Perfect Numbers

Perfect, Abundant & Deficient Numbers

The ancient Greeks classified numbers by comparing them to the sum of their proper divisors — all divisors except the number itself.

$s(n) = \sum_{\substack{d \mid n \\ d < n}} d$

Based on this sum, every positive integer $n > 1$ falls into one of three categories:

Perfect Numbers

A perfect number equals the sum of its proper divisors. The first four are: 6, 28, 496, 8128. Euler proved that all even perfect numbers have the form $2^{p-1}(2^p - 1)$ where $2^p - 1$ is a Mersenne prime. No odd perfect number has ever been found.

def proper_divisor_sum(n):
    return sum(i for i in range(1, n) if n % i == 0)

print(proper_divisor_sum(6))   # 6  → perfect
print(proper_divisor_sum(28))  # 28 → perfect

Classification

def is_perfect(n):
    return proper_divisor_sum(n) == n

def classify(n):
    s = proper_divisor_sum(n)
    if s == n:
        return "perfect"
    elif s > n:
        return "abundant"
    else:
        return "deficient"

print(classify(12))  # abundant
print(classify(7))   # deficient

Your Task

Implement proper_divisor_sum(n), is_perfect(n), and classify(n) which returns "perfect", "abundant", or "deficient".