Python Program to Check Perfect Number

A perfect number is a number that equals the sum of its positive divisors excluding itself; use sum_of_divisors = sum(i for i in range(1, n) if n % i == 0) and check sum_of_divisors == n to determine if n is perfect.

Examples

Input6

Output6 is a perfect number

Input28

Output28 is a perfect number

Input10

Output10 is not a perfect number

How to Think About It

To check if a number is perfect, find all numbers less than it that divide it evenly. Add these divisors together. If the total equals the original number, it is perfect; otherwise, it is not.

Algorithm

1

Get the input number n

2

Initialize sum_of_divisors to 0

3

For each number i from 1 to n-1, check if i divides n evenly

4

If yes, add i to sum_of_divisors

5

After the loop, compare sum_of_divisors with n

6

If equal, return that n is perfect; else, return not perfect

Code

python

n = int(input('Enter a number: '))
sum_of_divisors = 0
for i in range(1, n):
    if n % i == 0:
        sum_of_divisors += i
if sum_of_divisors == n:
    print(f'{n} is a perfect number')
else:
    print(f'{n} is not a perfect number')

Output

Enter a number: 6
6 is a perfect number

Dry Run

Let's trace the number 6 through the code

1

Input number

n = 6

2

Initialize sum

sum_of_divisors = 0

3

Check divisors from 1 to 5

i=1 divides 6, sum=1; i=2 divides 6, sum=3; i=3 divides 6, sum=6; i=4 no; i=5 no

4

Compare sum with n

sum_of_divisors = 6 equals n = 6

5

Print result

'6 is a perfect number'

i	n % i == 0	sum_of_divisors
1	True	1
2	True	3
3	True	6
4	False	6
5	False	6

Why This Works

Step 1: Find divisors

We find all numbers less than n that divide n evenly using n % i == 0.

Step 2: Sum divisors

We add these divisors to get their total sum.

Step 3: Compare sum to number

If the sum equals the original number, it means the number is perfect.

Alternative Approaches

Using list comprehension and sum function

python

n = int(input('Enter a number: '))
sum_of_divisors = sum(i for i in range(1, n) if n % i == 0)
if sum_of_divisors == n:
    print(f'{n} is a perfect number')
else:
    print(f'{n} is not a perfect number')

This method is shorter and uses Python's built-in functions for clarity but may be less clear for beginners.

Optimized divisor check up to n//2

python

n = int(input('Enter a number: '))
sum_of_divisors = 0
for i in range(1, n//2 + 1):
    if n % i == 0:
        sum_of_divisors += i
if sum_of_divisors == n:
    print(f'{n} is a perfect number')
else:
    print(f'{n} is not a perfect number')

Checking divisors only up to half of n reduces unnecessary checks, improving performance.

Complexity: O(n) time, O(1) space

Time Complexity

The program checks all numbers from 1 to n-1 to find divisors, so it runs in linear time O(n).

Space Complexity

It uses a fixed amount of extra space for variables, so space complexity is O(1).

Which Approach is Fastest?

The optimized approach checking divisors only up to n//2 is faster than checking all numbers up to n-1.

Approach	Time	Space	Best For
Basic loop to n-1	O(n)	O(1)	Simplicity and clarity
List comprehension	O(n)	O(1)	Concise code and uses constant space
Optimized loop to n//2	O(n/2) ~ O(n)	O(1)	Better performance for large numbers