Python Programs to Find Perfect Number

A Perfect Number is a positive integer equal to the sum of all its positive divisors, excluding the number itself.

Let’s say the input number is 6.

• The divisors of 6 are 1, 2, and 3.
• The sum of these divisors is 1 + 2 + 3 = 6.
• Since the sum is equal to the original number, 6 is a perfect number.

This article explores different ways to find perfect numbers in Python, with explanations and examples. We’ll delve into the logic behind finding divisors, summing them efficiently, and checking for perfect number properties.

This article explores different methods, including using the modulo operator, loops, list comprehensions, and math module.

1. How to Find a Perfect Number in Python

Code Example

def is_perfect_square(num):
    divisors_sum = 0

    # Loop through all numbers from 1 to num - 1
    for i in range(1, num):
        if num % i == 0:  # Check if 'i' is a divisor of 'num'
            divisors_sum += i  # Add the divisor to the sum

    # Check if the sum of divisors equals the number
    return divisors_sum == num

number = 6
print(is_perfect_square(number))     # True

number = 15
print(is_perfect_square(number))    # FalseCode language: Python (python)

2. Using math.sqrt function

Now, Let’s see how to find a perfect number using a math module.

The math.sqrt() is a Python built-in function that calculates the square root of a given number. For example, math.sqrt(25) returns 5.0.

In this method, the modulo operator (%) is used to find the divisor of a number. Here, instead of checking all numbers up to n, you only need to check up to the square root of n (√n).

This method improves efficiency by only checking divisors up to the square root of the number, leveraging the fact that divisors occur in pairs.

For Example,

• For number = 28:
  • The square root of 28 is approximately 5.29, so the loop checks divisors up to 5.
• Divisor pairs found:
  • For i = 2, number // i = 14. Add both 2 and 14 to the sum.
  • For i = 4, number // i = 7. Add both 4 and 7 to the sum.
• Total divisor sum: 1 + 2 + 14 + 4 + 7 = 28.
• Since the sum equals the original number, 28 is confirmed as a perfect number.

Code Example

import math

# Input number to check
num = 28

# Initialize the sum of divisors
divisors_sum = 1  # Start with 1, as it is a divisor of every number

# Check divisors up to the square root of the number
for i in range(2, int(math.sqrt(num)) + 1):
    if num % i == 0:  # Check if 'i' is a divisor of 'num'
        divisors_sum += i
        if i != num // i:  # Avoid adding the square root twice
            divisors_sum += num // i

# Check if the sum of divisors equals the number
is_perfect = divisors_sum == num

# Output the result
print(f"Is {num} a perfect number? -> {is_perfect}")Code language: Python (python)

Output:

Is 28 a perfect number? -> TrueCode language: Python (python)

Explanation

• Initialization: The sum of divisors starts at 1 because 1 is a divisor of every number.
• Square Root Check: The loop runs only up to the square root of num, calculated using math.sqrt(num). This reduces unnecessary checks for numbers larger than √n.
• Divisor Check: For each number i in the range, the modulo operator (%) checks if i is a divisor of num. If num % i == 0, it confirms that i divides num without leaving a remainder.
• Divisor Pairs: Whenever a divisor i is found, the corresponding divisor num // i is also calculated and added to the sum. This ensures all divisor pairs are included, except when i equals num // i (to avoid duplication).
• Comparison: After calculating the sum of divisors, it is compared with the original number. If they are equal, then it is a perfect number.

3. Using List Comprehension

List comprehension is a concise way to create Python lists by applying an expression to each item in an iterable, with optional filtering using a condition. It allows you to create a list of divisors efficiently.

The built-in sum() function is then used to calculate the total sum of the divisors in the list, enabling a concise and elegant way to determine if a number is perfect.

Code Example

# Input number to check
num = 28

# Calculate the sum of divisors using list comprehension
is_perfect = num == sum([i for i in range(1, num) if num % i == 0])

# Output the result
print(f"Is {num} a perfect number? -> {is_perfect}")Code language: Python (python)

Output:

Is 28 a perfect number? -> TrueCode language: Python (python)

Explanation

• List Comprehension: It generates a list of all divisors of num by iterating through numbers from 1 to num-1 and including only those that divide num with reminder 0, i.e., num % i == 0.
• sum() function: It calculates the total sum of the divisors.
• Comparison: After calculating the sum of divisors, it is compared with the original number. If they are equal, then it is a perfect number.

Summary

Each method has its unique advantages depending on the use case.

• For Loop:
  • Advantage: Simple and easy to understand.
  • Time Complexity: O(n) – The loop runs from 1 to num-1, performing a constant-time check for each number.
  • Space Complexity: O(1) – Only a few variables are used to store the sum and current divisor.
• Using math.sqrt function: Best Choice
  • Advantage: Fast method. Significantly reduces iterations by leveraging divisor pairs.
  • Time Complexity: O(√n) – The loop runs up to the square root of num.
  • Space Complexity: O(1) – Uses only a few variables for the sum and divisors.
• List Comprehension:
  • Advantage: Concise and Pythonic; combines looping and filtering in a single line.
  • Time Complexity: O(n) – Iterates through all numbers from 1 to num-1 to generate the list of divisors.
  • Space Complexity: O(n) – Creates a list of divisors, requiring additional memory to store the list.