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Python Program for Find sum of even factors of a number

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  • Difficulty Level : Medium
  • Last Updated : 09 Aug, 2022

Given a number n, the task is to find the even factor sum of a number. Examples:

Input : 30
Output : 48
Even dividers sum 2 + 6 + 10 + 30 = 48

Input : 18
Output : 26
Even dividers sum 2 + 6 + 18 = 26

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) 

If number is odd, then there are no even factors, so we simply return 0. If number is even, we use above formula. We only need to ignore 20. All other terms multiply to produce even factor sum. For example, consider n = 18. It can be written as 2132 and sun of all factors is (20 + 21)*(30 + 31 + 32). if we remove 20 then we get the Sum of even factors (2)*(1+3+32) = 26. To remove odd number in even factor, we ignore then 20 which is 1. After this step, we only get even factors. Note that 2 is the only even prime. 

python3




# Formula based Python3
# program to find sum
# of alldivisors of n.
import math
 
# Returns sum of all
# factors of n.
def sumofFactors(n) :
     
    # If n is odd, then
    # there are no even
    # factors.
    if (n % 2 != 0) :
        return 0
  
    # Traversing through
    # all prime factors.
    res = 1
    for i in range(2, (int)(math.sqrt(n)) + 1) :
         
        # While i divides n
        # print i and divide n
        count = 0
        curr_sum = 1
        curr_term = 1
        while (n % i == 0) :
            count= count + 1
  
            n = n // i
  
            # here we remove the
            # 2^0 that is 1. All
            # other factors
            if (i == 2 and count == 1) :
                curr_sum = 0
  
            curr_term = curr_term * i
            curr_sum = curr_sum + curr_term
         
        res = res * curr_sum
         
  
    # This condition is to
    # handle the case when
    # n is a prime number.
    if (n >= 2) :
        res = res * (1 + n)
  
    return res
 
 
# Driver code
n = 18
print(sumofFactors(n))
 
 
# This code is contributed by Nikita Tiwari.

Output

26

Method: Finding even factors sum of a given number using only for loop and if statements .

1. Iterate from the start range 1 to the given number to find the factors of a number using modulo division.

2. Finding even factors from the obtained factors by performing modulo division of a factor with 2. if the result of modulo division is equal to 0 then it should be considered as an even factor.

3. Adding all even factors and storing the result in s.

4. Print the sum of the even factors. 

Python3




# Python code
# To find the sum of even factors of a number
 
 
def evenfactorssum(n):
    s = 0
    for i in range(1, n+1):
        # finding factors of a given number
        if n % i == 0:
            # finding even factors of a given number
            if i % 2 == 0:
                # adding even factors of a given number
                s = s+# 2+6+10+30
                # printing the sum of even factors of a given number
    print(s)
 
 
# driver code
# input
n = 18
# the above input can also be given as
# n=int(input()) -> taking input from the user
evenfactorssum(n)
 
# this code is contributed by gangarajula laxmi

Output

26

Method: Using the list comprehension 

Python3




n=18
x=[i for i in range(1,n+1) if n%i==0 and i%2==0]
print(sum(x))

Output

26

Please refer complete article on Find sum of even factors of a number for more details!


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