4. Question 4 Fill in the empty function so that it returns the sum of all the divisors of a number, without including it. A divisor is a number that divides into another without a remainder.

answer.py

def sum_divisors(n):
  # Return the sum of all divisors of n, not including n
  i = 1
  sum = 0
  while i < n:
    if n % i == 0:
      sum += i
      i +=1
    else:
      i+=1
  return sum


print(sum_divisors(0))
# 0
print(sum_divisors(3)) # Should sum of 1
# 1
print(sum_divisors(36)) # Should sum of 1+2+3+4+6+9+12+18
# 55
print(sum_divisors(102)) # Should be sum of 2+3+6+17+34+51
# 114
print(is_power_of_two(1)) # Should be True
print(is_power_of_two(8)) # Should be True
print(is_power_of_two(9)) # Should be False
# DO NOT DELETE THIS COMMENT

def sum_divisors(n):
if n == 0:
return 0
i = 1
sum = 0
while i < n:
if n % i == 0:
sum +=i
i +=1
# Return the sum of all divisors of n, not including n
return sum

print(sum_divisors(0))

0

print(sum_divisors(3)) # Should sum of 1

1

print(sum_divisors(36)) # Should sum of 1+2+3+4+6+9+12+18

55

print(sum_divisors(102)) # Should be sum of 2+3+6+17+34+51

114

def sum_divisors(number):
# Initialize the appropriate variables
  divisor = 1
  total = 0

  # Avoid dividing by 0 and negative numbers 
  # in the while loop by exiting the function
  # if "number" is less than one
  if number < 1:
    return 0 

  # Complete the while loop
  while divisor < number :
    if number % divisor == 0:
      total += divisor
      
    # Increment the correct variable
    divisor += 1

  # Return the correct variable 
  return total