Problems with "while" looping

Question

I have to somehow do a function in which when I execute it I will be able to add all the divisors of a number, in the way shown below.

This is getting me crazy, I have been in the same problem for about an hour.

def sum_divisors(n):
 # Return the sum of all divisors of n, not including n
 divisor = 1
 while divisor < n:
 if n%divisor==0:
 return divisor
 divisor = divisor + 1
 else:
 divisor = divisor + 1
print(sum_divisors(6)) # Should be 1+2+3=6
print(sum_divisors(12)) # Should be 1+2+3+4+6=16

• Guy: Let me guess, the problem is this code is unreachable? (you should be the one to say it, not me).
• Blessed Geek: I think it is a tradition you should also upvote the answer you accepted.

Answer 1

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

Answer 2

In your fonction, you return instantly after finding a divisor. That's why your fonction doesnt work Try to put each n%divisor == 0 in a list ans return it AT the end of the while.

Or try to print it directly.

Answer 3

A simple google search would lead you to answers with a lot of explanations: sum of divisors in python

In case you think about some efficiency:

we need to check divisors till sqrt of a number

import math 
def sum_divisors(num) : 
 # Final result of summation of divisors 
 result = 0
 # find all divisors which divides 'num' 
 i = 2
 while i<= (math.sqrt(num)) : 
 # if 'i' is divisor of 'num' 
 if (num % i == 0) : 
 # if both divisors are same then 
 # add it only once else add both 
 if (i == (num / i)) : 
 result = result + i; 
 else : 
 result = result + (i + num//i); 
 i = i + 1
 # Add 1 to the result as 1 is also 
 # a divisor 
 return (result + 1); 
print(sum_divisors(6))
print(sum_divisors(12))

Answer 4

this here is weird:

if n%divisor==0:
 return divisor
 divisor = divisor + 1 //<= this is actually dead code, since is after the return statement...

on the other hand:

this here:

divisor = divisor + 1

works more like a counter but you are missing the accumulator...

you should do something like

accum = 0
while divisor < n:
 foo = n % divisor
 if foo == 0:
 accum = accum + divisor

Answer 5

def sum_divisors(n):

total = [0]
divisors = 1
while divisors < n:
 if n % divisors == 0:
 result.append(divisors)
 else:
 pass
 divisors += 1
return sum(total)

• Hoppeduppeanut: While this might answer the question, you should edit your answer to include an explanation of how this code block answers the question. This helps provide context and makes your answer much more useful to future readers.

Answer 6

You can use this simple while loop to print the sum of all the divisors of a number. you should use an accumulator to increment the temp.

def sum_divisors(n):
 sum = 0
 accum = 1
 while n != 0 and accum < n:
 if n % accum == 0:
 sum += accum
 accum += 1
 return sum
print(sum_divisors(6)) # prints 6
print(sum_divisors(12)) # prints 16

Answer 7

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

Answer 8

There should be a if condition at line 5, so that the required result will be obtained for 0 also

if(num==0):
return result;

Answer 9

def sum_divisors(n):
 sum = 0
 x = 1
 while n != 0 and x < n :
 
 if n % x == 0 :
 sum += x
 else:
 sum += 0
 x += 1 
 return sum

Answer 10

def sum_divisors(n):
 x = 1
 a = 0
 while n!=0 and x<n:
 if n%x == 0:
 a = a + x
 x += 1 
 #Return the sum of all divisors of n, not including n
 return a
print(sum_divisors(0)) #0
print(sum_divisors(3)) # Should sum of 1
print(sum_divisors(36)) # Should sum of 1+2+3+4+6+9+12+18
print(sum_divisors(102)) # Should be sum of 2+3+6+17+34+51

Answer 11

here's my code(didn't clean it up but it works well) :

def sum_divisors(n):
 sum = 0
 divisors = 1
 while divisors < n:
 if n == 0:
 return 0
 else:
 if n % divisors == 0:
 sum += divisors
 divisors += 1
 else:
 divisors += 1
 continue
 # 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