Hướng dẫn sum of all factors in python assignment expert - tổng tất cả các yếu tố trong chuyên gia chuyển nhượng python

Question

Một số nguyên được cho là một số hoàn hảo nếu tổng các yếu tố của nó, bao gồm 1 (nhưng không phải là chính số), bằng số. Ví dụ: 6 là một số hoàn hảo, vì 6 = 1

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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 return (True if sum == n and n!=1 else False) print("Below are all perfect numbers till 10000") n = 2 for n in range (10000): if isPerfect (n): print(n , " is a perfect number")1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak) We can notice that individual terms of above formula are Geometric Progressions (GP). We can rewrite the formula as. Sum of divisors = (p1a1+1 - 1)/(p1 -1) * (p2a2+1 - 1)/(p2 -1) * .................................. (pkak+1 - 1)/(pk -1)24def isPerfect( n ): sum = 1 i = 2 while i * i <= n: if n % i == 0: sum = sum + i + n/i i += 1 return (True if sum == n and n!=1 else False) print("Below are all perfect numbers till 10000") n = 2 for n in range (10000): if isPerfect (n): print(n , " is a perfect number")1 def isPerfect( n ): sum = 1 i = 2 while i * i <= n: if n % i == 0: sum = sum + i + n/i i += 1 return (True if sum == n and n!=1 else False) print("Below are all perfect numbers till 10000") n = 2 for n in range (10000): if isPerfect (n): print(n , " is a perfect number")4
728 Consider the number 18. Sum of factors = 1 + 2 + 3 + 6 + 9 + 18 Writing divisors as powers of prime factors. Sum of factors = (20)(30) + (21)(30) + (2^0)(31) + (21)(31) + (20)(3^2) + (2^1)(32) = (20)(30) + (2^0)(31) + (2^0)(32) + (21)(3^0) + (21)(31) + (21)(32) = (20)(30 + 31 + 32) + (21)(30 + 31 + 32) = (20 + 21)(30 + 31 + 32) If we take a closer look, we can notice that the above expression is in the form. (1 + p1) * (1 + p2 + p22) Where p1 = 2 and p2 = 3 and 18 = 213272
using Input : n = 30 Output : 72 Dividers sum 1 + 2 + 3 + 5 + 6 + 10 + 15 + 30 = 72 Input : n = 15 Output : 24 Dividers sum 1 + 3 + 5 + 15 = 2429
Làm thế nào để bạn tìm thấy tổng của tất cả các yếu tố của một số trong Python?
Làm thế nào để bạn tìm thấy tổng của tất cả các yếu tố của một số?
Tổng của tất cả các yếu tố của 20 là gì?
Làm cách nào để đếm các yếu tố trong Python?

+ 2 + 3. Viết một hàm hoàn hảo xác định xem số tham số có phải là một số hoàn hảo hay không. Sử dụng chức năng này trong một chương trình xác định và in tất cả các số hoàn hảo trong khoảng từ 1 đến 1000. In các yếu tố của từng số hoàn hảo để xác nhận rằng số này thực sự hoàn hảo.

def perfect(num):
    """ Function to check if parameter number is perfect.
    If so, return the list of factors."""

    factors = []
    for i in range(1, num):
        # Check if i is a factor of num
        if num % i == 0:
            factors.append(i)

    if sum(factors) != num:
        return False
    return factors


for num in range(1, 1000):
    if perfect(num):
        factors = perfect(num)
        print(num, '=', end=' ')
        print(*factors, sep=' + ')

Tìm hiểu thêm về sự giúp đỡ của chúng tôi với các bài tập: Python

Một số nguyên được cho là một con số hoàn hảo nếu các yếu tố của nó, bao gồm 1 (nhưng không phải là chính số), tổng cho số. Ví dụ: 6 là một số hoàn hảo, bởi vì 6 = 1 + 2 + 3. Viết một phương thức hoàn hảo xác định xem số tham số có phải là một số hoàn hảo hay không. Sử dụng phương pháp này trong một ứng dụng xác định và hiển thị tất cả các số hoàn hảo trong khoảng từ 1 đến 1000. Hiển thị các yếu tố của từng số hoàn hảo để xác nhận rằng số thực sự hoàn hảo.

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

Tìm hiểu thêm về sự giúp đỡ của chúng tôi với các bài tập: Python

Đưa ra một số n, nhiệm vụ là tìm tổng của tất cả các yếu tố.examples: & nbsp; & nbsp;
Examples :

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

Một giải pháp đơn giản là đi qua tất cả các giao diện và thêm chúng. & Nbsp; & nbsp;simple solution is to traverse through all divisors and add them.

C++

#include<bits/stdc++.h>

using namespace

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

0

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

2

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

8

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

9

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

9

0

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

6

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

1

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

3

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

5

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

7

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Java

8

9

#include<bits/stdc++.h>0 #include<bits/stdc++.h>1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6#include<bits/stdc++.h>3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

2

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

using0

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7using2using3using4

using5

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0 using3using8

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1 namespace1namespace2using8

4

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1 namespace8namespace9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

00

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

05namespace2using4

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

13

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

3

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

5

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

13

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

18

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

4

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

25using3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

27

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

31 #include<bits/stdc++.h>3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

33

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

34

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

39

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

40using8

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

43

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Python3

8

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

48

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

49

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

50

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

53

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54 using3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

57

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

58

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0 using3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

62

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54 namespace2

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

67

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

68

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

69

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7namespace9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

72

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1 ____

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

53

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

81

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

67

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54__

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

89

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

92

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

93

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

94

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

62

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

62

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

75

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

00

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

5

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

03

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

62

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

62

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

75

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

89

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

75

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

17

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

75

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

19

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

75

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

19

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

40

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

26

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

27

C#

using

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

29

#include<bits/stdc++.h>0 #include<bits/stdc++.h>1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6#include<bits/stdc++.h>3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

2

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

8

using5

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

1

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

4

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

53

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

6

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

13

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

3

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

5

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

13

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

18

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

4

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

31 #include<bits/stdc++.h>3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

33

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

83

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

4

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

90

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

PHP

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

94

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

95

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

2

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97using4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97

04

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

9

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

09

10

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

13

14

15

14

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97

28

14

30

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

14

37

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

09

44

14using8

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

5

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

09

51

14

53

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

09

53

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97

68

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97

71

72

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

2

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

27

76

JavaScript

77

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

95

79

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

8

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

9

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

88

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

09

10

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

13

14

15

14

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97

28

14

30

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

3

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

5

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

18

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

14

37

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

16

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

17

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

18

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2
72
09
72
44
72
14using8

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2
72
09
72
51
72
14
72
53
Input : n = 30 Output : 72 Dividers sum 1 + 2 + 3 + 5 + 6 + 10 + 15 + 30 = 72 Input : n = 15 Output : 24 Dividers sum 1 + 3 + 5 + 15 = 24
97 O(√n)
Auxiliary Space: O(1)

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

09

53

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97

68efficient solution is to use below formula.
Let p1, p2, … pk be prime factors of n. Let a1, a2, .. ak be highest powers of p1, p2, .. pk respectively that divide n, i.e., we can write n as n = (p1a1)*(p2a2)* … (pkak).

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97
72
71

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

72

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

2

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

97

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

27

C++

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

19

JavaScript

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

95

79

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

91

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

4

Đầu ra: & nbsp; & nbsp;

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

50

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

52

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

54

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

58

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Độ phức tạp về thời gian: O (√n) & nbsp; không gian phụ trợ: O (1)

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

65

Một giải pháp hiệu quả là sử dụng dưới đây công thức. & NBSP; Đặt P1, P2, PK PK là yếu tố chính của n. Đặt A1, A2, .. Ak là sức mạnh cao nhất của P1, P2, .. PK tương ứng chia n, tức là, chúng ta có thể viết n là n = (p1a1)* (p2a2)*

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Làm thế nào để công thức trên hoạt động? & Nbsp; & nbsp;

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

Vì vậy, nhiệm vụ làm giảm việc tìm kiếm tất cả các yếu tố chính và quyền hạn của họ. & Nbsp; & nbsp;

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

77

using namespace

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

0

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ): sum = 1 i = 2 while i * i <= n: if n % i == 0: sum = sum + i + n/i i += 1 return (True if sum == n and n!=1 else False) print("Below are all perfect numbers till 10000") n = 2 for n in range (10000): if isPerfect (n): print(n , " is a perfect number")1 Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak) We can notice that individual terms of above formula are Geometric Progressions (GP). We can rewrite the formula as. Sum of divisors = (p1a1+1 - 1)/(p1 -1) * (p2a2+1 - 1)/(p2 -1) * .................................. (pkak+1 - 1)/(pk -1)24def isPerfect( n ): sum = 1 i = 2 while i * i <= n: if n % i == 0: sum = sum + i + n/i i += 1 return (True if sum == n and n!=1 else False) print("Below are all perfect numbers till 10000") n = 2 for n in range (10000): if isPerfect (n): print(n , " is a perfect number")1 def isPerfect( n ): sum = 1 i = 2 while i * i <= n: if n % i == 0: sum = sum + i + n/i i += 1 return (True if sum == n and n!=1 else False) print("Below are all perfect numbers till 10000") n = 2 for n in range (10000): if isPerfect (n): print(n , " is a perfect number")4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

30

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

9

0

1

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

42

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

45

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

47

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

48

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

63

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

68

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

18using3using8

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

47

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

05namespace2using4

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

50

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

52

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

54

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

58

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

42namespace9using4

9

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

46using3

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

48

4

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

68

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

31 #include<bits/stdc++.h>3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

33

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

58

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

39

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

40using8

4

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

67

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Python3

8

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

72

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

49

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

74

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

76

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54 using3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

67

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

68

4

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

93

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54 using3

4

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

97

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54 using3

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

47

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

19

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

81

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

67

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54__

9

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

19

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

19

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

93

14

9

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

97

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

97

19

14

9

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

93

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

75

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

25

4

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

76

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

76

19

31

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

34namespace9

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

03

4

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

76

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

76

19

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7using3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

68

49

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

24

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

40using4

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

26

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

56

57

49using4

C#

using

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

29

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

31 #include<bits/stdc++.h>0 #include<bits/stdc++.h>1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6#include<bits/stdc++.h>3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

24

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

30

4

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

53

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

42

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

45

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

47

6

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

50

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

52

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

2

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

54

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

58

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7 #include<bits/stdc++.h>08

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

65

4

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

68

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

31 #include<bits/stdc++.h>3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

33

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

58

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

39

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

40using8

4#include<bits/stdc++.h>27

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

728 Consider the number 18. Sum of factors = 1 + 2 + 3 + 6 + 9 + 18 Writing divisors as powers of prime factors. Sum of factors = (20)(30) + (21)(30) + (2^0)(31) + (21)(31) + (20)(3^2) + (2^1)(32) = (20)(30) + (2^0)(31) + (2^0)(32) + (21)(3^0) + (21)(31) + (21)(32) = (20)(30 + 31 + 32) + (21)(30 + 31 + 32) = (20 + 21)(30 + 31 + 32) If we take a closer look, we can notice that the above expression is in the form. (1 + p1) * (1 + p2 + p22) Where p1 = 2 and p2 = 3 and 18 = 213272

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

94

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

49

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

74

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

76

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54 using3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

67

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

68

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

93

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54 using3

4

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

97

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54 using3

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

47

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

19

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

81

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

67

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54__

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

9

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

19

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

19

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

93

14

9

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

97

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

97

19

14

9

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

93

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

75

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

25

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

4

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

76

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

76

19

31

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

7

34namespace9

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

03

4

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

76

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (20)(30) + (21)(30) + (2^0)(31) +
                 (21)(31) + (20)(3^2) + (2^1)(32)
               = (20)(30) + (2^0)(31) + (2^0)(32) +
                 (21)(3^0) + (21)(31) + (21)(32)
               = (20)(30 + 31 + 32) + 
                 (21)(30 + 31 + 32)
               = (20 + 21)(30 + 31 + 32)
If we take a closer look, we can notice that the
above expression is in the form.
(1 + p1) * (1 + p2 + p22)
Where p1 = 2 and p2 = 3 and 18 = 2132

76

19

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7using3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

0

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

68

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

49

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

54

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

24

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

40using4

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

26

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

56

57

49using4

76

using Input : n = 30 Output : 72 Dividers sum 1 + 2 + 3 + 5 + 6 + 10 + 15 + 30 = 72 Input : n = 15 Output : 24 Dividers sum 1 + 3 + 5 + 15 = 2429

77

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

31 #include<bits/stdc++.h>0 #include<bits/stdc++.h>1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6using21

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6#include<bits/stdc++.h>3

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

24

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

4using28

4using30

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

30

4

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

50

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

52

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

54

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

58

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

6

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

4

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

6

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

7

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

53

4

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

65

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

1

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

42

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

9

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

16

using58

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p1a1+1 - 1)/(p1 -1) * 
                  (p2a2+1 - 1)/(p2 -1) *
                  ..................................
                  (pkak+1 - 1)/(pk -1)

18

9
def isPerfect( n ): sum = 1 i = 2 while i * i <= n: if n % i == 0: sum = sum + i + n/i i += 1 return (True if sum == n and n!=1 else False) print("Below are all perfect numbers till 10000") n = 2 for n in range (10000): if isPerfect (n): print(n , " is a perfect number")
1
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak) We can notice that individual terms of above formula are Geometric Progressions (GP). We can rewrite the formula as. Sum of divisors = (p1a1+1 - 1)/(p1 -1) * (p2a2+1 - 1)/(p2 -1) * .................................. (pkak+1 - 1)/(pk -1)
45

9
Sum of divisors = (1 + p1 + p12 ... p1a1) * (1 + p2 + p22 ... p2a2) * ............................................. (1 + pk + pk2 ... pkak) We can notice that individual terms of above formula are Geometric Progressions (GP). We can rewrite the formula as. Sum of divisors = (p1a1+1 - 1)/(p1 -1) * (p2a2+1 - 1)/(p2 -1) * .................................. (pkak+1 - 1)/(pk -1)
47
72
6 O(√n log n)
Auxiliary Space: O(1)

9

def isPerfect( n ):
    
    sum = 1
    
    i = 2
    while i * i <= n:
        if n % i == 0:
            sum = sum + i + n/i
        i += 1
    
     
    return (True if sum == n and n!=1 else False)
 
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
    if isPerfect (n):
        print(n , " is a perfect number")

5
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Further Optimization.
If there are multiple queries, we can use Sieve to find prime factors and their powers.
References:
https://www.math.upenn.edu/~deturck/m170/wk3/lecture/sumdiv.html
http://mathforum.org/library/drmath/view/71550.html

Làm thế nào để bạn tìm thấy tổng của tất cả các yếu tố của một số trong Python?

Làm thế nào để bạn tìm thấy các yếu tố của một số trong một vòng lặp trong Python ?..

Bước 1 - Lấy số đầu vào từ người dùng ..

Bước 2 - Lặp lại cho vòng lặp trên mỗi số từ 1 đến số đã cho ..

Bước 3 - Nếu trình lặp vòng lặp phân chia đồng đều số được cung cấp, tức là số % i == 0 in nó ..

Bước 4 - Kết quả in ..

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Công thức cho tổng của tất cả các yếu tố được đưa ra bởi;Tổng các yếu tố của n = [(xA+1-1)/x-1] × [(yb+1-1)/y-1] × [(zc+1-1)/z-1]Sum of factors of N = [(X^a⁺¹-1)/X-1] × [(Y^b⁺¹-1)/Y-1] × [(Z^c⁺¹-1)/Z-1]

Tổng của tất cả các yếu tố của 20 là gì?

Tổng của tất cả các yếu tố của 20 là gì?Vì tất cả các yếu tố của 20 là 1, 2, 4, 5, 10, 20, do đó, tổng của các yếu tố của nó là 1 + 2 + 4 + 5 + 10 + 20 = 42.42.