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I've been trying to create a function that checks for perfect numbers, I've used many of the resources I found here, but they don't seem to work for some reason. In my assignment, I'm required to find different types of numbers in a given range input, but this is the only function that doesn't work properly. This is what I have:

def is_perfect(a): #creates a perfect number checker
    sum=0
    for i in range(1,a):
        if a%1==0:
            sum += i
    if sum==a:
        return True
    else:
        return False
altalio
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3 Answers3

1

Change the line from a%1==0 to a%i==0 and your code will work perfectly. Because you've to check that the number is divisible from 1 to a and not only 1. It will return always True until it is not integer. Hence, it will keep adding all nos from 1 to a

0

A perfect number is any number that has the sum of it's devisors, excluding itself, equal to this number. Like the number six that is divided by 1, 2 and 3 and 1 + 2 + 3 = 6.

def is_perfect(number):
    sum_ = sum([x for x in range(1, number) if number % x == 0])
    return sum_ == number

is_perfect(6)  # Returns True
is_perfect(10) # Returns False
is_perfect(28) # Returns True

I called the variable that sum all the divisors with a underscore because sum is already a function keyword in Python

Danilo Matrangolo Marano
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def isPerfect( n ):
    sum = 1
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
     
    # If sum of divisors is equal to
    # n, then n is a perfect number
     
    return (True if sum == n and n!=1 else False)
Jules Spender
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