Question

A function calledsumOfFactorsthat takes n as a parameter and then returns the sum of the proper factors/divisors ofn; for example,the sum of factorsof 20= 10+5+4+2 = 21.Note that 1 is nota properfactor of any number.Sample command => output:(display (sumOfFactors20))=> 21Hint: Calla recursivefunction sum_helperfrom sumOfFactorsfunction as follows:(define (sumOfFactors n)(sum_helper n (-n 1)))Here sum_helperfunction takes two parameters: n, d and returns the sum of all factors of n whichare <= d; for e.g. (sum_helper 20 6) returnsall proper factors of 20 which are <=6, i.e.,it will returnsthe value of 5+4+2 = 11.

Answer 1

temp=set()

def sumOfFactors(n):

  factors=[]

  for i in range(2, n): #GETTING FACTORS OF n

       if n % i == 0:

           factors.append(i)

  d=int(input("Enter the value of d="))

  

  helper=sum_helper(n,d) #CALLING THE SUM_HELPER RECURSIVE FUNCTION

  sec_sum=0

  first_sum=0

  for i in (helper):

    sec_sum+=i  #ADDING THE FACTORS WHICH ARE LESS THAN d

  for i in factors:

    first_sum+=i   #ADDING THE FACTORS

  

  return (first_sum,sec_sum)

def sum_helper(n,d):

  if (d > 1):

    if (n % d == 0):

      global temp

      temp.add(d)

    sum_helper(n, d-1) #RECURSIVE FUNCTION

  return temp

sumOfFactors(20) #Calling the function, Change the value inside the bracket.

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SUMMARY:-

Since the programming language is not mentioned here is the python code which is implemented as per your need.
You can change the value of n in the last line "sumOfFactors(20)" by replacing 20.