numbers is printing for each dividen...

[XxChelseaxX]

# count the divisors of a number
def countDivisors(n):
    count = 0
    for possible in range(1, n):
        if n % possible == 0:
            count = count+possible

        # print Perfect if sum of divisors it equal to dividen
        if count == n:
             print n

any ideas on how to get this to only print once I am really confused.

[bvdet]

Please use code tags when posting code. Se out posting guidelines.

BV - Moderator

[bvdet]

Your problem is the indentation.

# count the divisors of a number
def countDivisors(n):
    count = 0
    for possible in range(1, n):
        if n % possible == 0:
            count += possible
 
    # print Perfect if sum of divisors it equal to dividen
    if count == n:
        print n

        
countDivisors(496)

Output:

>>> 496
>>>

[XxChelseaxX]

That didnt work

I trided that and it doesnt seem to work.

[bvdet]

What error do you receive when you run the code? Are you running it at the command prompt or in an IDE?

[XxChelseaxX]

Its not an error its just prints the number for each number it runs. So for exaple, 6 is a perfect number for 1,and 2 it prints nothing because its not perfect for 3, 4, and 5 It prints 6 because 4 and 5 dont devide in its still perfect. I want it to only print when it has the total count not each one, because after I fix this, I will make a loop and it should tell me all the perfect numbers from 1-1000.

[bvdet]

Your post led me to believe that you wanted to check if a number is perfect. If you are looking for perfect numbers below some limit, I would approach it a different way, but still brute force.

Function A returns a list of proper factors for a given number

def factors(num):
    .......................

Function B returns a list of perfect numbers below an upper limit

def proper_nums(upper):
    output = []
    for i in range(2, upper+1):
        if sum(factors(i)) == i:
            output.append(i)
    return output

print proper_nums(100000)

The output should be:

>>> [6, 28, 496, 8128]

To calculate the proper factors of a given number, you need only loop through the square root of the number.

for i in xrange(1, int(num**0.5)+1)

[XxChelseaxX]

Apparently I should not have taken this class. I have no idea what you just said. This is not like math, it is way harder and apparently my brain is not ment for this.I'll just drop the class LOL