Perfect Number in Python

June 29, 2023

Check Whether or Not the Number is a Perfect Number in Python

Given an integer input as a number, the objective is to check whether or not a number is a Perfect Number in Python Language. Therefore, we write a program to Check Whether or Not the Number is a Perfect Number in Python Language.

Example
Input : 28
Divisors : [1, 2, 4, 7, 14]
Sum = 1 + 2 + 4 + 7 + 14 = 28
Output : It's a Perfect Number

Perfect Number A Number that can be represented as the sum of all the factors of the number is known as a Perfect Number.
Let's Try and understand it better using an example

Example
Input : 28
Output : It's a Perfect Number
Explanation : Number = 28
28 = 1 + 2 + 14 + 4 + 7
as the number 28 has factors 1, 2, 4, 7 and 14.
We sum them up and check whether they match the original number.

As we can see from the above example, number 28 is a Perfect Number. Make sure you don't include the number itself as a factor.

Finding Prime Factors of a number

Check whether or not the number is Perfect Number in Python

Method 1: Using Simple Iteration I
Method 2: Using Simple Iteration II
Method 3: Using Simple Iteration III
Method 4: Using Recursion
Method 5: Using Factors

We’ll discuss the above-mentioned methods in detail in the upcoming sections.

Method 1: For Loop Iteration between [1, num]

For number num

Initialise sum = 0
Run an in ‘i’ iteration b/w [1, num]
For any i satisfying (num % i == 0)
If sum == num, its a perfect number
Add to the sum

Let’s implement the above mentioned logic in Python Language.

Python Code

Run

n = 28
sum = 0

for i in range(1, n):
    if n % i == 0:
        sum = sum + i

if sum == n:
    print("The number is a Perfect number")
else:
    print("The number is not a Perfect number")

Output

The number is a Perfect number

Method 2: While Loop Iteration between [1, num]

For number num

Initialise sum = 0
Run an in ‘i’ iteration b/w [1, num]
For any i satisfying (num % i == 0)
If sum == num, its a perfect number
Add to the sum

Run

num = 28
sum = 0

i = 1
while i < num:
    if num % i == 0:
        sum += i

    i += 1

if sum == num:
    print("The number is a Perfect number")
else:
    print("The number is not a Perfect number")

Output

The number is a Perfect number

Method 3: Iteration between [1, num/2+1]

This method uses fact that all the divisors of the number can be found in the range (1, num/2)

Example –

Divisors of 28 = {1, 2, 4, 7, 14}

Run

num = 28
sum = 0


for i in range(1, num//2 + 1):
    if num % i == 0:
        sum = sum + i

if sum == num:
    print("The number is a Perfect number")
else:
    print("The number is not a Perfect number")

Output

The number is a Perfect number

Method 4: Using recursion

We use recursion to find if the number is perfect or not.

Run

sum_n = 0

def getSumDivisors(num, i):
    global sum_n
    # since, all factors can be found will num/2
    # we will only check in this range
    if i <= num // 2:

        if num % i == 0:
            sum_n = sum_n + i

        i += 1

        # recursively call isPerfect
        getSumDivisors(num, i)

    # returns the sum
    # when i becomes > num/2
    return sum_n


num = 28

if getSumDivisors(num, 1) == num:
    print("The number is a Perfect number")
else:
    print("The number is not a Perfect number")

Output

The number is a Perfect number

Method 5: Factors come in pairs

This method uses observations that all factors come in pairs.

All Factors come in pairs For n = a * b (For each a there exists a unique b)

Example 1 : 100
(1,100), (2, 50), (4, 25), (5, 20), (10, 100)

Example 2 : 28
(1, 28), (2, 14), (4, 7)
Note : We will need to ignore pair of 1. As it will be the number itself.

Shorten the Loop We can shorten the loop running between [1, num] to [1, √num]

Since we will find all pairs before √num (n = sqrt(n) * sqrt(n))

Example: For 28, all pairs can be found before √28 = 5.2

Python Code

Run

n = 28
sump = 0

for i in range(1, int(pow(n, 0.5))):
    if n % i == 0:

        # For n : (1, n) will always be pair of divisor
        # acc to def., we must ignore adding num itself as divisor
        # when calculating for perfect number
        if i == 1:
            sump += i

        # Example For 100 (10,10)  will be one pair
        # But, we should add value to the sum just once
        elif i == n / i:
            sump += i

        # add both pairs as divisors
        # For any divisor i, n/i will also be a divisor
        else:
            sump += i + n / i

if sump == n:
    print("The number is a Perfect number")
else:
    print("The number is not a Perfect number")

# Time complexity: O(sqrt(N))
# Space complexity: O(1)

Output

The number is a Perfect number

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5 comments on “Perfect Number in Python”

mani
n=int(input())
a=0
i = 1
while i < n:
if (n % i == 0):
a+=i
i = i + 1
if a==n:
print("The number is a Perfect number")
else:
print("The number is not a Perfect number")

0

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PrepInsta Support
Hey !!
We’d suggest you to join our Discord Channel for all your technical queries.

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pamidi
n=int(input(“enter a number : “))
l=[i for i in range(1,n) if(n%i==0)]
if(sum(l)==n):
print(“Perfect number”)
else:
print(“not a perfect number”)

0

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Ahamed
n1 = int(input())
k = 0
for i in range(1, n1):
if n1%i == 0:
k = k + i
if k == n1:
print(‘Perfect number {}’.format(n1))
else:
print(‘Not perfect number{}’.format(n1))

0

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L_50_Vegi.Deekshita
a=int(input())
c=[]
s=0
for i in range(1,a):
if(a%i==0):
c.append(i)
for i in c:
s+=i
if(s==a):
print(“perect”)
else:
print(“not perfect”)

0

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